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Oct 08, 2025
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About This Presentation
It is a short and easy study material to score marks
Slide Content
1. Basic Concepts of Chemistry and Chemical Calculations
1. Define relative atomic mass
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified
atomic mass unit.
Relative atomic mass (Ar) =
Average mass of the atom
unified atomic mass
2. What do you understand by the term mole.
Mole is the SI unit to represent a specific amount of a substance.
Mole =
Weight of the substance
Molar mass of the substance
3. Define Avogadro Number
The total number of entities present in one mole of any substance is equalto 6.022 x 10
23
.
4. Define gram equivalent mass.
Gram equivalent mass of an element, compound or ion is the mass thatcombines or displaces
1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
5. What do you understand by the term oxidation number.
It is defined as the imaginary charge left on the atom when all other atoms of
the compound have been removed in their usual oxidation states that are assigned
according to set of rules.
6. Distinguish between oxidation and reduction.
Oxidation Reduction
1. Removal of electron 1. Addition of electron
2. Positive charge increases 2. Negative charge increases
3. Addition of oxygen 3. Removal of Oxygen
4. Removal of Hydrogen 4. Addition of hydrogen
7. Calculate the molar mass of the following compounds.
i) Urea [CO(NH2)2]
C 1 x 12 = 12
O 1 x 16 = 16
N 2 x 14 = 28
H 4 x 1 = 4
Molar mass = 60
ii) Acetone [CH3COCH3]
C 3 x 12 = 36
H 6x 1 = 6
O 1 x 16 = 16
Molar mass = 58
iii, Boric acid [H3BO3]
H 1 x 3 = 3
B 1 x 11 = 11
O 3 x 16 = 48
Molar mass = 62
iv) Sulphuric acid [H2SO4]
H 2x 1 = 2
S 1 x 32 = 32
O 4 x 16 = 64
Molar mass = 98 8. The density of carbon dioxide is equal to 1.965 kgm
-3
at 273 K and 1 atm pressure.
Calculate the molar mass of CO2.
Molecular mass = Density x Molar volume
Molar Volume of CO2 at 273K and 1 atm pressure = 2.24 x 10
-2
m
3
Density of CO2 = 1.965 kgm
-3
= 1.965 x 10
3
gm
-3
Molecular mass = 2.24 x 10
-2
m
3
x1.965 x 10
3
gm
-3
= 1.965 x 10
1
x 2.24
= 44 gm
-3
9. Which contains the greatest number of moles of oxygen atoms
i) 1 mole of ethanol ii) 1 mole of formic acid iii) 1 mole of H2O
Compounds No.of moles No.of Oxygen atoms
Ethanol C2H5OH 1 1 x 6.023 x 10
23
Formic acid HCOOH 1 2 x 6.023 x 10
23
Water H2O 1 1 x 6.023 x 10
23
1 mole of Formic acid contains the greatest number of Oxgygen atoms
10. Calculate the average atomic mass of naturally occurring magnesium using the
following data
Isotope Isotopic atomic mass Abundance (%)
Mg-24 23.99 78.99
Mg-26 24.99 10.00
Mg-25 25.98 11.01
Average atomic mass =
78.99 ??????23.99 + 10 ?????? 24.99 + (11.01 ?????? 25.98)
100
=
2430.9
100
= 24.31
Average atomic mass of Mg = 24.31
11. In a reaction X + Y + Z2 XYZ2identify the Limiting reagent if any, in the following
reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of Z2
(b) 1mole of x + 1 mole of y+3 moles of Z2
(c) 50 atoms of x + 25 atoms of y +50 molecules of Z2
(d) 2.5 moles of x + 5 moles of y+5 moles of Z2
Question Number of moles of
reactants
allowed to react
Number of moles of reactants
consumed during reaction
Limiting
Reagent
X Y Z2 X Y Z2
a 200
atoms
200
atoms
50
molecules
50 atoms 50 atoms 50
molecules
Z2
b 1mole 1mole 3moles 1mole 1mole 1mole X & Y
c 50
atoms
25
atoms
50
molecules
25 atoms 25 atoms 25
molecules
Y
d 2.5
moles
5 moles 5 moles 2.5 moles 2.5 moles 2.5 moles X
12. Mass of one atom of an element is 6.645 x 10
-23
g. How many moles of element are
there in 0.320 kg.
Given: mass of one atom = 6.645 × 10
–23
g
∴ Mass of 1 mole of atom = 6.645 x 10
-23
x 6.023 x 10
23
= 40 g
∴ Number of moles of element in 0.320 kg = =
1 ����
40 ??????
x 0.320Kg =
1 ����
40 ??????
x 320 g = 8moles
1. Define relative atomic mass
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified
atomic mass unit.
Relative atomic mass (Ar) =
Average mass of the atom
unified atomic mass
2. What do you understand by the term mole.
Mole is the SI unit to represent a specific amount of a substance.
Mole =
Weight of the substance
Molar mass of the substance
3. Define Avogadro Number
The total number of entities present in one mole of any substance is equalto 6.022 x 10
23
.
4. Define gram equivalent mass.
Gram equivalent mass of an element, compound or ion is the mass thatcombines or displaces
1.008 g hydrogen or 8 g oxygen or 35.5 g chlorine.
5. What do you understand by the term oxidation number.
It is defined as the imaginary charge left on the atom when all other atoms of
the compound have been removed in their usual oxidation states that are assigned
according to set of rules.
6. Distinguish between oxidation and reduction.
Oxidation Reduction
1. Removal of electron 1. Addition of electron
2. Positive charge increases 2. Negative charge increases
3. Addition of oxygen 3. Removal of Oxygen
4. Removal of Hydrogen 4. Addition of hydrogen
7. Calculate the molar mass of the following compounds.
i) Urea [CO(NH2)2]
C 1 x 12 = 12
O 1 x 16 = 16
N 2 x 14 = 28
H 4 x 1 = 4
Molar mass = 60
ii) Acetone [CH3COCH3]
C 3 x 12 = 36
H 6x 1 = 6
O 1 x 16 = 16
Molar mass = 58
iii, Boric acid [H3BO3]
H 1 x 3 = 3
B 1 x 11 = 11
O 3 x 16 = 48
Molar mass = 62
iv) Sulphuric acid [H2SO4]
H 2x 1 = 2
S 1 x 32 = 32
O 4 x 16 = 64
Molar mass = 98 8. The density of carbon dioxide is equal to 1.965 kgm
-3
at 273 K and 1 atm pressure.
Calculate the molar mass of CO2.
Molecular mass = Density x Molar volume
Molar Volume of CO2 at 273K and 1 atm pressure = 2.24 x 10
-2
m
3
Density of CO2 = 1.965 kgm
-3
= 1.965 x 10
3
gm
-3
Molecular mass = 2.24 x 10
-2
m
3
x1.965 x 10
3
gm
-3
= 1.965 x 10
1
x 2.24
= 44 gm
-3
9. Which contains the greatest number of moles of oxygen atoms
i) 1 mole of ethanol ii) 1 mole of formic acid iii) 1 mole of H2O
Compounds No.of moles No.of Oxygen atoms
Ethanol C2H5OH 1 1 x 6.023 x 10
23
Formic acid HCOOH 1 2 x 6.023 x 10
23
Water H2O 1 1 x 6.023 x 10
23
1 mole of Formic acid contains the greatest number of Oxgygen atoms
10. Calculate the average atomic mass of naturally occurring magnesium using the
following data
Isotope Isotopic atomic mass Abundance (%)
Mg-24 23.99 78.99
Mg-26 24.99 10.00
Mg-25 25.98 11.01
Average atomic mass =
78.99 ??????23.99 + 10 ?????? 24.99 + (11.01 ?????? 25.98)
100
=
2430.9
100
= 24.31
Average atomic mass of Mg = 24.31
11. In a reaction X + Y + Z2 XYZ2identify the Limiting reagent if any, in the following
reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of Z2
(b) 1mole of x + 1 mole of y+3 moles of Z2
(c) 50 atoms of x + 25 atoms of y +50 molecules of Z2
(d) 2.5 moles of x + 5 moles of y+5 moles of Z2
Question Number of moles of
reactants
allowed to react
Number of moles of reactants
consumed during reaction
Limiting
Reagent
X Y Z2 X Y Z2
a 200
atoms
200
atoms
50
molecules
50 atoms 50 atoms 50
molecules
Z2
b 1mole 1mole 3moles 1mole 1mole 1mole X & Y
c 50
atoms
25
atoms
50
molecules
25 atoms 25 atoms 25
molecules
Y
d 2.5
moles
5 moles 5 moles 2.5 moles 2.5 moles 2.5 moles X
12. Mass of one atom of an element is 6.645 x 10
-23
g. How many moles of element are
there in 0.320 kg.
Given: mass of one atom = 6.645 × 10
–23
g
∴ Mass of 1 mole of atom = 6.645 x 10
-23
x 6.023 x 10
23
= 40 g
∴ Number of moles of element in 0.320 kg = =
1 ����
40 ??????
x 0.320Kg =
1 ����
40 ??????
x 320 g = 8moles
13. What is the empirical formula of the following ?
i) Fructose C6H12O6found in honey
ii) ii) Caffeine (C8H10N4O2) a substance found in tea and coffee.
Compound Molecular formula Emphirical formula
Fructose C6H12O6 CH2O
Caffeine C8H10N4O2 C4H5N2O
14. How many moles of ethane is required to produce 44 g of CO2 (g) after combustion.
Balanced equation for the combustion of ethane
2C2H6 + 7O2 4CO2 + 6H2O
To produce 4 moles of CO2, 2 moles of ethane is required
44g of CO2 is equal to 1 mole of CO2
To produce 1 mole of CO2 required = 2 / 4
= 0.5 mole of ethane
15. The reaction between aluminium and ferric oxide can generate temperatures up to
3273 K and is used in welding metals. (Atomic mass of Al = 27 u
Atomic mass of 0 = 16 u)
2Al + Fe2O3 Al2O3 + 2Fe; If, in this process, 324 g of aluminium is allowed to react
with 1.12 kg of ferric oxide.
i) Calculate the mass of Al2O3 formed.
ii) How much of the excess reagent is left at the end of the reaction?
2Al + Fe2O3 Al2O3 + 2Fe
Reactants Products
Al Fe2O3 Al2O3 Fe
Amount of reactant allowed
to react
324g 1.12Kg = 1120g
Number of moles allowed
toreact
324
27
= 12
1120
160
= 7
Stoichiometric Co-efficient 2 1 1 2
Number of moles
consumedduring reaction
12 6
Number of moles of reactant
unreacted and number
ofmoles of product formed
- 1 6 12
Molar mass of Al2O3
Molar mass of Al2O3= (2 x 27) + (3 x 16) = 102
Mass of Al2O3 formed = molexMolar mass = 6 x 102 = 612g
Excess reagent = Fe2O3
Molar mass of Fe2O3= (2 x 56) + (3 x 16) = 160
Mass of Fe2O3 = mole x Molar mass = 1 x 160 = 160g
16. Calculate the empirical and molecular formula of a compound containing 76.6%
carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Element Percenta
ge
Atomic
mass
Relative number
of atoms
Simple
ratio
Whole
number
C 76.6 12
76.6
12
= 6.38
6.38
1.06
= 6
6
H 6.38 1
6.38
1
= 6.38
6.38
1.06
= 6
6
O 17.02 16
17.02
16
= 1.06
1.06
1.06
= 1
1
Empirical formula = C6H6O
Molar mass = 2 x vapour density
= 2 x 47 = 94
n =
Molar mass
calculated empirical formula mass
= ( 6 x 12) + ( 6 x 1) + ( 1 x 16) = 94
n =
94
94
= 1
Molecular formula= (C6H6O)n = (C6H6O)1 = C6H6O
17. Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and O= 69.5%
calculate the molecular formula of the compound if all the hydrogen in the compound
is present in combination with oxygen as water of crystallization. (molecular mass of
the compound is 322
Element Percentage Atomic mass Relative number of
atoms
Simple ratio Whole
number
Na 14.31 23
14.31
23
= 0.62
0.62
0.31
= 2
2
S 9.97 32
9.97
32
= 0.31
0.31
0.31
= 1
1
H 6.22 1
6.22
1
= 6.2
6.2
0.31
= 20
20
O 69.5 16 69.5
16
= 4.34
4.34
0.31
= 14
14
Empirical formula = Na2SH20O14
Empirical formula mass = ( 2 x 23 ) + (1 x 32) + ( 20 x 1) + (14 x 16)
= 322
Molar mass = 322
n =
Molar mass
calculated empirical formula mass
n =
322
322
= 1
Molecular formula = (Na2SH20O14)n = (Na2SH20O14)1
= Na2SH20O14
Since all the hydrogen in the compound present as water
Molecular formula = Na2SO4 .10H2O
i) Fructose C6H12O6found in honey
ii) ii) Caffeine (C8H10N4O2) a substance found in tea and coffee.
Compound Molecular formula Emphirical formula
Fructose C6H12O6 CH2O
Caffeine C8H10N4O2 C4H5N2O
14. How many moles of ethane is required to produce 44 g of CO2 (g) after combustion.
Balanced equation for the combustion of ethane
2C2H6 + 7O2 4CO2 + 6H2O
To produce 4 moles of CO2, 2 moles of ethane is required
44g of CO2 is equal to 1 mole of CO2
To produce 1 mole of CO2 required = 2 / 4
= 0.5 mole of ethane
15. The reaction between aluminium and ferric oxide can generate temperatures up to
3273 K and is used in welding metals. (Atomic mass of Al = 27 u
Atomic mass of 0 = 16 u)
2Al + Fe2O3 Al2O3 + 2Fe; If, in this process, 324 g of aluminium is allowed to react
with 1.12 kg of ferric oxide.
i) Calculate the mass of Al2O3 formed.
ii) How much of the excess reagent is left at the end of the reaction?
2Al + Fe2O3 Al2O3 + 2Fe
Reactants Products
Al Fe2O3 Al2O3 Fe
Amount of reactant allowed
to react
324g 1.12Kg = 1120g
Number of moles allowed
toreact
324
27
= 12
1120
160
= 7
Stoichiometric Co-efficient 2 1 1 2
Number of moles
consumedduring reaction
12 6
Number of moles of reactant
unreacted and number
ofmoles of product formed
- 1 6 12
Molar mass of Al2O3
Molar mass of Al2O3= (2 x 27) + (3 x 16) = 102
Mass of Al2O3 formed = molexMolar mass = 6 x 102 = 612g
Excess reagent = Fe2O3
Molar mass of Fe2O3= (2 x 56) + (3 x 16) = 160
Mass of Fe2O3 = mole x Molar mass = 1 x 160 = 160g
16. Calculate the empirical and molecular formula of a compound containing 76.6%
carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Element Percenta
ge
Atomic
mass
Relative number
of atoms
Simple
ratio
Whole
number
C 76.6 12
76.6
12
= 6.38
6.38
1.06
= 6
6
H 6.38 1
6.38
1
= 6.38
6.38
1.06
= 6
6
O 17.02 16
17.02
16
= 1.06
1.06
1.06
= 1
1
Empirical formula = C6H6O
Molar mass = 2 x vapour density
= 2 x 47 = 94
n =
Molar mass
calculated empirical formula mass
= ( 6 x 12) + ( 6 x 1) + ( 1 x 16) = 94
n =
94
94
= 1
Molecular formula= (C6H6O)n = (C6H6O)1 = C6H6O
17. Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and O= 69.5%
calculate the molecular formula of the compound if all the hydrogen in the compound
is present in combination with oxygen as water of crystallization. (molecular mass of
the compound is 322
Element Percentage Atomic mass Relative number of
atoms
Simple ratio Whole
number
Na 14.31 23
14.31
23
= 0.62
0.62
0.31
= 2
2
S 9.97 32
9.97
32
= 0.31
0.31
0.31
= 1
1
H 6.22 1
6.22
1
= 6.2
6.2
0.31
= 20
20
O 69.5 16 69.5
16
= 4.34
4.34
0.31
= 14
14
Empirical formula = Na2SH20O14
Empirical formula mass = ( 2 x 23 ) + (1 x 32) + ( 20 x 1) + (14 x 16)
= 322
Molar mass = 322
n =
Molar mass
calculated empirical formula mass
n =
322
322
= 1
Molecular formula = (Na2SH20O14)n = (Na2SH20O14)1
= Na2SH20O14
Since all the hydrogen in the compound present as water
Molecular formula = Na2SO4 .10H2O
18. Define molecular mass
Molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit.
Relative molecular mass of hydrogen molecule (H2) = 2 x 1.008 = 2.016u
19. Define Oxidation number?
The residual chargewhich its atom has (or) appears to have when all other atoms from the molecule are
removed as ions..
20. What is the difference between molecular mass andmolar mass? Calculate the
molecular mass and molar mass for carbon monoxide.
Molecular mass Molar mass
The ratio of the mass of the molecule to the unified
atomic mass unit
The Mass of the one mole of a substance
It can be calculated by adding the relative atomic
masses of its constituent atoms
It is equal to the sum of the relative atomic
masses of the substance
Unit ; u Unit : g.mol
-1
Molecular mass carbonmonooxide is 28u Molar mass of carbon mono oxide is
28 g.mol
-1
21. What is Limiting Reagents ?
The productyield will be determined by the reactantthat is completely consumed. It limits the
further reaction from taking place.
22. Calculate the equivalent mass of H2SO4
Equivalent mass of H2SO4. =
Molar mass of H2SO4.
Basicity of H2SO4.
Molar mass of H2SO4. = (2 x 1) + (1 x 32) + (4 x 16) = 98
Basicity of H2SO4. = 2
Equivalent mass of H2SO4. =
98
2
= 49
23. Calculate Oxidation number of oxygen in H2O2
(2 x 1) + 2x = 0
2x = -2
X = -1
24. Define Disproportionation reaction (Autoredox reactions)
In some redox reactions, the same compound can undergo both oxidation and reduction. In such
reactions, the oxidation state of one and the same element is both increased and decreased.
Eg : 2H2O2 2H2O + O2
25. Define basicity. Find the basicity of ortho-phosphoric acid.
Basicity : No. of moles of ionisable H
+
ions present in 1 mole of the acid
Basicity of ortho-phosphoric acid is 3
26. Define acidity. Find the acidity of Calciumhydroxide.
Acidity :- no. of moles of ionisable OH– ion present in 1 mole of the base
Acidity of Calciumhydroxide is 2
27. What are Combination reactions?
Redox reactions in which two substances combine to form a single compound are called combination reaction
Eg : C + O2 CO2
2 . Quantum Mechanical Model of Atom
1. Which quantum number reveal information about the shape, energy, orientation
and size of orbitals?
Property Quantum number
Reveal information about Energyand Size Principle quantum number
Reveal information about Shape Azimuthal quantum number
Orientation Magnetic quantum number
2. How many orbitals are possible for n =4?
If n= 4 l = 0,1,2,3 (s,p,d,f)
If l =0 m =0 (4s orbital) = 1 orbital
If l =1 m = -1, 0,1 = 3 orbitals
If l = 2 m = -2, -1, 0, 1, 2 = 5 orbitals
If l = 3 m = -3, -2, -1, 0, 1, 2, 3 = 7 orbitals
Total number of orbitals are possible =16 orbtals
3. How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular
nodes?.
Orbital n l
radial nodes
n-(l +1)
Angularnodes
l
2s 2 0 1 0
4p 4 1 2 1
5d 5 2 2 2
4f 4 3 0 3
4. The stabilisation of a half filled d - orbital is more pronounced than that of the
p-orbital why?
In d- orbital 10 exchanges are possible but in p-orbital 3 exchanges are possible.
5. Consider the following electronic arrangements for the d
5
configuration.
(i) which of these represents the ground state
(ii) which configuration has the maximum exchange energy.
(i) which of these represents the ground state
(ii) which configuration has the maximum exchange energy.
6. State and explain Pauli's exclusion principle.
"No two electrons in an atom can have the same set of values of all four quantum numbers."
Eg. For the electron in Helium
Valus of Quantum number Ist electron IInd electron
n 1 1
l 0 0
m 0 0
s +½ -½
7. Define orbital? what are the n and l values for 3px and 4dx2-y2 electron?
Orbital is a three dimensional space which the probability of finding the electron is maximum
3pxorbital n = 3 &l = -1
4dx2-y2orbital n = 4 & l = -2
8. Explain briefly the time independent schrodinger wave equation?
The time independent Schrodinger equation can be expressed as,
is called Hamiltonian operator
Molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit.
Relative molecular mass of hydrogen molecule (H2) = 2 x 1.008 = 2.016u
19. Define Oxidation number?
The residual chargewhich its atom has (or) appears to have when all other atoms from the molecule are
removed as ions..
20. What is the difference between molecular mass andmolar mass? Calculate the
molecular mass and molar mass for carbon monoxide.
Molecular mass Molar mass
The ratio of the mass of the molecule to the unified
atomic mass unit
The Mass of the one mole of a substance
It can be calculated by adding the relative atomic
masses of its constituent atoms
It is equal to the sum of the relative atomic
masses of the substance
Unit ; u Unit : g.mol
-1
Molecular mass carbonmonooxide is 28u Molar mass of carbon mono oxide is
28 g.mol
-1
21. What is Limiting Reagents ?
The productyield will be determined by the reactantthat is completely consumed. It limits the
further reaction from taking place.
22. Calculate the equivalent mass of H2SO4
Equivalent mass of H2SO4. =
Molar mass of H2SO4.
Basicity of H2SO4.
Molar mass of H2SO4. = (2 x 1) + (1 x 32) + (4 x 16) = 98
Basicity of H2SO4. = 2
Equivalent mass of H2SO4. =
98
2
= 49
23. Calculate Oxidation number of oxygen in H2O2
(2 x 1) + 2x = 0
2x = -2
X = -1
24. Define Disproportionation reaction (Autoredox reactions)
In some redox reactions, the same compound can undergo both oxidation and reduction. In such
reactions, the oxidation state of one and the same element is both increased and decreased.
Eg : 2H2O2 2H2O + O2
25. Define basicity. Find the basicity of ortho-phosphoric acid.
Basicity : No. of moles of ionisable H
+
ions present in 1 mole of the acid
Basicity of ortho-phosphoric acid is 3
26. Define acidity. Find the acidity of Calciumhydroxide.
Acidity :- no. of moles of ionisable OH– ion present in 1 mole of the base
Acidity of Calciumhydroxide is 2
27. What are Combination reactions?
Redox reactions in which two substances combine to form a single compound are called combination reaction
Eg : C + O2 CO2
2 . Quantum Mechanical Model of Atom
1. Which quantum number reveal information about the shape, energy, orientation
and size of orbitals?
Property Quantum number
Reveal information about Energyand Size Principle quantum number
Reveal information about Shape Azimuthal quantum number
Orientation Magnetic quantum number
2. How many orbitals are possible for n =4?
If n= 4 l = 0,1,2,3 (s,p,d,f)
If l =0 m =0 (4s orbital) = 1 orbital
If l =1 m = -1, 0,1 = 3 orbitals
If l = 2 m = -2, -1, 0, 1, 2 = 5 orbitals
If l = 3 m = -3, -2, -1, 0, 1, 2, 3 = 7 orbitals
Total number of orbitals are possible =16 orbtals
3. How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular
nodes?.
Orbital n l
radial nodes
n-(l +1)
Angularnodes
l
2s 2 0 1 0
4p 4 1 2 1
5d 5 2 2 2
4f 4 3 0 3
4. The stabilisation of a half filled d - orbital is more pronounced than that of the
p-orbital why?
In d- orbital 10 exchanges are possible but in p-orbital 3 exchanges are possible.
5. Consider the following electronic arrangements for the d
5
configuration.
(i) which of these represents the ground state
(ii) which configuration has the maximum exchange energy.
(i) which of these represents the ground state
(ii) which configuration has the maximum exchange energy.
6. State and explain Pauli's exclusion principle.
"No two electrons in an atom can have the same set of values of all four quantum numbers."
Eg. For the electron in Helium
Valus of Quantum number Ist electron IInd electron
n 1 1
l 0 0
m 0 0
s +½ -½
7. Define orbital? what are the n and l values for 3px and 4dx2-y2 electron?
Orbital is a three dimensional space which the probability of finding the electron is maximum
3pxorbital n = 3 &l = -1
4dx2-y2orbital n = 4 & l = -2
8. Explain briefly the time independent schrodinger wave equation?
The time independent Schrodinger equation can be expressed as,
is called Hamiltonian operator
is the wave function
E is the energy of the system
The above schrodinger wave equation does not contain time as a variableand is referred to as
time independent Schrodinger wave equation.
9. Determine the values of all the four quantum numbers of the 8th electron in
O- atom and 15th electron in Cl atom and the last electron in chromium.
Atom Electron n l m s
O 8
th
2 1 -1 +½
Cl 15 3 1 +1 +½
Cr Last electron(24
th
) 4 2 +2 +½
10 For each of the following, give the sub level designation, the allowable m values and
the number of orbitals
i) n = 4, l = 2, ii) n = 5, l = 3 iii) n = 7, l = 0
n l sub level
designation
m values Number of orbitals
4 2 4d +2, +1, 0 ,-1, -2 5 – d orbitals
5 3 5f +3, +2, +1, 0 ,-1, -2, -3 7 – 4f Orbitals
7 0 7s 0 1– 7s orbital
11 Give the electronic configuration of Mn
2+
and Cr
3+
.
Electronic configuration of Mn
2+
is 1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
Electronic configuration of Cr
3+
is 1s
2
2s
2
2p
6
3s
2
3p
6
3d
3
12
.
Describe the Aufbau principle
In the ground state of the atoms, the orbitals are filled in the order of their increasing
energies.which is in accordance with (n+l) rule.
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
8s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s
13
.
An atom of an element contains 35 electrons and 45 neutrons. Deduce
i) The number of protons
The number of protons = The number of electrons = 35
ii) The electronic configuration for the element
1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
5
iii) All the four quantum numbers for the last electron
Last electron present in 4Pyorbital
n l m s
4 1 0 -½
14
.
How that the circumference of the Bohr orbit for the hydrogen atom is an integral
multiple of the de Broglie wave length associated with the electron revolving around the
nucleus.
According to the de Broglie concept, the electron that revolves around the nucleus exhibits
both particle and wave character. In order for the electron wave to exist in phase, the circumference of
the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron
wave is out of phase.Circumference of the orbit = nλ
2πr = nλ
2πr = nh/mv
Rearranging, mvr = nh/2π
Angular momentum = nh/2π
15
.
Identify the missing quantum numbers and the sub energy level
n l m Sub energy level
4 2 0 4d
3 1 0 3p
5 1 +1, 0, -1 5p
3 2 -2 3d
16 How many unpaired electrons are present in the ground state of Fe
3+
(Z = 26),
Mn
2+
(Z =25) andAr (Z = 180)
Fe(Z=26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
6
Fe
3+
1s
2
2s
2
2p
6
3s
2
3p
6
4s
0
3d
5
3d
5
Number of unpaired electrons= 5
Mn (Z=26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
5
Mn
2+
1s
2
2s
2
2p
6
3s
2
3p
6
4s
0
3d
5
3d
5
Number of unpaired electrons = 5
Ar(Z=18) 1s
2
2s
2
2p
6
3s
2
3p
6
3p
6
No unpaired electrons in it
E is the energy of the system
The above schrodinger wave equation does not contain time as a variableand is referred to as
time independent Schrodinger wave equation.
9. Determine the values of all the four quantum numbers of the 8th electron in
O- atom and 15th electron in Cl atom and the last electron in chromium.
Atom Electron n l m s
O 8
th
2 1 -1 +½
Cl 15 3 1 +1 +½
Cr Last electron(24
th
) 4 2 +2 +½
10 For each of the following, give the sub level designation, the allowable m values and
the number of orbitals
i) n = 4, l = 2, ii) n = 5, l = 3 iii) n = 7, l = 0
n l sub level
designation
m values Number of orbitals
4 2 4d +2, +1, 0 ,-1, -2 5 – d orbitals
5 3 5f +3, +2, +1, 0 ,-1, -2, -3 7 – 4f Orbitals
7 0 7s 0 1– 7s orbital
11 Give the electronic configuration of Mn
2+
and Cr
3+
.
Electronic configuration of Mn
2+
is 1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
Electronic configuration of Cr
3+
is 1s
2
2s
2
2p
6
3s
2
3p
6
3d
3
12
.
Describe the Aufbau principle
In the ground state of the atoms, the orbitals are filled in the order of their increasing
energies.which is in accordance with (n+l) rule.
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p
8s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s
13
.
An atom of an element contains 35 electrons and 45 neutrons. Deduce
i) The number of protons
The number of protons = The number of electrons = 35
ii) The electronic configuration for the element
1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
5
iii) All the four quantum numbers for the last electron
Last electron present in 4Pyorbital
n l m s
4 1 0 -½
14
.
How that the circumference of the Bohr orbit for the hydrogen atom is an integral
multiple of the de Broglie wave length associated with the electron revolving around the
nucleus.
According to the de Broglie concept, the electron that revolves around the nucleus exhibits
both particle and wave character. In order for the electron wave to exist in phase, the circumference of
the orbit should be an integral multiple of the wavelength of the electron wave. Otherwise, the electron
wave is out of phase.Circumference of the orbit = nλ
2πr = nλ
2πr = nh/mv
Rearranging, mvr = nh/2π
Angular momentum = nh/2π
15
.
Identify the missing quantum numbers and the sub energy level
n l m Sub energy level
4 2 0 4d
3 1 0 3p
5 1 +1, 0, -1 5p
3 2 -2 3d
16 How many unpaired electrons are present in the ground state of Fe
3+
(Z = 26),
Mn
2+
(Z =25) andAr (Z = 180)
Fe(Z=26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
6
Fe
3+
1s
2
2s
2
2p
6
3s
2
3p
6
4s
0
3d
5
3d
5
Number of unpaired electrons= 5
Mn (Z=26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
5
Mn
2+
1s
2
2s
2
2p
6
3s
2
3p
6
4s
0
3d
5
3d
5
Number of unpaired electrons = 5
Ar(Z=18) 1s
2
2s
2
2p
6
3s
2
3p
6
3p
6
No unpaired electrons in it
17 Explain the meaning of the symbol 4f
2
.Write all the four quantum numbers
for these electrons.
4f
2
no. of electrons
f-orbital
Principal energy level
Electrons n l m s
1
st
e
-
4 3 -3 +½
2
nd
e
-
4 3 -2 -½
18
.
Which has stable electronic configuration ? Ni
2+
or Fe
3+
Ni ( Z = 28) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8
Ni
2+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
8
Fe ( Z = 26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
6
Fe
3+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
If d – orbital is half filled, Fe
3+
is more stable. So is more stable Fe
3+
than Ni
2+
19
.
What are the conclusions of Rutherford’s experiment?
Rutherford bombarded a thin gold foil with a stream of fast moving -particles.
It was observed that most of the -particles passed through the foil.
Some of them were deflected through a small angle.
Based on these observations, he proposed that in an atom there is a tiny positively charged
nucleus and the electrons are moving around the nucleus with high speed.
20
.
Write the Limitation of Rutherford’s atom model.
Does not explain the distribution of electrons around the nucleus and theirenergies.
21
.
Write the quation for the radius of the n
th
orbit and the energy of the electron using
Bohr's atom model.
Radius ofthe n
th
orbitelectronEn =
−����.� �
�
�
�
KJ mol
-1
Energy of the electronrn =
�.��� �
�
�
A
0
22 What are the limitations of Bohr’s atom model?
The Bohr‟s atom model is applicable only to species having one electron.
It was unable to explain the splitting of spectral lines in the presence of magnetic field
(Zeeman’s effect or an electrical field ( Stark effect).
Bohr‟s theory was unable to explain why the electron is restricted to revolve around the nucleus in
a fixed orbit in which the angular momentum of the electron is equal to
nh
2
23 Derive de-Broglie equation
E = h(according to the Planck‟s quantum theory) ...(1)
where is the frequency of the wave and h is Planck‟s constant.
E = mc
2
(according to Einstein equation) ...(2)
where m is the mass of photon and c is the velocity of light.
From equations (1) and (2), we get
h = mc
2
...(3)
Butv =
�
h
�
= mc
2
or =
h
mC
……… ( 4 )
the velocity “c” of the photon is replaced by the velocity v of the material particle. =
h
mv
……… ( 5 ) p = mv
=
h
p
……… ( 6 ) p - the momentum of the particle
24 Explain Heisenberg uncertainty principle.
It is impossible to measure simultaneously both the position and velocity (or momentum) of a
microscopic particle with absolute accuracy or certainty.”
Mathematically, uncertainty principle can be put as follows. x.p > h/4
25 What is dual property of matter?
All material particles possessed both wave character as well as particle character.
26 Describe about Bohr atom model.
The energies of electrons are quantized
The electron is revolving around the nucleus in a certain fixed circular path called stationary orbit.
Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron
must be equal to an integral multiple of h/2
mr =
�ℎ
2??????
where n = 1,2,3 ….
As long as an electron revolves in the fixed stationary orbit, it doesn‟t lose its energy.
However, when an electron jumps from higher energy state (E2) to a lower energy state (E1),
the excess energy is emitted as radiation. The frequency of the emitted radiation is
E2 – E1 = h
=
�1− �1
ℎ
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy
orbit to a higher energy orbit.
The radius of the nth orbit and the energy of the electron revolving in the nth orbit were
derived.
The results are as follows:
27 What are the Main features of the quantum mechanical model of atom
The energy of electrons in atoms is quantized
The solutions of Schrodinger wave equation gives the allowed energy levels (orbits).
According to Heisenberg uncertainty principle, the exact position and momentum of an
electron can not be determined with absolute accuracy.Orbital is a three dimensional space in
which the probability of finding the electron is maximum.
The solution of Schrodinger wave equation for the allowed energies of an atom gives the wave
function ψ, which represents an atomic orbital.
The probability of finding the electron in a small volume dxdydz around a point (x,y,z) is
proportional to |ψ(x,y,z)|
2
dxdydz |ψ(x,y,z)|
2
is known as probability density and is always
positive.
28 Explain J.J.Thomson’s atom model.
J. J. Thomson‟s cathode ray experiment revealed that atoms consist of negatively charged
particles called electrons.
He proposed that atom is a positively charged sphere in which the electrons are embedded like
the seeds in the watermelon.
29 What is the condition for the electron wave to exist in phase
The circumference of the orbit should be an integral multiple of the wavelength of the electron wave.
Circumference of the orbit = nλ
2πr = nλ
2
.Write all the four quantum numbers
for these electrons.
4f
2
no. of electrons
f-orbital
Principal energy level
Electrons n l m s
1
st
e
-
4 3 -3 +½
2
nd
e
-
4 3 -2 -½
18
.
Which has stable electronic configuration ? Ni
2+
or Fe
3+
Ni ( Z = 28) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
8
Ni
2+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
8
Fe ( Z = 26) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
6
Fe
3+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
5
If d – orbital is half filled, Fe
3+
is more stable. So is more stable Fe
3+
than Ni
2+
19
.
What are the conclusions of Rutherford’s experiment?
Rutherford bombarded a thin gold foil with a stream of fast moving -particles.
It was observed that most of the -particles passed through the foil.
Some of them were deflected through a small angle.
Based on these observations, he proposed that in an atom there is a tiny positively charged
nucleus and the electrons are moving around the nucleus with high speed.
20
.
Write the Limitation of Rutherford’s atom model.
Does not explain the distribution of electrons around the nucleus and theirenergies.
21
.
Write the quation for the radius of the n
th
orbit and the energy of the electron using
Bohr's atom model.
Radius ofthe n
th
orbitelectronEn =
−����.� �
�
�
�
KJ mol
-1
Energy of the electronrn =
�.��� �
�
�
A
0
22 What are the limitations of Bohr’s atom model?
The Bohr‟s atom model is applicable only to species having one electron.
It was unable to explain the splitting of spectral lines in the presence of magnetic field
(Zeeman’s effect or an electrical field ( Stark effect).
Bohr‟s theory was unable to explain why the electron is restricted to revolve around the nucleus in
a fixed orbit in which the angular momentum of the electron is equal to
nh
2
23 Derive de-Broglie equation
E = h(according to the Planck‟s quantum theory) ...(1)
where is the frequency of the wave and h is Planck‟s constant.
E = mc
2
(according to Einstein equation) ...(2)
where m is the mass of photon and c is the velocity of light.
From equations (1) and (2), we get
h = mc
2
...(3)
Butv =
�
h
�
= mc
2
or =
h
mC
……… ( 4 )
the velocity “c” of the photon is replaced by the velocity v of the material particle. =
h
mv
……… ( 5 ) p = mv
=
h
p
……… ( 6 ) p - the momentum of the particle
24 Explain Heisenberg uncertainty principle.
It is impossible to measure simultaneously both the position and velocity (or momentum) of a
microscopic particle with absolute accuracy or certainty.”
Mathematically, uncertainty principle can be put as follows. x.p > h/4
25 What is dual property of matter?
All material particles possessed both wave character as well as particle character.
26 Describe about Bohr atom model.
The energies of electrons are quantized
The electron is revolving around the nucleus in a certain fixed circular path called stationary orbit.
Electron can revolve only in those orbits in which the angular momentum (mvr) of the electron
must be equal to an integral multiple of h/2
mr =
�ℎ
2??????
where n = 1,2,3 ….
As long as an electron revolves in the fixed stationary orbit, it doesn‟t lose its energy.
However, when an electron jumps from higher energy state (E2) to a lower energy state (E1),
the excess energy is emitted as radiation. The frequency of the emitted radiation is
E2 – E1 = h
=
�1− �1
ℎ
Conversely, when suitable energy is supplied to an electron, it will jump from lower energy
orbit to a higher energy orbit.
The radius of the nth orbit and the energy of the electron revolving in the nth orbit were
derived.
The results are as follows:
27 What are the Main features of the quantum mechanical model of atom
The energy of electrons in atoms is quantized
The solutions of Schrodinger wave equation gives the allowed energy levels (orbits).
According to Heisenberg uncertainty principle, the exact position and momentum of an
electron can not be determined with absolute accuracy.Orbital is a three dimensional space in
which the probability of finding the electron is maximum.
The solution of Schrodinger wave equation for the allowed energies of an atom gives the wave
function ψ, which represents an atomic orbital.
The probability of finding the electron in a small volume dxdydz around a point (x,y,z) is
proportional to |ψ(x,y,z)|
2
dxdydz |ψ(x,y,z)|
2
is known as probability density and is always
positive.
28 Explain J.J.Thomson’s atom model.
J. J. Thomson‟s cathode ray experiment revealed that atoms consist of negatively charged
particles called electrons.
He proposed that atom is a positively charged sphere in which the electrons are embedded like
the seeds in the watermelon.
29 What is the condition for the electron wave to exist in phase
The circumference of the orbit should be an integral multiple of the wavelength of the electron wave.
Circumference of the orbit = nλ
2πr = nλ
30 Write a notes on Principal quantum number
It is denoted by the symbol „n‟
The „n‟ can have the values 1,2,3….
Shell K L M N
„n‟ value 1 2 3 4
The maximum number of electrons in a shell can be calculated by the formula 2n
2
.
The energy of the electron is given by
En =
−1312.8Z
2
n
2
KJmol
-1
.
The distance of the electron nucleus is given by rn =
0.529 n
2
�
A
0
.
31 Write a notes on Azimuthal quantum number
It is denoted by the symbol „l‟
l = (n-1) ,the „l‟ can have the values 0, 1,2,3….
Sub Shell (orbitals) s p d f
„n‟ value 0 1 2 3
The maximum number of electrons in a orbital can be calculated by the formula 2(2l+1).
It is used to calculate the orbitals angular momentum by using
32 Write a notes on Magnetic quantum number
It is denoted by the symbol „m‟
Its value ranging from –l to + l through 0
i.e, if l =1 m = -1, 0 and +1
The values of „l‟ represent different orientation of orbitals in space
The Zeeman effect experimental justification of this quantum number.
33 Write a notes on Spin quantum number
It is denoted by the symbol „s‟
The electron in an atom revolves around the nucleus and also spins in a clockwise direction or
in anti-clockwise direction.
Spin direction clockwise anti-clockwise
‘s’ value +½ -½
34
.
Explain the shapes of s-orbital.
For s-orbital l = 0 and hence, m can have only one value,i.e., m = 0.
This means that the probability of finding the electron in s-orbital is the
same in all directions at a particular distance.
In other words s-orbitals are spherically symmetrical.
35
.
Explain the shapes of p-orbitals.
p-orbitals, l = 1 and the corresponding m values are 1, 0
&+1.
The three different „m‟ values indicates that there are
three different possible orientations as px, py and pz and the angular distribution along
the x, y and z axis respectively.
36
.
Explain the shapes of d-orbitals.
For „d‟ orbital l = 2 and the corresponding m values
are -2, -1, 0,+1,+2.
The shape of the d orbital looks like a 'clover leaf '.
The five m values give rise to five d orbitals
namely dxy, dyz, dzx,dx2-y2 and dz2.
The 3d orbitals contain two nodal plane
37
.
Explain the shapes of f-orbitals
For 'f ' orbital, l = 3 and the m values are -3, -2,-1, 0, +1, +2, +3
Seven f orbitals fz
3
, fxz
2
, fyz
2
, fxyz, fz(x
2
−y
2
), fx(x
2
−3y
2
), fy(3x
2
−y
2
),
There are 3 nodal planes in the f-orbitals. 38 Define Exchange energy
If two or more electrons with thesame spin are present in degenerate orbitals, there is a
possibility for exchanging their positions.
During exchange process the energy is released and the released energy is called exchange
energy.
For example, in chromium the electronic configuration is [Ar]3d
5
4s
1
. The 3d orbital ishalf
filled and there are ten possible exchanges.
39 Which is the actual electronic configuration of Cr (Z=24) why?
Cr (Z=24) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
5
Cr with 3d
5
is half filled and it will be morestable.
40 Which is the actual electronic configuration of Cu (Z=29) why?
Cu (Z=29) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
10
Cu with 3d
10
is completely filled and it will be more stable
41 The energies of the same orbital decreases with an increase in the atomic number. Justify
this statement.
Atomic number is inversely proportional to the energy of the same orbital.
For example the energy of the 2s orbital of Lithium atom is greater than that of Sodium.
42 State hund’s rule of maximum multiplicity
It states that electron pairing in the degenerate orbitals does not take place until all the
available orbitals contain one electron each.
Eg:- Electronic configuration of Carbon ↿⇂ ↿⇂ ↿ ↿
1s
2
2s
2
2px
1
2py
1
43 What are degenerate orbitals?
Three different orientations in space that are possible for a p-orbital. All the three p - orbitals
namely px, py and pz have same energies and are called degenerate orbitals.
In the presence of magnetic or electric filed the degeneracy is lost.
44 What are Ground state and Excited states?
Ground state: - The electron in the hydrogen atom occupies the 1s orbital that has the
lowestenergy.
Excited states : -When this electron gains some energy, it moves to the higher energy orbitals
such as 2s, 2p etc….
45 What is Effective nuclear charge?
The net charge experienced by the electron is called effective nuclear charge.
The effective nuclear charge depends on the shape of the orbitals.
The order of the effective nuclear charge felt by a electron in an orbital within the given shell is
s > p > d > f.
Greater the effective nuclear charge, greater is the stability of the orbital.
46 What is radial node (or) radial surface ?
The region where this probability density function reduces to zero is called nodal surface or a
radial node.
ns-orbital has (n–1) nodes
np and nd-orbital has (n–(l + 1) nodes
47 How many unpaired electrons are present in the ground state of Cr
3+
( Z = 24) and
Ne(Z=10)
Cr (Z=24) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
5
Cr
3+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
3
it contains 3 unpaired electrons
Ne (Z=10) 1s
2
2s
2
2p
6
no unpaired electrons in it
48 Define Quantum number and what are its types?
The electron in a atom can explained by Quantum numbers
Types of quantum numbers
1, Principal quantum number 2, Azimuthal quantum number 3, magnetic quantum number and 4,
Spin quantum number
It is denoted by the symbol „n‟
The „n‟ can have the values 1,2,3….
Shell K L M N
„n‟ value 1 2 3 4
The maximum number of electrons in a shell can be calculated by the formula 2n
2
.
The energy of the electron is given by
En =
−1312.8Z
2
n
2
KJmol
-1
.
The distance of the electron nucleus is given by rn =
0.529 n
2
�
A
0
.
31 Write a notes on Azimuthal quantum number
It is denoted by the symbol „l‟
l = (n-1) ,the „l‟ can have the values 0, 1,2,3….
Sub Shell (orbitals) s p d f
„n‟ value 0 1 2 3
The maximum number of electrons in a orbital can be calculated by the formula 2(2l+1).
It is used to calculate the orbitals angular momentum by using
32 Write a notes on Magnetic quantum number
It is denoted by the symbol „m‟
Its value ranging from –l to + l through 0
i.e, if l =1 m = -1, 0 and +1
The values of „l‟ represent different orientation of orbitals in space
The Zeeman effect experimental justification of this quantum number.
33 Write a notes on Spin quantum number
It is denoted by the symbol „s‟
The electron in an atom revolves around the nucleus and also spins in a clockwise direction or
in anti-clockwise direction.
Spin direction clockwise anti-clockwise
‘s’ value +½ -½
34
.
Explain the shapes of s-orbital.
For s-orbital l = 0 and hence, m can have only one value,i.e., m = 0.
This means that the probability of finding the electron in s-orbital is the
same in all directions at a particular distance.
In other words s-orbitals are spherically symmetrical.
35
.
Explain the shapes of p-orbitals.
p-orbitals, l = 1 and the corresponding m values are 1, 0
&+1.
The three different „m‟ values indicates that there are
three different possible orientations as px, py and pz and the angular distribution along
the x, y and z axis respectively.
36
.
Explain the shapes of d-orbitals.
For „d‟ orbital l = 2 and the corresponding m values
are -2, -1, 0,+1,+2.
The shape of the d orbital looks like a 'clover leaf '.
The five m values give rise to five d orbitals
namely dxy, dyz, dzx,dx2-y2 and dz2.
The 3d orbitals contain two nodal plane
37
.
Explain the shapes of f-orbitals
For 'f ' orbital, l = 3 and the m values are -3, -2,-1, 0, +1, +2, +3
Seven f orbitals fz
3
, fxz
2
, fyz
2
, fxyz, fz(x
2
−y
2
), fx(x
2
−3y
2
), fy(3x
2
−y
2
),
There are 3 nodal planes in the f-orbitals. 38 Define Exchange energy
If two or more electrons with thesame spin are present in degenerate orbitals, there is a
possibility for exchanging their positions.
During exchange process the energy is released and the released energy is called exchange
energy.
For example, in chromium the electronic configuration is [Ar]3d
5
4s
1
. The 3d orbital ishalf
filled and there are ten possible exchanges.
39 Which is the actual electronic configuration of Cr (Z=24) why?
Cr (Z=24) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
5
Cr with 3d
5
is half filled and it will be morestable.
40 Which is the actual electronic configuration of Cu (Z=29) why?
Cu (Z=29) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
10
Cu with 3d
10
is completely filled and it will be more stable
41 The energies of the same orbital decreases with an increase in the atomic number. Justify
this statement.
Atomic number is inversely proportional to the energy of the same orbital.
For example the energy of the 2s orbital of Lithium atom is greater than that of Sodium.
42 State hund’s rule of maximum multiplicity
It states that electron pairing in the degenerate orbitals does not take place until all the
available orbitals contain one electron each.
Eg:- Electronic configuration of Carbon ↿⇂ ↿⇂ ↿ ↿
1s
2
2s
2
2px
1
2py
1
43 What are degenerate orbitals?
Three different orientations in space that are possible for a p-orbital. All the three p - orbitals
namely px, py and pz have same energies and are called degenerate orbitals.
In the presence of magnetic or electric filed the degeneracy is lost.
44 What are Ground state and Excited states?
Ground state: - The electron in the hydrogen atom occupies the 1s orbital that has the
lowestenergy.
Excited states : -When this electron gains some energy, it moves to the higher energy orbitals
such as 2s, 2p etc….
45 What is Effective nuclear charge?
The net charge experienced by the electron is called effective nuclear charge.
The effective nuclear charge depends on the shape of the orbitals.
The order of the effective nuclear charge felt by a electron in an orbital within the given shell is
s > p > d > f.
Greater the effective nuclear charge, greater is the stability of the orbital.
46 What is radial node (or) radial surface ?
The region where this probability density function reduces to zero is called nodal surface or a
radial node.
ns-orbital has (n–1) nodes
np and nd-orbital has (n–(l + 1) nodes
47 How many unpaired electrons are present in the ground state of Cr
3+
( Z = 24) and
Ne(Z=10)
Cr (Z=24) 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
3d
5
Cr
3+
1s
2
2s
2
2p
6
3s
2
3p
6
3d
3
it contains 3 unpaired electrons
Ne (Z=10) 1s
2
2s
2
2p
6
no unpaired electrons in it
48 Define Quantum number and what are its types?
The electron in a atom can explained by Quantum numbers
Types of quantum numbers
1, Principal quantum number 2, Azimuthal quantum number 3, magnetic quantum number and 4,
Spin quantum number
49 Write a notes on Zeeman efeect
The spitting of spectral lines in the presence of magnetic field.
50 Write a notes on Stark efeect
The spitting of spectral lines in the presence of electrical field.
51 Calculate the orbital angular momentum for ‘d’ and ‘f ’ orbital.
Angular momentum for „d‟orbital : l = 2
mvr =
h
2
� (�+1) =
h
2
2 (2+1) =
3 h
2
Angular momentum for „f‟orbital : l = 3
mvr =
h
2
� (�+1) =
h
2
3 (3+1) =
3 h
52 In degenerate orbitals, why do the completely filled and half filled configurations are more
stable than the partially filled configuration ?
Due to high exchange energy
53 Calculate the maximum number of electrons that can be accommodated in L shell
Maximum number of electrons in L shell = 2n
2
(n = 2) 2 x (2)
2
= 8
3 .PERIODIC CLASSIFICATION OF ELEMENTS
1. How Lavoisier classified the elements
Acid making elements Sulphur
Gas like elements Oxygen
Metallic element Gold
Earthly elements Lime
2. State Dobereier’s law of Triads
The atomic weight of the middle element nearly equal to the mean of the atomic weight of the
remaining two elements.
Elements like chlorine, Bromine and Iodine have same chemical properties in the group. These
are called as triads.
3. State the Newland’s law of Octaves
On arranging the elements in the increasing order of atomic weights, the property of every
eighth elements is similar to the property of the first element.
4. State the Lothar Meyer Periodic law
Lother-Meyer plotted atomic volumes versus atomic weights of elements and obtained a curve.
He pointed out that elements occupying similar positions in the curve possessed similar
properties.
5. State the Mendeleev’s Periodic law
The properties of the elements are the periodic functions of the atomic weights.
6. Describe about the merits of Mendeleev’s periodic table.
The comparative studies of elements were made easier.
The table shows the relationship in properties of elements in a group.
At the time of Mendeleev, about 70 elements were known and thus blank spaces were left for unknown
elements which helped further discoveries.
Both Gallium (Ga) in III group and Germanium (Ge) in IV group were unknown at the time by
Mendeleev predicted their existence and properties.
7. Give the Anomalies of Mendeleev’s Periodic table
Elements with same properties are placed in different groups
Elements with different properties are placed in same groups.
Elements with higher atomic weights are placed before the lower atomic weights.
8. State Modern Periodic law
The Physical and chemical properties of the elements are the periodicfunctions of the atomic
numbers.
9. Define Periodicity
The repetition of the Physical and the chemical properties at regular intervals are called
periodicity. 10. Write a notes on first group elements
Elements of 1
st
group are called alkali metals.
Their Electronic configuration is ns
1
These metals readily lose their outermost electrons to form +1 ion.
These metals are soft with low melting and boiling points.
11. Write a note on period and group in long form of the periodic Table
The long form of the periodic table consists of horizontal rows called periods and vertical
columns called groups.
There are seven periods and each period starts with a different principal quantum number.
A group consists of a series of elements having similar configuration of the outer energy shell.
There are eighteen vertical columns in long from of the periodic table.
12. Write the electronic configuration of Alkali metals and Alkali earth metals.
Electronic configuration of Alkali metals is ns
1
Electronic configuration of Alkali earth metals is ns
2
13. Write the electronic configuration of Transition elements.
Electronic configuration of Transition elements is (n-1)d
1-10
ns
1-2
14. Write the characteristic properties of transition elements
Theyare all metals.
They mostly form colored ions and exhibit variable valency.
They have high boiling and melting points
They are good conductor of heat and electricity
They are used as catalyst
15. Writethe electronic configuration of Lanthanides and actinides
Electronic configuration of Lanthanides is 4f
1-14
5d
0-1
6s
2
Electronic configuration of Actinides is 5f
0-14
6d
0-2
7s
2
16. Write the characteristic properties of p-Block Elements (or) Representative Elements or
Main Group Elements.
The elements of group 13 to 18 are called P-block elements
Their general electron configuration is ns
2
np
1-6
They form covalent compounds
They have high electron affinity and Ionization energy values
The nonmetallic character increases as we move from left to right across a period and metallic
character increases as we go down the group
17. Define atomic radius
The distance between the center of the nucleus and the outer mostshell containing the valence
electron called as atomic radius.
Along the Period: It Decrease along the period.
Along the Group: It Increases along the group
18. Define Covalent radius
The half of the inter nuclear distance between two identical atoms linked by a single covalent bond
is called as Covalent radius
19. Covalent radius is smaller than its corresponding atomic radius. Justify this statement.
The formation of covalent bond involves the overlapping of atomic orbitalsand it reduces the expected
internuclear distance
20. Explain about periodic variation of atomic radius across a period and along a group.
Along the Period :It Decrease along the period.
Reason
As we move along the period the valence electrons are added to the same shell.
So the Nuclear charge increases,
And the attractive force between the valence electron and thenucleus increases
Along the Group: It Increases along the group.
Reason:As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases
The spitting of spectral lines in the presence of magnetic field.
50 Write a notes on Stark efeect
The spitting of spectral lines in the presence of electrical field.
51 Calculate the orbital angular momentum for ‘d’ and ‘f ’ orbital.
Angular momentum for „d‟orbital : l = 2
mvr =
h
2
� (�+1) =
h
2
2 (2+1) =
3 h
2
Angular momentum for „f‟orbital : l = 3
mvr =
h
2
� (�+1) =
h
2
3 (3+1) =
3 h
52 In degenerate orbitals, why do the completely filled and half filled configurations are more
stable than the partially filled configuration ?
Due to high exchange energy
53 Calculate the maximum number of electrons that can be accommodated in L shell
Maximum number of electrons in L shell = 2n
2
(n = 2) 2 x (2)
2
= 8
3 .PERIODIC CLASSIFICATION OF ELEMENTS
1. How Lavoisier classified the elements
Acid making elements Sulphur
Gas like elements Oxygen
Metallic element Gold
Earthly elements Lime
2. State Dobereier’s law of Triads
The atomic weight of the middle element nearly equal to the mean of the atomic weight of the
remaining two elements.
Elements like chlorine, Bromine and Iodine have same chemical properties in the group. These
are called as triads.
3. State the Newland’s law of Octaves
On arranging the elements in the increasing order of atomic weights, the property of every
eighth elements is similar to the property of the first element.
4. State the Lothar Meyer Periodic law
Lother-Meyer plotted atomic volumes versus atomic weights of elements and obtained a curve.
He pointed out that elements occupying similar positions in the curve possessed similar
properties.
5. State the Mendeleev’s Periodic law
The properties of the elements are the periodic functions of the atomic weights.
6. Describe about the merits of Mendeleev’s periodic table.
The comparative studies of elements were made easier.
The table shows the relationship in properties of elements in a group.
At the time of Mendeleev, about 70 elements were known and thus blank spaces were left for unknown
elements which helped further discoveries.
Both Gallium (Ga) in III group and Germanium (Ge) in IV group were unknown at the time by
Mendeleev predicted their existence and properties.
7. Give the Anomalies of Mendeleev’s Periodic table
Elements with same properties are placed in different groups
Elements with different properties are placed in same groups.
Elements with higher atomic weights are placed before the lower atomic weights.
8. State Modern Periodic law
The Physical and chemical properties of the elements are the periodicfunctions of the atomic
numbers.
9. Define Periodicity
The repetition of the Physical and the chemical properties at regular intervals are called
periodicity. 10. Write a notes on first group elements
Elements of 1
st
group are called alkali metals.
Their Electronic configuration is ns
1
These metals readily lose their outermost electrons to form +1 ion.
These metals are soft with low melting and boiling points.
11. Write a note on period and group in long form of the periodic Table
The long form of the periodic table consists of horizontal rows called periods and vertical
columns called groups.
There are seven periods and each period starts with a different principal quantum number.
A group consists of a series of elements having similar configuration of the outer energy shell.
There are eighteen vertical columns in long from of the periodic table.
12. Write the electronic configuration of Alkali metals and Alkali earth metals.
Electronic configuration of Alkali metals is ns
1
Electronic configuration of Alkali earth metals is ns
2
13. Write the electronic configuration of Transition elements.
Electronic configuration of Transition elements is (n-1)d
1-10
ns
1-2
14. Write the characteristic properties of transition elements
Theyare all metals.
They mostly form colored ions and exhibit variable valency.
They have high boiling and melting points
They are good conductor of heat and electricity
They are used as catalyst
15. Writethe electronic configuration of Lanthanides and actinides
Electronic configuration of Lanthanides is 4f
1-14
5d
0-1
6s
2
Electronic configuration of Actinides is 5f
0-14
6d
0-2
7s
2
16. Write the characteristic properties of p-Block Elements (or) Representative Elements or
Main Group Elements.
The elements of group 13 to 18 are called P-block elements
Their general electron configuration is ns
2
np
1-6
They form covalent compounds
They have high electron affinity and Ionization energy values
The nonmetallic character increases as we move from left to right across a period and metallic
character increases as we go down the group
17. Define atomic radius
The distance between the center of the nucleus and the outer mostshell containing the valence
electron called as atomic radius.
Along the Period: It Decrease along the period.
Along the Group: It Increases along the group
18. Define Covalent radius
The half of the inter nuclear distance between two identical atoms linked by a single covalent bond
is called as Covalent radius
19. Covalent radius is smaller than its corresponding atomic radius. Justify this statement.
The formation of covalent bond involves the overlapping of atomic orbitalsand it reduces the expected
internuclear distance
20. Explain about periodic variation of atomic radius across a period and along a group.
Along the Period :It Decrease along the period.
Reason
As we move along the period the valence electrons are added to the same shell.
So the Nuclear charge increases,
And the attractive force between the valence electron and thenucleus increases
Along the Group: It Increases along the group.
Reason:As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases
21. Explain the Nomenclature of Elements using IUPAC rules
The name was derived directly from the atomic number of the new element using the
following numerical roots.
Notation for IUPAC Nomenclature of elements
Digit 0 1 2 3 4 5 6 7 8 9
Root nil un bi tri quat pent hex sept oct enn
Abbreviation n u b t q p h s o e
The numerical roots corresponding to the atomic number are put together and ium‟is added
as suffix
The final „n‟ of „enn‟ is omitted when it is written before „nil‟ (enn + nil = enil) similarly the final
„i' of „bi‟ and „tri‟ is omitted when it written before „ium‟ (bi + ium = bium; tri + ium = trium)
The symbol of the new element is derived from the first letter of the numerical roots.The
following table illustrates these facts.
Atomic
number
Temporary
Name
Temporary
Symbol
Name of the element Symbol
101 Unnilunium Unu Mendelevium Md
102 Unnilbium Unb Nobelium No
103 Unniltrium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordiium Rf
105 Unnilpentium Unp Dubnium Db
106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
22. Define Shielding effect
The inner shell electrons act as a shield between the nucleus and the valence electrons, this
effect is called as Shielding effect.
23. Define Effective nuclear charge
The net nuclear charge experienced by the valence electron in the outer most shell is called as
effective nuclear charge .
24 How to calculate the Effective nuclear charge?
Effective Nuclear charge Zeff = Z - S
Z = atomic number ; S= Shielding constant
25. Define ionic radius
The distance between the center of the nucleus of a ion and the outermost shell containing the
valence electron called as ionic radius.
Along the Period :It Decrease along the period.
Along the Group:It Increases along the group
26. Explain the Pauling method of calculating Ionic radius
In the Pauling method the Ionic radius was Calculated by using the Inter Ionic Distance
Example for Isoelectronic :
Na
+
= 2, 8 ;F
= 2, 8
r( C
+
) + r(A
-
) = d(C+ - A- ) ------ 1
Effective Nucelar charge Zeff = Z – S
1
r( C +) α -------------- -------- 2
Zeff( C +)
1
r( A -) α -------------- -------- 3
Zeff( A -)
r( C +) Zeff ( A -)
-------- = ---------- ------- 4
r( A -) Zeff ( C +)
where
r( C
+
) = ionic radius of the cation ; r( A- ) = ionic radius of the anion
d(C
+
- A
) = inter ionic distance
Z eff( A
) = Effective nuclear charge of the anion
Zeff( C
+
) = Effective nuclear charge of the cation
Using Eqn(1) and (4) Ionic radius can be calculated.
27. The internuclear distance of d
(C-Cl) is experimentally found as 1.78A
0
and the value of r(Cl)
is 0.99A
0
and calculate the atomic radius of carbon.
r
(Cl) + r
(C) = r
(C-Cl)
r
(C) = r
(C-Cl) - r
(Cl)
r
(C) = 1.78 - 0.99
r
(C) = 0.79A
0
28. Explain about periodic variation of ionic radius across a period and along a group.
Along the Period :It Decrease along the period.
Reason
As we move along the period the valence electrons are added to the same shell.
So the Nuclear charge increases,
And the attractive force between the valence electron and thenucleus increases
Along the Group: It Increases along the group.
Reason
As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases
29.
Define ionization energy.
Ionization energy is the amount of energy required to remove an looselybounded electron from
the outermost shell of an atom. Unit is eV
Along the Group : It decreases along the group.
Along the Period :It Increases along the period
30. How would you explain the fact that the second ionization potential is always higher than
first ionization potential.
The total number of electrons is less in a cation than a neutralatom
So the effective nuclear charge of the cation is greater than theatom
The order is IE1 < IE2 < IE3 < ………
31. Explain the periodic trend of Ionization energy .
Along the Group : It decreases along the group.
Reason
As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases.
Hence the nuclear charge decreases and the ionization alsodecreases.
Along the period : It increases along the period
Reason
As we move along the period the valence electrons are added to thesame shell.
So the Nuclear charge increases, And the attraction between the valence electron and the
nucleusincreases
Hence more energy is required to remove the valence electron, soIonization energy increases
The name was derived directly from the atomic number of the new element using the
following numerical roots.
Notation for IUPAC Nomenclature of elements
Digit 0 1 2 3 4 5 6 7 8 9
Root nil un bi tri quat pent hex sept oct enn
Abbreviation n u b t q p h s o e
The numerical roots corresponding to the atomic number are put together and ium‟is added
as suffix
The final „n‟ of „enn‟ is omitted when it is written before „nil‟ (enn + nil = enil) similarly the final
„i' of „bi‟ and „tri‟ is omitted when it written before „ium‟ (bi + ium = bium; tri + ium = trium)
The symbol of the new element is derived from the first letter of the numerical roots.The
following table illustrates these facts.
Atomic
number
Temporary
Name
Temporary
Symbol
Name of the element Symbol
101 Unnilunium Unu Mendelevium Md
102 Unnilbium Unb Nobelium No
103 Unniltrium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordiium Rf
105 Unnilpentium Unp Dubnium Db
106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
22. Define Shielding effect
The inner shell electrons act as a shield between the nucleus and the valence electrons, this
effect is called as Shielding effect.
23. Define Effective nuclear charge
The net nuclear charge experienced by the valence electron in the outer most shell is called as
effective nuclear charge .
24 How to calculate the Effective nuclear charge?
Effective Nuclear charge Zeff = Z - S
Z = atomic number ; S= Shielding constant
25. Define ionic radius
The distance between the center of the nucleus of a ion and the outermost shell containing the
valence electron called as ionic radius.
Along the Period :It Decrease along the period.
Along the Group:It Increases along the group
26. Explain the Pauling method of calculating Ionic radius
In the Pauling method the Ionic radius was Calculated by using the Inter Ionic Distance
Example for Isoelectronic :
Na
+
= 2, 8 ;F
= 2, 8
r( C
+
) + r(A
-
) = d(C+ - A- ) ------ 1
Effective Nucelar charge Zeff = Z – S
1
r( C +) α -------------- -------- 2
Zeff( C +)
1
r( A -) α -------------- -------- 3
Zeff( A -)
r( C +) Zeff ( A -)
-------- = ---------- ------- 4
r( A -) Zeff ( C +)
where
r( C
+
) = ionic radius of the cation ; r( A- ) = ionic radius of the anion
d(C
+
- A
) = inter ionic distance
Z eff( A
) = Effective nuclear charge of the anion
Zeff( C
+
) = Effective nuclear charge of the cation
Using Eqn(1) and (4) Ionic radius can be calculated.
27. The internuclear distance of d
(C-Cl) is experimentally found as 1.78A
0
and the value of r(Cl)
is 0.99A
0
and calculate the atomic radius of carbon.
r
(Cl) + r
(C) = r
(C-Cl)
r
(C) = r
(C-Cl) - r
(Cl)
r
(C) = 1.78 - 0.99
r
(C) = 0.79A
0
28. Explain about periodic variation of ionic radius across a period and along a group.
Along the Period :It Decrease along the period.
Reason
As we move along the period the valence electrons are added to the same shell.
So the Nuclear charge increases,
And the attractive force between the valence electron and thenucleus increases
Along the Group: It Increases along the group.
Reason
As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases
29.
Define ionization energy.
Ionization energy is the amount of energy required to remove an looselybounded electron from
the outermost shell of an atom. Unit is eV
Along the Group : It decreases along the group.
Along the Period :It Increases along the period
30. How would you explain the fact that the second ionization potential is always higher than
first ionization potential.
The total number of electrons is less in a cation than a neutralatom
So the effective nuclear charge of the cation is greater than theatom
The order is IE1 < IE2 < IE3 < ………
31. Explain the periodic trend of Ionization energy .
Along the Group : It decreases along the group.
Reason
As we move down the group the valence electrons are added intonew shells.
As a result the distance between the nucleus and the valenceelectrons increases.
Hence the nuclear charge decreases and the ionization alsodecreases.
Along the period : It increases along the period
Reason
As we move along the period the valence electrons are added to thesame shell.
So the Nuclear charge increases, And the attraction between the valence electron and the
nucleusincreases
Hence more energy is required to remove the valence electron, soIonization energy increases
32. Why Beryllium has high Ionisation energy than Boron ?
Element Electronic configuration
Beryllium ( Be ) Be
4
= 1s
2
2s
2
Boron (B ) B
5
= 1s
2
2s
2
2p
1
Beryllium has high Nuclear Charge
Beryllium has Stable Fully filled ns
2
electronic configuration
33. Why Nitrogen has high Ionisation energy than oxygen ?
Element ம் Electronic configuration
Nitrogen ( N ) N
7
=1s
2
,2s
2
,2p
3
Oxygen (O ) O
8
=1s
2
,2s
2
,2p
4
Nitrogen has Stable Half filled np
3
electronic configuration
The force of attraction between the nucleus and the outermostelectron is very high in Nitrogen
34. Why Lithium has high Ionisation energy than Sodium ?
Element ம் Electronic configuration
Lithium ( Li ) Li
3
=1s
2
,2s
1
Sodium (Na ) Na
11
=1s
2
,2s
2
,2p
6
,3s
1
The size of the Lithium is smaller than Sodium.
The ionization energy increases with decreasing atomic sizes.
35. Define electron affinity
Electron affinity is defined as the amount of energy released when a electron is added to the valence
shell of a atom.
36. Explain the periodic trend of Electron affinity
Along the Group : It decreases along the group.
As we move down the group the nuclear charge decreases
The atomic size increases.
Along the period : It increases along the period
Reason
As we move along the period the nuclear charge increases
The atomic size decreases.
The attraction between the valence electron and the nucleusincreases
37. Why the Electron affinity of Alkali earth metal is zero
The outer Electronic configuration of Alkali earth metal is ns
2
These elements has filled stable electronic configuration.
These elements will not accept electrons
38. Why Nitrogen has Zero electron affinity ?
Electronic configuration of Nitrogen (N
7
) = 1s
2
2s
2
2p
3
Nitrogen has Stable Half filled np
3
electronic configuration
If a electron is added it will disturb the stable electronic configuration.
So it will not accept electrons.
39. Why the 17th group ( Halogens ) have high electron affinity ?
Halogens have a unstable np
5
electronic configuration.
By gaining one electron it becomes a Stable Fully filled np
6
electronicconfiguration
Hence it accepts one electron and become a stable Noble gasconfiguration.
40. Define electronegativity
Electro negativity is a tendency of a element present is covalent molecule to attract the shared
pair ofelectrons towards itself.
41. Explain the Pauling’s method of calculating the Electro Negativity
According the Pauling the EN value of Hydrogen is 2.1 and forFluorine is 4.0
He calculated by using the Formula
( XA – XB ) = 0.182 EA-B - ( EA-A × EB-B )
½
Where EA-B, EA-A and EB-B are the bond energy of AB, A2 and B2molecules. 42. Explain the periodic trend of Electronegativity
Along the Group : It decreases along the group.
As we move down the group the nuclear charge decreases
The atomic size increases.
Along the period : It increases along the period
Reason
As we move along the period the nuclear charge increases
The atomic size decreases.
The attraction between the valence electron and the nucleusincreases
43. Define Diagonal relationship
The similarities in the properties between the diagonally present elements are called as Diagonal
relationship.
Eg Li and Mg have same properties.
Example lithium and magnesium have similar properties
Li Be B C
Na Mg Al Si
44. K
+
and Cl
-
ions are isoelectronic ions, compare the effective nuclear charge value for these
ions.
Electronic configuration of K
+
ion = 1s
2
,2s
2
,2p
6
,3s
2
3p
6
Effctive nuclear charge of K+ ion = Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
2
3p
6
,
(2 x 1) + ( 8 x 0.85) + (7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (7 x 0.35) = 11.25
Z
*
eff= 19 – 11.25 = 7.75
Electronic configuration of Cl
-
ion = 1s
2
,2s
2
,2p
6
,3s
2
3p
6
Effctive nuclear charge of Cl
-
ion = Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
2
3p
6
,
( 2 x 1) + ( 8 x 0.85) + (7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (7 x 0.35) = 11.25
Z
*
eff= = 17 – 11.25 = 5.75
K
+
and Cl
-
ions are isoelectronic ions, but the atomic number of potassium is higher than chlorine. so
the effective nuclear charge of K
+
ion is higher than that of Cl
-
ion.
45. Calculate the effective nuclear charge of sodium atom and sodium ion.
Electronic configuration of Na atom = 1s
2
,2s
2
,2p
6
,3s
1
Effctive nuclear charge of Na atom Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
1
(2 x 1) + ( 8 x 0.85) + (0 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (0 x 0.35) = 8.8
Effctive nuclear charge of Na atom = Z
*
eff= 11 – 8.8 = 2.2
Electronic configuration of Na
+
ion = 1s
2
,2s
2
,2p
6
Effctive nuclear charge of Na
+
ion Z
*
eff = Z – S
Element Electronic configuration
Beryllium ( Be ) Be
4
= 1s
2
2s
2
Boron (B ) B
5
= 1s
2
2s
2
2p
1
Beryllium has high Nuclear Charge
Beryllium has Stable Fully filled ns
2
electronic configuration
33. Why Nitrogen has high Ionisation energy than oxygen ?
Element ம் Electronic configuration
Nitrogen ( N ) N
7
=1s
2
,2s
2
,2p
3
Oxygen (O ) O
8
=1s
2
,2s
2
,2p
4
Nitrogen has Stable Half filled np
3
electronic configuration
The force of attraction between the nucleus and the outermostelectron is very high in Nitrogen
34. Why Lithium has high Ionisation energy than Sodium ?
Element ம் Electronic configuration
Lithium ( Li ) Li
3
=1s
2
,2s
1
Sodium (Na ) Na
11
=1s
2
,2s
2
,2p
6
,3s
1
The size of the Lithium is smaller than Sodium.
The ionization energy increases with decreasing atomic sizes.
35. Define electron affinity
Electron affinity is defined as the amount of energy released when a electron is added to the valence
shell of a atom.
36. Explain the periodic trend of Electron affinity
Along the Group : It decreases along the group.
As we move down the group the nuclear charge decreases
The atomic size increases.
Along the period : It increases along the period
Reason
As we move along the period the nuclear charge increases
The atomic size decreases.
The attraction between the valence electron and the nucleusincreases
37. Why the Electron affinity of Alkali earth metal is zero
The outer Electronic configuration of Alkali earth metal is ns
2
These elements has filled stable electronic configuration.
These elements will not accept electrons
38. Why Nitrogen has Zero electron affinity ?
Electronic configuration of Nitrogen (N
7
) = 1s
2
2s
2
2p
3
Nitrogen has Stable Half filled np
3
electronic configuration
If a electron is added it will disturb the stable electronic configuration.
So it will not accept electrons.
39. Why the 17th group ( Halogens ) have high electron affinity ?
Halogens have a unstable np
5
electronic configuration.
By gaining one electron it becomes a Stable Fully filled np
6
electronicconfiguration
Hence it accepts one electron and become a stable Noble gasconfiguration.
40. Define electronegativity
Electro negativity is a tendency of a element present is covalent molecule to attract the shared
pair ofelectrons towards itself.
41. Explain the Pauling’s method of calculating the Electro Negativity
According the Pauling the EN value of Hydrogen is 2.1 and forFluorine is 4.0
He calculated by using the Formula
( XA – XB ) = 0.182 EA-B - ( EA-A × EB-B )
½
Where EA-B, EA-A and EB-B are the bond energy of AB, A2 and B2molecules. 42. Explain the periodic trend of Electronegativity
Along the Group : It decreases along the group.
As we move down the group the nuclear charge decreases
The atomic size increases.
Along the period : It increases along the period
Reason
As we move along the period the nuclear charge increases
The atomic size decreases.
The attraction between the valence electron and the nucleusincreases
43. Define Diagonal relationship
The similarities in the properties between the diagonally present elements are called as Diagonal
relationship.
Eg Li and Mg have same properties.
Example lithium and magnesium have similar properties
Li Be B C
Na Mg Al Si
44. K
+
and Cl
-
ions are isoelectronic ions, compare the effective nuclear charge value for these
ions.
Electronic configuration of K
+
ion = 1s
2
,2s
2
,2p
6
,3s
2
3p
6
Effctive nuclear charge of K+ ion = Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
2
3p
6
,
(2 x 1) + ( 8 x 0.85) + (7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (7 x 0.35) = 11.25
Z
*
eff= 19 – 11.25 = 7.75
Electronic configuration of Cl
-
ion = 1s
2
,2s
2
,2p
6
,3s
2
3p
6
Effctive nuclear charge of Cl
-
ion = Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
2
3p
6
,
( 2 x 1) + ( 8 x 0.85) + (7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (7 x 0.35) = 11.25
Z
*
eff= = 17 – 11.25 = 5.75
K
+
and Cl
-
ions are isoelectronic ions, but the atomic number of potassium is higher than chlorine. so
the effective nuclear charge of K
+
ion is higher than that of Cl
-
ion.
45. Calculate the effective nuclear charge of sodium atom and sodium ion.
Electronic configuration of Na atom = 1s
2
,2s
2
,2p
6
,3s
1
Effctive nuclear charge of Na atom Z
*
eff = Z – S
1s
2
, 2s
2
,2p
6
, 3s
1
(2 x 1) + ( 8 x 0.85) + (0 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 1) + ( 8 x 0.85) + (0 x 0.35) = 8.8
Effctive nuclear charge of Na atom = Z
*
eff= 11 – 8.8 = 2.2
Electronic configuration of Na
+
ion = 1s
2
,2s
2
,2p
6
Effctive nuclear charge of Na
+
ion Z
*
eff = Z – S
1s
2
, s
2
,2p
6
(2 x 0.85) + ( 7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 0.85) + ( 7 x 0.35) = 4.15
Effctive nuclear charge of Na
+
ion = Z
*
eff = 11 – 4.15 = 6.85
46. Compare the periodic property of the elements on the left side and right side of the
periodic table
Periodic properties Left side elements Right side elements
Atomic radius High Low
Ionic radius High Low
Ionization energy Low High
Electron Affinity Low High
Electro Negativity Low High
47. Explain the 1
st
group elements are Alkali in nature.
For example 1st group element sodium react with oxygen to give sodium oxide. and further
sodium oxide react with water to give strong base sodium hydroxide.
4 Na + O2 2 Na2O
Na2O + H2O 2NaOH(strong base)
48. Explain the Halogen are Acidic in nature.
For example chlorine react with oxygen to give Cl2O7, and further react with water to give
strong acid called perchloric acid.
2 Cl2 + 7 O2 2 Cl2O7
Cl2O7 + H2O 2 HClO4 (strong acid)
49. Beryllium hydroxide as amphoteric in nature – Explain.
Beryllium hydroxide reacts with both acid and base as it is amphoteric in nature
Be(OH)2 + HCl BeCl2 + 2H2O
Be(OH)2 + 2 NaOH Na2BeO2 + 2H2O
50. Define Isoelectronic ions.
Ions having same number of electrons and same electronic configuration are called as
Isoelectronic ions
Example for Isoelectronic:
Na
+
= 1s
2
,2s
2
,2p
6
; F
= 1s
2
,2s
2
,2p
6
51. Why halogens act as oxidizing agents ?
Halogens have high electron negativity and electron affinity values.
Halogens have a unstable np
5
electronic configuration.
By gaining one electron it becomes a Stable Fully filled np
6
electronicconfiguration
Hence it accepts one electron and become a stable Noble gasconfiguration
4 .HYDROGEN
1. Explain why hydrogen is not placed with the halogen in the periodic table.
Hydrogen resembles alkali (group 1 elements) as well as Halogens (Group 17 elements )
Hydrogen resembles more alkali metals than halogens
Electron affinity value of Hydrogen is much less than halogens.
The tendency to form hydride ion is low compound to that of halogens.
In most of its compound hydrogen exists in +1 oxidation state.
2. An the cube at 0
0
C is placed in some liquid water at 0
0
C, the ice cube sinks - Why ?
Ice has low density than water.
At 0
0
C Ice and Water are equilibrium
So liquid water at 0
0
C, the ice cube sinks 3. Discuss the three types of Covalent hydrides.
Sl.N
o
Covalent hydrides Example
1 Electron Precise Hydride CH4
2 Electron rich hydride NH3
3 Electron deficient hydride B2H6
4. Predict which of the following hydrides is a gas on a solid (a) HCl (b) NaH. Give your
reason.
(a) HCl : - At room temperature HCl is a colourless gas and the solution of HCl in water is called
Hydrochloric acid and it is in liquid state.
(b) NaH :- Sodium hydride (NaH) is an ionic compound and it has octahedral crystal structure
5. Write the expected formulas for the hydrides of 4th period elements. What is the trend in
the formulas? In what way the first two numbers of the series different from the others
The hydrides of 4th period elements
Sl.No Group Type Example
1 1 & 2 Ionic hydrides KH & CaH2
2 3 to 12 Metallic hydrides TiH1.5-1.8
3 13 Electron deficient hydride GaH3
4 14 Electron Precise Hydride CH4
5 15 Electron rich hydride NH3
First two numbers of the series are Ionic hydrides.
6. Write chemical equation for the following reactions.
i) Reaction of hydrogen with tungsten (VI) oxide on heating.
WO3 + 3 H2 W + 3 H2O
ii) Hydrogen gas and chlorine gas.
H2 + Cl2 2HCl
7. Complete the following chemical reactions and classify them in to (a) hydrolysis (b)redox (c)
hydration reactions.
1) KMnO4 + H2O2
KMnO4 + H2O2 MnO2 + KOH + H2O + O2 (Redox reaction)
2) CrCl3 + H2O
CrCl3 + H2O [Cr(H2O)6]Cl3 (Hydrationreaction)
3) CaO + H2O
CaO + H2O Ca(OH)2 (Hydrolysisreaction)
8. Hydrogen peroxide can function as an oxidising agent as well as reducing agent.
Substantiate this statement with suitable examples.
Oxidation takes place in acidic medium
2FeSO4 + H2SO4 + H2O2 Fe2(SO4)3 + 2H2O
Reduction takes place in basic medium
2KMnO4 + 3H2O2 2MnO2+ 2KOH + 2H2O +3O2
9. Do you think that heavy water can be used for drinking purposes ?
It is not completely safe to drink.
The bio chemical reaction in our cells are affected by the difference in the mass of hydrogen atoms.
10. What is water-gas shift reaction ?
The mixture of carbon mono oxide and hydrogen are called Water Gas.
When carbon mono oxide of Water gas can be converted into carbon dioxide by mixing the gas
mixture with more steam (H2O).
CO + H2O
Cu Fe 400
0
C
CO2 + H2
2
, s
2
,2p
6
(2 x 0.85) + ( 7 x 0.35)
(Select the electron for which the value of S is to be calculated)
s = (2 x 0.85) + ( 7 x 0.35) = 4.15
Effctive nuclear charge of Na
+
ion = Z
*
eff = 11 – 4.15 = 6.85
46. Compare the periodic property of the elements on the left side and right side of the
periodic table
Periodic properties Left side elements Right side elements
Atomic radius High Low
Ionic radius High Low
Ionization energy Low High
Electron Affinity Low High
Electro Negativity Low High
47. Explain the 1
st
group elements are Alkali in nature.
For example 1st group element sodium react with oxygen to give sodium oxide. and further
sodium oxide react with water to give strong base sodium hydroxide.
4 Na + O2 2 Na2O
Na2O + H2O 2NaOH(strong base)
48. Explain the Halogen are Acidic in nature.
For example chlorine react with oxygen to give Cl2O7, and further react with water to give
strong acid called perchloric acid.
2 Cl2 + 7 O2 2 Cl2O7
Cl2O7 + H2O 2 HClO4 (strong acid)
49. Beryllium hydroxide as amphoteric in nature – Explain.
Beryllium hydroxide reacts with both acid and base as it is amphoteric in nature
Be(OH)2 + HCl BeCl2 + 2H2O
Be(OH)2 + 2 NaOH Na2BeO2 + 2H2O
50. Define Isoelectronic ions.
Ions having same number of electrons and same electronic configuration are called as
Isoelectronic ions
Example for Isoelectronic:
Na
+
= 1s
2
,2s
2
,2p
6
; F
= 1s
2
,2s
2
,2p
6
51. Why halogens act as oxidizing agents ?
Halogens have high electron negativity and electron affinity values.
Halogens have a unstable np
5
electronic configuration.
By gaining one electron it becomes a Stable Fully filled np
6
electronicconfiguration
Hence it accepts one electron and become a stable Noble gasconfiguration
4 .HYDROGEN
1. Explain why hydrogen is not placed with the halogen in the periodic table.
Hydrogen resembles alkali (group 1 elements) as well as Halogens (Group 17 elements )
Hydrogen resembles more alkali metals than halogens
Electron affinity value of Hydrogen is much less than halogens.
The tendency to form hydride ion is low compound to that of halogens.
In most of its compound hydrogen exists in +1 oxidation state.
2. An the cube at 0
0
C is placed in some liquid water at 0
0
C, the ice cube sinks - Why ?
Ice has low density than water.
At 0
0
C Ice and Water are equilibrium
So liquid water at 0
0
C, the ice cube sinks 3. Discuss the three types of Covalent hydrides.
Sl.N
o
Covalent hydrides Example
1 Electron Precise Hydride CH4
2 Electron rich hydride NH3
3 Electron deficient hydride B2H6
4. Predict which of the following hydrides is a gas on a solid (a) HCl (b) NaH. Give your
reason.
(a) HCl : - At room temperature HCl is a colourless gas and the solution of HCl in water is called
Hydrochloric acid and it is in liquid state.
(b) NaH :- Sodium hydride (NaH) is an ionic compound and it has octahedral crystal structure
5. Write the expected formulas for the hydrides of 4th period elements. What is the trend in
the formulas? In what way the first two numbers of the series different from the others
The hydrides of 4th period elements
Sl.No Group Type Example
1 1 & 2 Ionic hydrides KH & CaH2
2 3 to 12 Metallic hydrides TiH1.5-1.8
3 13 Electron deficient hydride GaH3
4 14 Electron Precise Hydride CH4
5 15 Electron rich hydride NH3
First two numbers of the series are Ionic hydrides.
6. Write chemical equation for the following reactions.
i) Reaction of hydrogen with tungsten (VI) oxide on heating.
WO3 + 3 H2 W + 3 H2O
ii) Hydrogen gas and chlorine gas.
H2 + Cl2 2HCl
7. Complete the following chemical reactions and classify them in to (a) hydrolysis (b)redox (c)
hydration reactions.
1) KMnO4 + H2O2
KMnO4 + H2O2 MnO2 + KOH + H2O + O2 (Redox reaction)
2) CrCl3 + H2O
CrCl3 + H2O [Cr(H2O)6]Cl3 (Hydrationreaction)
3) CaO + H2O
CaO + H2O Ca(OH)2 (Hydrolysisreaction)
8. Hydrogen peroxide can function as an oxidising agent as well as reducing agent.
Substantiate this statement with suitable examples.
Oxidation takes place in acidic medium
2FeSO4 + H2SO4 + H2O2 Fe2(SO4)3 + 2H2O
Reduction takes place in basic medium
2KMnO4 + 3H2O2 2MnO2+ 2KOH + 2H2O +3O2
9. Do you think that heavy water can be used for drinking purposes ?
It is not completely safe to drink.
The bio chemical reaction in our cells are affected by the difference in the mass of hydrogen atoms.
10. What is water-gas shift reaction ?
The mixture of carbon mono oxide and hydrogen are called Water Gas.
When carbon mono oxide of Water gas can be converted into carbon dioxide by mixing the gas
mixture with more steam (H2O).
CO + H2O
Cu Fe 400
0
C
CO2 + H2
11. Justify the position of hydrogen in the periodic table ?
Hydrogen has similaritieswith alkali metals as well as the halogens
It is difficult to find the right position inthe periodic table.
However, in most of itscompounds hydrogen exists in +1 oxidationstate.
Therefore, it is reasonable to placethe hydrogen in group 1 along with alkalimetals as shown in the
latest periodic table published by IUPAC
12. What are isotopes? Write the names of isotopes of hydrogen
Hydrogen has three naturally occurring isotopes,
Protium (1H
1
or H), Deuterium (1H
2
or D) and Tritium (1H
3
orT).
13. Give the uses of heavy water
It is used as Moderators in Nuclear reactor.
It is used as tracer element to study the mechanisms of organic reactions.
It is used as coolant in nuclear reactors to absorb the heat.
14. Explain the exchange reactions of deuterium.
CH4 + 2 D2 CD4 + 2 H2
2 NH3 + 3 D2 2 ND3 + 3 H2.
15. How do you convert para hydrogen into ortho hydrogen ?
By using catalyst like Iron
By passing electric discharge
By heating at 800
0
C
By mixing with paramagnetic molecules like oxygen
By mixing with atomic hydrogen
16. Mention the uses of deuterium.
It is used as tracer element to study the mechanisms of organic reactions.
High speed deuterium is used in Artificial radio activity
It is used in the preparation of heavy water which used as Moderators in Nuclear reactor
17. Explain preparation of hydrogen using electrolysis.
Hydrogen is prepared by the electrolysis of water containing small amountof NaOH.
Anode : Nickel
Cathode : Iron
Anode:- 2OH
-
-
H2O+½O2+2e
-
Cathode:- 2 H2O+2 e
-
2OH
-
+H2
Over all reaction H2O H2 + ½ O2
18. A groups metal (A) which is present in common salt reacts with (B) to give compound (C)
in which hydrogen is present in –1 oxidation state. (B) on reaction with a gas (C) to give
universal solvent (D). The compound (D) on reacts with (A) to give ((F)), a strong base.
Identify A, B, C, D,E and F.
Explain the reactions.
2Na + H2 2NaH
(A) (B) (C)
2H2 + O2 2H2O
(A) (D) (E)
2Na + 2H2O 2NaOH + H2O
(F)
Compounds Molecular Formula Name
A Na Sodium
B H2 Hydrogen
C NaH Sodium Hydride
D O2 Oxygen E H2O Water
F NaOH Sodium Hydroxide
19. An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies
group number 16 and period number 2 to give compound (B) is used as a modulator in
nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C3H6
to give (D). Identify A, B, C and D.
2D2 + O2 2D2O
(A) (B)
3D2 + CH3-CH = CH2 CH3-CHD-CH2D + 3H2
(A) (C) (D)
Compounds Molecular Formula Name
A D2 Deuterium
B D2O Heavy Water
C CH3-CH=CH2 Propene
D CH3-CHD-CH2D Deutride propene
20. NH3 has exceptionally high melting point and boiling point as compared to those of the
hydrides of the remaining element of group 15 - Explain.
Nitrogen has more electronegative value other than 15
th
group element.
NH3 has inter molecular Hydrogen bonding.
So It has high melting point and boiling point.
21. Why interstitial hydrides have a lower density than the parent metal.
Most of the hydrides are non-stoichiometric with variable composition
(TiH1.5-1.8 and PdH0.6-0.8)
So it have lower density than parent metal
22. How do you expect the metallic hydrides to be useful for hydrogen storage ?
In metallic hydrides, Hydrogen is absorbed as hydrogen atoms due to adsorption of hydrogen
atoms the metal lattice expands and becomes unstable.
Thus when metallic hydride is heated it decomposes to form hydrogen and finally divided metal.
23. Compare the structures of H2O and H2O2.
H2O H2O2
Bent structure
H-O-Hbond angle is104.5
0
O
H 104.5
0
H
The hydrogen atoms would lie on the pages of a partly opened
book,and the oxygen atoms along the spine.
H2O2 has a non-polar structure. Themolecular dimensions in
the gas phase andsolid phase differas shown in fig
24. Arrange NH3, H2O and HF in the order of increasing magnitude of hydrogen bandingand
explain the basis for your arrangement.
Increasing magnitude of Hydrogen bonding NH3 < H2O < HF
Increasing order of their electro negativities are N < O < F
Other questions
1. Explain the Lab preparation of hydrogen
Hydrogen is prepared by heating Zinc with dilute Acids
Zn + 2HCl ZnCl2 + H2
H
0.95A
94.8
o
H
115.5
o
0.98 A
101.9
o
1.47A
H
1.46A
90.2
H
H
H
H
o
Hydrogen has similaritieswith alkali metals as well as the halogens
It is difficult to find the right position inthe periodic table.
However, in most of itscompounds hydrogen exists in +1 oxidationstate.
Therefore, it is reasonable to placethe hydrogen in group 1 along with alkalimetals as shown in the
latest periodic table published by IUPAC
12. What are isotopes? Write the names of isotopes of hydrogen
Hydrogen has three naturally occurring isotopes,
Protium (1H
1
or H), Deuterium (1H
2
or D) and Tritium (1H
3
orT).
13. Give the uses of heavy water
It is used as Moderators in Nuclear reactor.
It is used as tracer element to study the mechanisms of organic reactions.
It is used as coolant in nuclear reactors to absorb the heat.
14. Explain the exchange reactions of deuterium.
CH4 + 2 D2 CD4 + 2 H2
2 NH3 + 3 D2 2 ND3 + 3 H2.
15. How do you convert para hydrogen into ortho hydrogen ?
By using catalyst like Iron
By passing electric discharge
By heating at 800
0
C
By mixing with paramagnetic molecules like oxygen
By mixing with atomic hydrogen
16. Mention the uses of deuterium.
It is used as tracer element to study the mechanisms of organic reactions.
High speed deuterium is used in Artificial radio activity
It is used in the preparation of heavy water which used as Moderators in Nuclear reactor
17. Explain preparation of hydrogen using electrolysis.
Hydrogen is prepared by the electrolysis of water containing small amountof NaOH.
Anode : Nickel
Cathode : Iron
Anode:- 2OH
-
-
H2O+½O2+2e
-
Cathode:- 2 H2O+2 e
-
2OH
-
+H2
Over all reaction H2O H2 + ½ O2
18. A groups metal (A) which is present in common salt reacts with (B) to give compound (C)
in which hydrogen is present in –1 oxidation state. (B) on reaction with a gas (C) to give
universal solvent (D). The compound (D) on reacts with (A) to give ((F)), a strong base.
Identify A, B, C, D,E and F.
Explain the reactions.
2Na + H2 2NaH
(A) (B) (C)
2H2 + O2 2H2O
(A) (D) (E)
2Na + 2H2O 2NaOH + H2O
(F)
Compounds Molecular Formula Name
A Na Sodium
B H2 Hydrogen
C NaH Sodium Hydride
D O2 Oxygen E H2O Water
F NaOH Sodium Hydroxide
19. An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies
group number 16 and period number 2 to give compound (B) is used as a modulator in
nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C3H6
to give (D). Identify A, B, C and D.
2D2 + O2 2D2O
(A) (B)
3D2 + CH3-CH = CH2 CH3-CHD-CH2D + 3H2
(A) (C) (D)
Compounds Molecular Formula Name
A D2 Deuterium
B D2O Heavy Water
C CH3-CH=CH2 Propene
D CH3-CHD-CH2D Deutride propene
20. NH3 has exceptionally high melting point and boiling point as compared to those of the
hydrides of the remaining element of group 15 - Explain.
Nitrogen has more electronegative value other than 15
th
group element.
NH3 has inter molecular Hydrogen bonding.
So It has high melting point and boiling point.
21. Why interstitial hydrides have a lower density than the parent metal.
Most of the hydrides are non-stoichiometric with variable composition
(TiH1.5-1.8 and PdH0.6-0.8)
So it have lower density than parent metal
22. How do you expect the metallic hydrides to be useful for hydrogen storage ?
In metallic hydrides, Hydrogen is absorbed as hydrogen atoms due to adsorption of hydrogen
atoms the metal lattice expands and becomes unstable.
Thus when metallic hydride is heated it decomposes to form hydrogen and finally divided metal.
23. Compare the structures of H2O and H2O2.
H2O H2O2
Bent structure
H-O-Hbond angle is104.5
0
O
H 104.5
0
H
The hydrogen atoms would lie on the pages of a partly opened
book,and the oxygen atoms along the spine.
H2O2 has a non-polar structure. Themolecular dimensions in
the gas phase andsolid phase differas shown in fig
24. Arrange NH3, H2O and HF in the order of increasing magnitude of hydrogen bandingand
explain the basis for your arrangement.
Increasing magnitude of Hydrogen bonding NH3 < H2O < HF
Increasing order of their electro negativities are N < O < F
Other questions
1. Explain the Lab preparation of hydrogen
Hydrogen is prepared by heating Zinc with dilute Acids
Zn + 2HCl ZnCl2 + H2
H
0.95A
94.8
o
H
115.5
o
0.98 A
101.9
o
1.47A
H
1.46A
90.2
H
H
H
H
o
2. Give the preparation of Deuterium by the electrolysis of heavy water
The dissociation of water faster than heavy water
When water is electrolyzed hydrogen is liberated faster than deuterium
The electrolysis is continued till the solution rich in heavy water and finally gives deuterium.
2D2O
���������������
2D2 + O2
3. Explain Ortho and Para hydrogen
In hydrogen molecule, if the two nuclei rotates in the same direction is called as Ortho hydrogen
In a hydrogen molecule, if the two nuclei rotates in the opposite direction is called as Para
hydrogen
Orthohydrogen Para hydrogen
4. Give the difference ( Compare ) between Ortho and para hydrogen
Sl.No Ortho hydrogen Para hydrogen
1 Both the nuclei rotates in the samedirection Both the nuclei rotates in theopposite direction
2 75% at room temperature 25% at room temperature
3 It is more stable It is less stable
4 It has a net magnetic moment It has Zero magnetic moment
5. Give the preparation of Tritium
Tritium is prepared by bombarding Lithium with slow neutrons.
3Li
6
+ 0n
1
1T
3
+ 2He
4
6. Which molecule react with hydrogen to cause explosive reaction and write the equation.
Hydrogen reacts with oxygen to give water. This is an explosive reaction and releases lot of
energy. This is used in fuelcells to generate electricity.
2H2 + O2 2H2O
7. Which metal react with hydrogen to give ionic hydride?
It also has a tendency to react with reactive metals such as lithium, sodium and calcium to give
ionic hydrides in which theoxidation state of hydrogen is -1.
2Na + H2 2NaH
2Li + H2 2LiH
8. Explain Hydrogen act as a reducing agent
In the presence of finely divided nickel, it adds to the unsaturated organic compounds to form
saturated compounds.
CH CH
��/??????
�
CH2 = CH2
��/??????
�
CH3 - CH3
Acetylene Ethylene Ethane
9. List the uses of Hydrogen
Liquid hydrogen is used as Rocket Fuel
Hydrogen is used for preparing Fertilizer and explosives
It is used as catalyst for the preparation of Vanaspathi.
10. What happens Red hot Iron react with steam ? and write the equation
Steam with Red hot Iron :- Steam passed over red hot iron results in the formation of iron oxide
with the release of hydrogen.
3 Fe + 4H2O Fe3O4 + H2
11. What happens Carbon react with steam ? and write the equation
Carbon will react with steam to givewater gas
C + H2O CO + H2
12. What happens Chlorine react with water ? and write the equation
Chlorine react with water to give hydrochloric acid and hypochlorous acid.
Cl2 + 2H2O 2HCl + 2HOCl 13. What happens Fluorine react with water ? and write the equation
Fluorine react with water to give oxygen and hydrogen fluoride.
2F2 + 2H2O 4HF + O2
14. Write chemical reaction to show the amphoteric nature of water
It has the ability to accept as well as donate protons and hence it can act as an acid or a base.
NH3 + H
+
-OH
NH4
+
+ OH
(acid) (donate proton(H
+
))
H
+
- Cl
+ H2O H3O
+
+ Cl
(base) (accept proton(H
+
))
15. What causes the Temporary hardness and permanent Hardness of water?
Temporary hardness is due to the presence of Bicarbonates ofMagnesium and Calcium.
Permanent hardness is due to the presence of Chlorides and Sulphates of Magnesium and
Calcium
16. Explain the removing of Temporary hardness of water ?
Boiling :-It can be removed by boiling and filtration.
Ca(HCO3)2 CaCO3 + CO2 + H2O
Mg(HCO3)2 MgCO3 + CO2 + H2O
MgCO3 + H2O Mg(OH)2 + CO2
Clark’s method: -Calculated amount of lime is added to hard waterand filtered-off.
Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O
Mg(HCO3)2 + Ca(OH)2 CaCO3 + Mg(OH)2 + 2H2O
17. Explain the removing of permanent hardness of water?
It can be removed by adding washing soda (Na2CO3)
CaCl2 + Na2CO3 CaCO3 + 2NaCl
MgSO4 + Na2CO3 MgCO3 + Na2SO4
18. Explain the Ion Exchange method of of Softening hard water.
In this method the hard water is passed through an ion exchange Zeolites.
The Zeolites are hydrated Sodium Alumino Silicates (NaO_Al2O3_xSiO2_yH2O)
The Zeolites contains porous structure of mono valent sodium ions, which is exchanged with
Calcium and Magnesium ions in water.
Na2-Z + Ca
2+
2Na
+
+ Ca – Z
Na2- Z + Mg
2+
2Na
+
+ Mg– Z
The Zeolites can be reused by treating with sodium chloride.
Ca -Z +2NaCl CaCl2 + Na2-Z (Zeolites )
19. Explain the exchange reaction of Heavy water
NaOH + D – OD NaOD + H – OD
HCl + D – OD DCl + H – OD
NH4Cl + 4D – OD ND4Cl + 4H – OD
20. Givethe action of Heavy water with the following
a)Aluminium Carbide , b)Calcium Carbide and
c) Calcium Phosphide d)Magnesium Nitride
a)Aluminium Carbide
Al4C3 + 12 D2O 4 Al(OD)3+ 3 CD4
b)Calcium Carbide
CaC2 + 2D–OD Ca(OD)2 + C2D2
The dissociation of water faster than heavy water
When water is electrolyzed hydrogen is liberated faster than deuterium
The electrolysis is continued till the solution rich in heavy water and finally gives deuterium.
2D2O
���������������
2D2 + O2
3. Explain Ortho and Para hydrogen
In hydrogen molecule, if the two nuclei rotates in the same direction is called as Ortho hydrogen
In a hydrogen molecule, if the two nuclei rotates in the opposite direction is called as Para
hydrogen
Orthohydrogen Para hydrogen
4. Give the difference ( Compare ) between Ortho and para hydrogen
Sl.No Ortho hydrogen Para hydrogen
1 Both the nuclei rotates in the samedirection Both the nuclei rotates in theopposite direction
2 75% at room temperature 25% at room temperature
3 It is more stable It is less stable
4 It has a net magnetic moment It has Zero magnetic moment
5. Give the preparation of Tritium
Tritium is prepared by bombarding Lithium with slow neutrons.
3Li
6
+ 0n
1
1T
3
+ 2He
4
6. Which molecule react with hydrogen to cause explosive reaction and write the equation.
Hydrogen reacts with oxygen to give water. This is an explosive reaction and releases lot of
energy. This is used in fuelcells to generate electricity.
2H2 + O2 2H2O
7. Which metal react with hydrogen to give ionic hydride?
It also has a tendency to react with reactive metals such as lithium, sodium and calcium to give
ionic hydrides in which theoxidation state of hydrogen is -1.
2Na + H2 2NaH
2Li + H2 2LiH
8. Explain Hydrogen act as a reducing agent
In the presence of finely divided nickel, it adds to the unsaturated organic compounds to form
saturated compounds.
CH CH
��/??????
�
CH2 = CH2
��/??????
�
CH3 - CH3
Acetylene Ethylene Ethane
9. List the uses of Hydrogen
Liquid hydrogen is used as Rocket Fuel
Hydrogen is used for preparing Fertilizer and explosives
It is used as catalyst for the preparation of Vanaspathi.
10. What happens Red hot Iron react with steam ? and write the equation
Steam with Red hot Iron :- Steam passed over red hot iron results in the formation of iron oxide
with the release of hydrogen.
3 Fe + 4H2O Fe3O4 + H2
11. What happens Carbon react with steam ? and write the equation
Carbon will react with steam to givewater gas
C + H2O CO + H2
12. What happens Chlorine react with water ? and write the equation
Chlorine react with water to give hydrochloric acid and hypochlorous acid.
Cl2 + 2H2O 2HCl + 2HOCl 13. What happens Fluorine react with water ? and write the equation
Fluorine react with water to give oxygen and hydrogen fluoride.
2F2 + 2H2O 4HF + O2
14. Write chemical reaction to show the amphoteric nature of water
It has the ability to accept as well as donate protons and hence it can act as an acid or a base.
NH3 + H
+
-OH
NH4
+
+ OH
(acid) (donate proton(H
+
))
H
+
- Cl
+ H2O H3O
+
+ Cl
(base) (accept proton(H
+
))
15. What causes the Temporary hardness and permanent Hardness of water?
Temporary hardness is due to the presence of Bicarbonates ofMagnesium and Calcium.
Permanent hardness is due to the presence of Chlorides and Sulphates of Magnesium and
Calcium
16. Explain the removing of Temporary hardness of water ?
Boiling :-It can be removed by boiling and filtration.
Ca(HCO3)2 CaCO3 + CO2 + H2O
Mg(HCO3)2 MgCO3 + CO2 + H2O
MgCO3 + H2O Mg(OH)2 + CO2
Clark’s method: -Calculated amount of lime is added to hard waterand filtered-off.
Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O
Mg(HCO3)2 + Ca(OH)2 CaCO3 + Mg(OH)2 + 2H2O
17. Explain the removing of permanent hardness of water?
It can be removed by adding washing soda (Na2CO3)
CaCl2 + Na2CO3 CaCO3 + 2NaCl
MgSO4 + Na2CO3 MgCO3 + Na2SO4
18. Explain the Ion Exchange method of of Softening hard water.
In this method the hard water is passed through an ion exchange Zeolites.
The Zeolites are hydrated Sodium Alumino Silicates (NaO_Al2O3_xSiO2_yH2O)
The Zeolites contains porous structure of mono valent sodium ions, which is exchanged with
Calcium and Magnesium ions in water.
Na2-Z + Ca
2+
2Na
+
+ Ca – Z
Na2- Z + Mg
2+
2Na
+
+ Mg– Z
The Zeolites can be reused by treating with sodium chloride.
Ca -Z +2NaCl CaCl2 + Na2-Z (Zeolites )
19. Explain the exchange reaction of Heavy water
NaOH + D – OD NaOD + H – OD
HCl + D – OD DCl + H – OD
NH4Cl + 4D – OD ND4Cl + 4H – OD
20. Givethe action of Heavy water with the following
a)Aluminium Carbide , b)Calcium Carbide and
c) Calcium Phosphide d)Magnesium Nitride
a)Aluminium Carbide
Al4C3 + 12 D2O 4 Al(OD)3+ 3 CD4
b)Calcium Carbide
CaC2 + 2D–OD Ca(OD)2 + C2D2
c) Calcium Phosphide
Ca3P2+ 6D–OD 3Ca(OD)2 + 2PD3
d) Magnesium Nitride
Mg3N2 + 6D–OD 3 Mg(OD)2 + 2 ND3
21. How is hydrogen peroxide prepared on industrial scale?
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of 2-alkyl anthraquinol.
22. What is 30% solution of hydrogen peroxide? (Or) what is 100 volume of hydro peroxide?
30 % solution of hydrogen peroxide is called as „100-volume‟ hydrogenperoxide.
At STP on heating 1 ml of hydrogen peroxide liberated 100 ml of oxygen
2H2O2 2H2O + O2
23. List the uses of Hydrogen peroxide
It is used as Antiseptic
It is used to bleach paper and textile
It is used in water treatment to oxidize pollutant in water.
24. What is hydrogen bonding and explain its types
When hydrogen is covalently bonded to a highly electronegative atom such as fluorine, oxygen
and nitrogen, the bond is polarized.
The polarized hydrogen forms a weak electrostatic force of interaction with another
electronegative atom.
Intra molecular hydrogen bonding:-The hydrogen bonding which occur within the
molecule is called as Intra molecular hydrogen bonding. Eg. Ortho Nitro Phenol
Inter molecular hydrogen bonding :- The hydrogen bonding which occur between two or
more molecules of same or different types is called as Inter molecular hydrogen bonding.
Eg.Water.
25. Give the industrial preparation of hydrogen
Hydrogen is prepared by mixing methane with steam at 900
0
C and35 atm pressures in the presence
of Nickel catalyst. CH4 + O2
Ni 35atm 900
0
C
CO2 + 2H2
26. How Hydrogen peroxide is used to restore the colour of old paintings
Hydrogen peroxide is used to restore the white colour of the old paintings
Hydrogen sulphide in air reacts with the white pigment to form a black colored lead sulphide.
Hydrogen peroxide reacts lead sulphide to give white coloured lead sulphate
27.
What are the importance of Hydrogen bonding
It plays an important role in bio molecules like proteins.
It plays a important role in the structure of DNA,
It holds the two helical Nucleic acid chains of the DNA together.
28. Why hydrogen peroxide is store in Plastic bottles not in Glass bottles
It dissolves the Alkali metals present in glass.
It undergoes a Catalyzed disproportionation reaction.
So it is stored in Plastic bottles.
29. Complete the following equation. Na2O2 + ? Na2SO4 + H2O2
Na2O2 + H2SO4 Na2SO4 + H2O2 5 . ALKALI AND ALKALINE EARTH METALS
1. Why sodium hydroxide is much more water soluble than sodium chloride?
NaOH
���
Na
+
+ OH
-
This reaction is an exothermic reaction.
The heat evolved increases the stability.
So sodium hydroxide crystals are readily dissolved in water
2. Define Efflorescence
The spontaneous loss of water by a hydrated salt is called Efflorescence.
3. Give the preparation of Sodium Carbonate by Solvay process.
In the Solvay process, sodium carbonate is prepared from Ammonia.
The ammonia is recovered by adding Calcium Hydroxide.
2NH3 + H2O + CO2 (NH4)2CO3(ammonium carbonate)
(NH4)2CO3 + H2O + CO2 2NH4HCO3(ammonium bicarbonate)
2NH4HCO3 + 2NaCl 2NH4Cl + 2NaHCO3(sodium bicarbonate)
2NaHCO3 Na2CO3 + H2O + CO2
(sodium carbonate)
4. An alkali metal (x) forms a hydrated sulphate, X2SO4. 10H2O. Is the metal more likely to be
sodium (or) potassium.
The metal is more likely be sodium. So X is Na2SO4.10H2O. it is otherwise called as Glauber’s salt
5. Write balanced chemical equation for each of the following chemical reactions.
(i) Lithium metal with nitrogen gas N2 + 6Li 2Li3N (Lithium nitride)
(ii) heating solid sodium bicarbonate
NaHCO3
∆
CO2 + H2O + Na2CO3
(Sodium carbonate)
(iii) Rubidum with oxgen gas Rb + O2 RbO2(Rubidium oxide)
(iv) solid potassium hydroxide with CO2 KOH + CO2 KHCO3(Potassium bicarbonate)
(v) heating calcium carbonate
CaCO3
∆
CO2 +CaO(Quick lime)
(vi) heating calcium with oxygen Ca + O22CaO (Calcium oxide)
6. Explain the diagonal relation ( Similarities ) between Beryllium and Aluminum
Both beryllium chloride and aluminium chloride forms dimeric structure with chloride bridges.
Both beryllium and Aluminium forms complexes. BeF4
-2
and AlF6
-3
Beryllium carbide and Aluminium carbide give methane on hydrolysis
Both beryllium and Aluminiumrenders passive with Nitric acid
Both beryllium and Aluminium Hydroxides are Amphoteric in nature.
7. Give the systematic names for the following
(i) milk of magnesia Magnesium hydroxide
(ii) lye Sodium hydroxide
(iii) lime Calcium hydroxide
(iv) Caustic potash Potassium hydroxide
(v) washing soda Sodium carbonate
(vi) soda ash Sodium carbonate
(v) trona Trisodium hydrogen carbonate dihydtrate
8 Substantiate Lithium fluoride has the lowest solubility among group one metal fluorides.
Lithium fluoride is covalent compound. So it is less soluble in water.
9. List the used of Plaster of Paris
In Building industry it is used as Plasters
It is used in the treatment of Bone fracture and Sprains
It is used for making Statues
It is used in Dentistry
Ca3P2+ 6D–OD 3Ca(OD)2 + 2PD3
d) Magnesium Nitride
Mg3N2 + 6D–OD 3 Mg(OD)2 + 2 ND3
21. How is hydrogen peroxide prepared on industrial scale?
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of 2-alkyl anthraquinol.
22. What is 30% solution of hydrogen peroxide? (Or) what is 100 volume of hydro peroxide?
30 % solution of hydrogen peroxide is called as „100-volume‟ hydrogenperoxide.
At STP on heating 1 ml of hydrogen peroxide liberated 100 ml of oxygen
2H2O2 2H2O + O2
23. List the uses of Hydrogen peroxide
It is used as Antiseptic
It is used to bleach paper and textile
It is used in water treatment to oxidize pollutant in water.
24. What is hydrogen bonding and explain its types
When hydrogen is covalently bonded to a highly electronegative atom such as fluorine, oxygen
and nitrogen, the bond is polarized.
The polarized hydrogen forms a weak electrostatic force of interaction with another
electronegative atom.
Intra molecular hydrogen bonding:-The hydrogen bonding which occur within the
molecule is called as Intra molecular hydrogen bonding. Eg. Ortho Nitro Phenol
Inter molecular hydrogen bonding :- The hydrogen bonding which occur between two or
more molecules of same or different types is called as Inter molecular hydrogen bonding.
Eg.Water.
25. Give the industrial preparation of hydrogen
Hydrogen is prepared by mixing methane with steam at 900
0
C and35 atm pressures in the presence
of Nickel catalyst. CH4 + O2
Ni 35atm 900
0
C
CO2 + 2H2
26. How Hydrogen peroxide is used to restore the colour of old paintings
Hydrogen peroxide is used to restore the white colour of the old paintings
Hydrogen sulphide in air reacts with the white pigment to form a black colored lead sulphide.
Hydrogen peroxide reacts lead sulphide to give white coloured lead sulphate
27.
What are the importance of Hydrogen bonding
It plays an important role in bio molecules like proteins.
It plays a important role in the structure of DNA,
It holds the two helical Nucleic acid chains of the DNA together.
28. Why hydrogen peroxide is store in Plastic bottles not in Glass bottles
It dissolves the Alkali metals present in glass.
It undergoes a Catalyzed disproportionation reaction.
So it is stored in Plastic bottles.
29. Complete the following equation. Na2O2 + ? Na2SO4 + H2O2
Na2O2 + H2SO4 Na2SO4 + H2O2 5 . ALKALI AND ALKALINE EARTH METALS
1. Why sodium hydroxide is much more water soluble than sodium chloride?
NaOH
���
Na
+
+ OH
-
This reaction is an exothermic reaction.
The heat evolved increases the stability.
So sodium hydroxide crystals are readily dissolved in water
2. Define Efflorescence
The spontaneous loss of water by a hydrated salt is called Efflorescence.
3. Give the preparation of Sodium Carbonate by Solvay process.
In the Solvay process, sodium carbonate is prepared from Ammonia.
The ammonia is recovered by adding Calcium Hydroxide.
2NH3 + H2O + CO2 (NH4)2CO3(ammonium carbonate)
(NH4)2CO3 + H2O + CO2 2NH4HCO3(ammonium bicarbonate)
2NH4HCO3 + 2NaCl 2NH4Cl + 2NaHCO3(sodium bicarbonate)
2NaHCO3 Na2CO3 + H2O + CO2
(sodium carbonate)
4. An alkali metal (x) forms a hydrated sulphate, X2SO4. 10H2O. Is the metal more likely to be
sodium (or) potassium.
The metal is more likely be sodium. So X is Na2SO4.10H2O. it is otherwise called as Glauber’s salt
5. Write balanced chemical equation for each of the following chemical reactions.
(i) Lithium metal with nitrogen gas N2 + 6Li 2Li3N (Lithium nitride)
(ii) heating solid sodium bicarbonate
NaHCO3
∆
CO2 + H2O + Na2CO3
(Sodium carbonate)
(iii) Rubidum with oxgen gas Rb + O2 RbO2(Rubidium oxide)
(iv) solid potassium hydroxide with CO2 KOH + CO2 KHCO3(Potassium bicarbonate)
(v) heating calcium carbonate
CaCO3
∆
CO2 +CaO(Quick lime)
(vi) heating calcium with oxygen Ca + O22CaO (Calcium oxide)
6. Explain the diagonal relation ( Similarities ) between Beryllium and Aluminum
Both beryllium chloride and aluminium chloride forms dimeric structure with chloride bridges.
Both beryllium and Aluminium forms complexes. BeF4
-2
and AlF6
-3
Beryllium carbide and Aluminium carbide give methane on hydrolysis
Both beryllium and Aluminiumrenders passive with Nitric acid
Both beryllium and Aluminium Hydroxides are Amphoteric in nature.
7. Give the systematic names for the following
(i) milk of magnesia Magnesium hydroxide
(ii) lye Sodium hydroxide
(iii) lime Calcium hydroxide
(iv) Caustic potash Potassium hydroxide
(v) washing soda Sodium carbonate
(vi) soda ash Sodium carbonate
(v) trona Trisodium hydrogen carbonate dihydtrate
8 Substantiate Lithium fluoride has the lowest solubility among group one metal fluorides.
Lithium fluoride is covalent compound. So it is less soluble in water.
9. List the used of Plaster of Paris
In Building industry it is used as Plasters
It is used in the treatment of Bone fracture and Sprains
It is used for making Statues
It is used in Dentistry
10. Beryllium halides are Covalent whereas magnesium halides are ionic why?
Beryllium ion (Be
2+
) size is small and it is involved in equal sharing of electrons with halogens to
form covalent bond.
Magnesium ion (Mg
2+
) size is big and it is involved in transfer of electrons to form ionic bond.
11. Alkaline earth metal (A), belongs to 3rd period reacts with oxygen and nitrogen to
form compound (B) and (C) respectively. It undergo metal displacement reaction
with AgNO3 solution to form compound (D).
An alkaline earth metal (A) belongs to third period is Magnesium (Mg)
Magnesium reacts with oxygen to give Magnesium oxide. (MgO) (B)
2Mg + O2 2MgO
Magnesium reacts with nitrogen ti form Magnesium nitride Mg3N2(C)
3Mg + N2 Mg3N2
Magnesium undergoes metal displacement reaction with AgNO3 solution to form Magnesium
nitrate Mg(NO3)2 (D).
Mg + 2AgNO3 Mg(NO3)2 + 2Ag
Compound Molecular Formula Name
A Mg Magnesium
B MgO Magnesium oxide
C Mg3N2 Magnesium nitride
D Mg(NO3)2 Magnesium nitrate
12. Write balanced chemical equation for the following processes
(a) heating calcium in oxygen
2Ca + O2
∆
2CaO (Calcium oxide)
(b) heating calcium carbonate
CaCO3
∆
CO2 +CaO(Calcium oxide)
(c) evaporating a solution of calcium
hydrogen carbonate
CaHCO3
∆
H2O + CO2 + CaCO3
(Calcium carbonate)
(d) heating calcium oxide with carbon
CaO + C
∆
CO2+ CaC2(Calcium carbide)
13. Explain the important common features of Group 2 elements
S.No Properties of Alkali earth metals Explanation
1. Electronic configuration ns
2
2. Atomic and ionic radius Smaller than Alkali metals, on moving down
the group, the radii increases.
3. Oxidation state +2
4. Ionization enthalpy Higher than Alkali metals
5. Hydration enthalpies Decreases as we go down the group.
6. Electro negativity Decreases as we go down the group.
14. Why Alkaline metals are harder than alkali metals
Alkaline earth metals are smaller in size
They have high density.
So they form strong metallic bonds.
15. Give the preparation of Plaster of Paris
When gypsum is heated at 393K , Plaster of Paris is formed.
CaSO4 . 2 H2O
����
CaSO4 .½ H2O +1½ H2O
Plaster of Paris
16. List the uses of Gypsum
It is used for making Plaster Boards, which is used for walls and ceilings
It is used to make Plaster of Paris, which is used in Sculptures
It used in making surgical and ortho pedic casts
In agriculture, it is used as soil conditioner and fertilizer.
It is used in tooth paste, Shampoos and hair products It is used to make Portland cement.
17. Give the Biological importance of Magnesium and Calcium Magnesium
Magnesium plays a important role in Bio-chemical reactions catalyzed by enzymes
Magnesium is essential for DNA synthesis
Magnesium is essential for the stability and functioning of the DNA
Magnesium is used for balancing the electrolyte in the body
Magnesium is present in Chlorophyll and play a main role in Photosynthesis
Deficiency of Mg caused Neuro -Muscular irritation.
18. Which would you expect to have a higher melting point, magnesium oxide or magnesium
fluoride? Explain your reasoning.
Mg
2+
and O
2-
ions have charges of +2 and -2 respectively
Oxygen ion is smaller than Fluoride ion.
The smaller ionic radii, the smaller the bond length in MgO and the bond is strongest than MgF2
Due to more strong bond nature in MgO, it has high melting point then MgF2
19. What are similarities ( Diagonal relation ) between Lithium and Magnesium
Both lithium and Magnesium are Hard
Both lithium and Magnesium reacts slowly with water.
Both forms Nitrides with nitrogen. Eg. Li3N and Mg3N2.
Both do not form Super oxides.
Both forms only Oxides like Li2O and MgO.
Their carbonates decompose on heating giving CO2 gas and Oxides.
Both do not forms Bicarbonates
20. Compare the properties of Beryllium and other elements ( How Beryllium differs from rest
of the member of the family )
S.No Beryllium Other elements
1 Forms covalent compounds Forms Ionic compounds
2 High melting and Boiling point Low melting and Boiling point
3 Do not reacts with water Reacts with water
4 Do not reacts with hydrogen directly Reacts with hydrogen directly
5 Do not reacts with Halogen directly Reacts with Halogen directly
6 Do not attacked by acid Attacked by acids
7 Hydroxides are Amphoteric Hydroxides are Basic
ADDITIONAL QUESTIONS
1. Write the ionization value of Alkali metals.
The ionization value of alkali metalsdecreases as we go down the group.
Because as we go down the group the size of the atom is increases and screening effect also increases.
2. Which colour give when alkali metals heated in Bunsen flame
Element Colour
1, Lithium Crimson red
2, Sodium Yellow
3, potassium Lilac
4, Rubidium Reddish violet
5, Cesium Blue
3. Why the second ionisation enthalpy of Alkali metals are very high ?
When a electron is removed from the alkali metals, it gives a monovalentcation with fully filled
stable np
6
electronic configuration. So it is verydifficult to remove the electrons .
4. Why lithium salts are more soluble than the other metals in group 1 ?
Lithium is more soluble because of high solvation and Smaller sizeoflithium ion.
5. Why Alkali metal have smaller value of Electro negativity than the other elements in the
respective period?
When they react with other elements, they usually produce ionic compounds.
Beryllium ion (Be
2+
) size is small and it is involved in equal sharing of electrons with halogens to
form covalent bond.
Magnesium ion (Mg
2+
) size is big and it is involved in transfer of electrons to form ionic bond.
11. Alkaline earth metal (A), belongs to 3rd period reacts with oxygen and nitrogen to
form compound (B) and (C) respectively. It undergo metal displacement reaction
with AgNO3 solution to form compound (D).
An alkaline earth metal (A) belongs to third period is Magnesium (Mg)
Magnesium reacts with oxygen to give Magnesium oxide. (MgO) (B)
2Mg + O2 2MgO
Magnesium reacts with nitrogen ti form Magnesium nitride Mg3N2(C)
3Mg + N2 Mg3N2
Magnesium undergoes metal displacement reaction with AgNO3 solution to form Magnesium
nitrate Mg(NO3)2 (D).
Mg + 2AgNO3 Mg(NO3)2 + 2Ag
Compound Molecular Formula Name
A Mg Magnesium
B MgO Magnesium oxide
C Mg3N2 Magnesium nitride
D Mg(NO3)2 Magnesium nitrate
12. Write balanced chemical equation for the following processes
(a) heating calcium in oxygen
2Ca + O2
∆
2CaO (Calcium oxide)
(b) heating calcium carbonate
CaCO3
∆
CO2 +CaO(Calcium oxide)
(c) evaporating a solution of calcium
hydrogen carbonate
CaHCO3
∆
H2O + CO2 + CaCO3
(Calcium carbonate)
(d) heating calcium oxide with carbon
CaO + C
∆
CO2+ CaC2(Calcium carbide)
13. Explain the important common features of Group 2 elements
S.No Properties of Alkali earth metals Explanation
1. Electronic configuration ns
2
2. Atomic and ionic radius Smaller than Alkali metals, on moving down
the group, the radii increases.
3. Oxidation state +2
4. Ionization enthalpy Higher than Alkali metals
5. Hydration enthalpies Decreases as we go down the group.
6. Electro negativity Decreases as we go down the group.
14. Why Alkaline metals are harder than alkali metals
Alkaline earth metals are smaller in size
They have high density.
So they form strong metallic bonds.
15. Give the preparation of Plaster of Paris
When gypsum is heated at 393K , Plaster of Paris is formed.
CaSO4 . 2 H2O
����
CaSO4 .½ H2O +1½ H2O
Plaster of Paris
16. List the uses of Gypsum
It is used for making Plaster Boards, which is used for walls and ceilings
It is used to make Plaster of Paris, which is used in Sculptures
It used in making surgical and ortho pedic casts
In agriculture, it is used as soil conditioner and fertilizer.
It is used in tooth paste, Shampoos and hair products It is used to make Portland cement.
17. Give the Biological importance of Magnesium and Calcium Magnesium
Magnesium plays a important role in Bio-chemical reactions catalyzed by enzymes
Magnesium is essential for DNA synthesis
Magnesium is essential for the stability and functioning of the DNA
Magnesium is used for balancing the electrolyte in the body
Magnesium is present in Chlorophyll and play a main role in Photosynthesis
Deficiency of Mg caused Neuro -Muscular irritation.
18. Which would you expect to have a higher melting point, magnesium oxide or magnesium
fluoride? Explain your reasoning.
Mg
2+
and O
2-
ions have charges of +2 and -2 respectively
Oxygen ion is smaller than Fluoride ion.
The smaller ionic radii, the smaller the bond length in MgO and the bond is strongest than MgF2
Due to more strong bond nature in MgO, it has high melting point then MgF2
19. What are similarities ( Diagonal relation ) between Lithium and Magnesium
Both lithium and Magnesium are Hard
Both lithium and Magnesium reacts slowly with water.
Both forms Nitrides with nitrogen. Eg. Li3N and Mg3N2.
Both do not form Super oxides.
Both forms only Oxides like Li2O and MgO.
Their carbonates decompose on heating giving CO2 gas and Oxides.
Both do not forms Bicarbonates
20. Compare the properties of Beryllium and other elements ( How Beryllium differs from rest
of the member of the family )
S.No Beryllium Other elements
1 Forms covalent compounds Forms Ionic compounds
2 High melting and Boiling point Low melting and Boiling point
3 Do not reacts with water Reacts with water
4 Do not reacts with hydrogen directly Reacts with hydrogen directly
5 Do not reacts with Halogen directly Reacts with Halogen directly
6 Do not attacked by acid Attacked by acids
7 Hydroxides are Amphoteric Hydroxides are Basic
ADDITIONAL QUESTIONS
1. Write the ionization value of Alkali metals.
The ionization value of alkali metalsdecreases as we go down the group.
Because as we go down the group the size of the atom is increases and screening effect also increases.
2. Which colour give when alkali metals heated in Bunsen flame
Element Colour
1, Lithium Crimson red
2, Sodium Yellow
3, potassium Lilac
4, Rubidium Reddish violet
5, Cesium Blue
3. Why the second ionisation enthalpy of Alkali metals are very high ?
When a electron is removed from the alkali metals, it gives a monovalentcation with fully filled
stable np
6
electronic configuration. So it is verydifficult to remove the electrons .
4. Why lithium salts are more soluble than the other metals in group 1 ?
Lithium is more soluble because of high solvation and Smaller sizeoflithium ion.
5. Why Alkali metal have smaller value of Electro negativity than the other elements in the
respective period?
When they react with other elements, they usually produce ionic compounds.
6. Explain the flame colour and the Spectra of Alkali metals
When heated the valence electrons are excited to the higher energy level
When drops back to the ground state the energy is emitted as light in the visible region.
7. Compare the properties of lithium and other elements
S.No. Lithium Other elements
1
Hard, high melting and boiling
point
Soft, low melting and boiling point
2 Less reactive More reactive
3 Reacts with nitrogen to give Li3N No reaction
4 Reacts with bromine slowly Reacts violently
Compounds are less soluble in water Compounds are Highly soluble in Water
5
Lithium nitrate decomposes to
give Oxides
Decomposes to give Nitrite
8. Why Lithium does not form peroxide and super oxide?
Least reactive,
Forms peroxides with great difficulty and its higher oxides are unstable
9. Reactivity of alkali metals with halogens increases down the group – give the reason?
Reason :- because of corresponding decrease in ionization enthalpy
10. Lithium halide has covalent character, but other Alkali metal halides has ionic character –
justify your answer.
The size of the Lithium atom is very small compare to other alkali metal atom.
So Lithium halide has covalent character.
11. Why the Alkali metals give deep blue colour in Ammonia solution
The blue colour is because the ammoniated electron absorbs energy in thevisible region of the
light.
( The concentrated solution is Bronze colour and diamagnetic )
M
+
+ e
−
+ NH3 MNH 2+ ½H2
12. Explain the Alkali metal act as a reducing agent
Alkali metals can lose their valence electron readily hence they act as goodreducing agents.
M(s) M
+
(g) + e
–
13. Give the uses of Lithium
Lithium used for making alloys.Eg: Li-Al alloy is used for making Aircraft parts.
Lithium carbonate is used in medicines
Lithium is used for making electro chemical cells
14. What is Soda Ash ( Give the action of heat of Sodium Carbonate).
When Sodium carbonate is heated, it loses a water molecule to becomemono hydrate.
On heating at 373K it becomes a white powder called as Soda Ash
Na2CO3.10H2O
∆
Na2CO3.H2O + 9H2O
Na2CO3.H2O
∆
Na2CO3+ H2O
15. List the uses of Sodium Carbonate (or) Washing soda
Sodium carbonate is called as Washing soda, which used in Laundry.
It is used in Volumetric analysis
It is used in water treatment to convert hard water to soft water.
It is used to prepare Glass, Paper and Paints
16. Explain the preparation of sodium chloride?
Sodium chloride is isolated by evaporation from sea water
Crude sodium chloride can be obtained by crystallization of brine solution which contains sodium
sulphate, calcium sulphate, calcium chloride and magnesium chloride as impurities.
Pure sodium chloride can be obtained from crude salt as follows.
Firstly removal of insoluble impurities by filtration from the crude salt solution with minimum
amount of water. Sodium chloride can be crystallised by passing HCl gas into this solution. Calcium and
magnesium chloride, being more soluble than sodium chloride, remain in solution.
17. List the use of Sodium chloride
It is used as common salt
It is used to prepare NaOH and Sodium Carbonate.
18. Give the preparation of Sodium hydroxide.
Sodium hydroxide is prepared by the electrolysis of Brine solution byusing Castner - Kellner
cell.
Cathode - Mercury
Anode - Carbon rod
Sodium metal is liberated at the cathode and reacts with mercury to form Sodium Amalgam
Sodium amalgam is treated with water to give Sodium Hydroxide.
Chorine is liberated at the anode
At the cathode
Na
+
+ e
-
Na-amalgam
At the anode
Cl
-
½ Cl2 + e
-
2 Na-amalgam + 2 H2O 2 NaOH + 2 Hg + H2
19. Give the used of Sodium hydroxide.
It is used as laboratory reagent.
It is used in refining Petroleum
It is used to prepare soap and paper.
It is used in textile industry
20. Give the preparation of Backing soda (or) Sodium bicarbonate
Backing soda is prepared by saturating a solution of sodium carbonatewith carbon dioxide.
Na2CO3 + CO2 + H2O 2NaHCO3
21. List the used of Backing soda (or) Sodium bicarbonate.
It is used in Fire Extinguisher
It is used as aAntiseptic for skin infections
It used in Baking.
22. Give the preparation of beryllium chloride.
BeO + C + Cl2
���–��� �
BeCl2 + CO
23. Give the preparation of beryllium fluoride.
BeO + C + F2
���–��� �
BeF2 + CO
24. Give the preparation of beryllium hydride
2BeCl2 + LiAlH4 2BeH2 + LiCl + AlCl3
25. List the uses of Beryllium
It is used as radiation windows in X-ray tubes and X-ray detector
It used as sample holders in X-Ray emission studies.
It used to built Beam pipes in accelerators
It used in Detectors
26. List the uses of Magnesium
It is used to remove sulphur from Iron and steel
It is used in Refining of Titanium in Kroll Process
It is used in Printing Industry
Magnesium allow in used in Aero plane construction
It is used as Desiccant
27. List the uses of Calcium
It is used a reducing agent in the metallurgy of Uranium.
It is used for making cement
It is used for making Fertilizers
It is used in Vaccum Tubes
When heated the valence electrons are excited to the higher energy level
When drops back to the ground state the energy is emitted as light in the visible region.
7. Compare the properties of lithium and other elements
S.No. Lithium Other elements
1
Hard, high melting and boiling
point
Soft, low melting and boiling point
2 Less reactive More reactive
3 Reacts with nitrogen to give Li3N No reaction
4 Reacts with bromine slowly Reacts violently
Compounds are less soluble in water Compounds are Highly soluble in Water
5
Lithium nitrate decomposes to
give Oxides
Decomposes to give Nitrite
8. Why Lithium does not form peroxide and super oxide?
Least reactive,
Forms peroxides with great difficulty and its higher oxides are unstable
9. Reactivity of alkali metals with halogens increases down the group – give the reason?
Reason :- because of corresponding decrease in ionization enthalpy
10. Lithium halide has covalent character, but other Alkali metal halides has ionic character –
justify your answer.
The size of the Lithium atom is very small compare to other alkali metal atom.
So Lithium halide has covalent character.
11. Why the Alkali metals give deep blue colour in Ammonia solution
The blue colour is because the ammoniated electron absorbs energy in thevisible region of the
light.
( The concentrated solution is Bronze colour and diamagnetic )
M
+
+ e
−
+ NH3 MNH 2+ ½H2
12. Explain the Alkali metal act as a reducing agent
Alkali metals can lose their valence electron readily hence they act as goodreducing agents.
M(s) M
+
(g) + e
–
13. Give the uses of Lithium
Lithium used for making alloys.Eg: Li-Al alloy is used for making Aircraft parts.
Lithium carbonate is used in medicines
Lithium is used for making electro chemical cells
14. What is Soda Ash ( Give the action of heat of Sodium Carbonate).
When Sodium carbonate is heated, it loses a water molecule to becomemono hydrate.
On heating at 373K it becomes a white powder called as Soda Ash
Na2CO3.10H2O
∆
Na2CO3.H2O + 9H2O
Na2CO3.H2O
∆
Na2CO3+ H2O
15. List the uses of Sodium Carbonate (or) Washing soda
Sodium carbonate is called as Washing soda, which used in Laundry.
It is used in Volumetric analysis
It is used in water treatment to convert hard water to soft water.
It is used to prepare Glass, Paper and Paints
16. Explain the preparation of sodium chloride?
Sodium chloride is isolated by evaporation from sea water
Crude sodium chloride can be obtained by crystallization of brine solution which contains sodium
sulphate, calcium sulphate, calcium chloride and magnesium chloride as impurities.
Pure sodium chloride can be obtained from crude salt as follows.
Firstly removal of insoluble impurities by filtration from the crude salt solution with minimum
amount of water. Sodium chloride can be crystallised by passing HCl gas into this solution. Calcium and
magnesium chloride, being more soluble than sodium chloride, remain in solution.
17. List the use of Sodium chloride
It is used as common salt
It is used to prepare NaOH and Sodium Carbonate.
18. Give the preparation of Sodium hydroxide.
Sodium hydroxide is prepared by the electrolysis of Brine solution byusing Castner - Kellner
cell.
Cathode - Mercury
Anode - Carbon rod
Sodium metal is liberated at the cathode and reacts with mercury to form Sodium Amalgam
Sodium amalgam is treated with water to give Sodium Hydroxide.
Chorine is liberated at the anode
At the cathode
Na
+
+ e
-
Na-amalgam
At the anode
Cl
-
½ Cl2 + e
-
2 Na-amalgam + 2 H2O 2 NaOH + 2 Hg + H2
19. Give the used of Sodium hydroxide.
It is used as laboratory reagent.
It is used in refining Petroleum
It is used to prepare soap and paper.
It is used in textile industry
20. Give the preparation of Backing soda (or) Sodium bicarbonate
Backing soda is prepared by saturating a solution of sodium carbonatewith carbon dioxide.
Na2CO3 + CO2 + H2O 2NaHCO3
21. List the used of Backing soda (or) Sodium bicarbonate.
It is used in Fire Extinguisher
It is used as aAntiseptic for skin infections
It used in Baking.
22. Give the preparation of beryllium chloride.
BeO + C + Cl2
���–��� �
BeCl2 + CO
23. Give the preparation of beryllium fluoride.
BeO + C + F2
���–��� �
BeF2 + CO
24. Give the preparation of beryllium hydride
2BeCl2 + LiAlH4 2BeH2 + LiCl + AlCl3
25. List the uses of Beryllium
It is used as radiation windows in X-ray tubes and X-ray detector
It used as sample holders in X-Ray emission studies.
It used to built Beam pipes in accelerators
It used in Detectors
26. List the uses of Magnesium
It is used to remove sulphur from Iron and steel
It is used in Refining of Titanium in Kroll Process
It is used in Printing Industry
Magnesium allow in used in Aero plane construction
It is used as Desiccant
27. List the uses of Calcium
It is used a reducing agent in the metallurgy of Uranium.
It is used for making cement
It is used for making Fertilizers
It is used in Vaccum Tubes
It is used in dehydrating oils
It is used to prepare Plaster of Paris.
28. List the uses of Strontium
Sr-90 is used in Cancer Treatment
It is used in Dating Rocks
It is used as Radioactive tracer in determining the source of ancient coins.
Sr-87 / Sr-86 ratio is used in Criminal Forensic science.
29. List the uses of Barium
It is used in Metallurgy and Radiology
It is used as De-Oxidizer in refining Copper
Ba-133 is used in the calibration of Gamma rays detectors in nuclear chemistry
It is used to remove Oxygen in Television and Electronic tubes.
30. List the uses of Radium
It is used as self luminous paint in watches, clocks, Dials and Aircraft switches.
31. Explain the Amphoteric nature of Beryllium Hydroxide.
Beryllium Hydroxide reacts both with Acid and Base.
Be(OH)2 + 2HCl BeCl2 + 2H2O
Be(OH)2 + 2NaOH Na2BeO2 + 2H2O
32. Magnesium nitrate crystal is hexa hydrate, whereas barium nitrate crystal is anhydrous–
why?
Size of Magnesium is smaller than Barium
decreasing tendency to form hydrates with increasing size.
33. What is Quick lime give the preparation
Cao - Calcium Oxide is called as Quick lime.
It is prepare by heating lime stone at 1070K.
CaCO3 ⇋ CaO +CO2
34. Mention the uses of Calcium oxide
To manufacture cement, mortar and glass.
In the manufacture of sodium carbonate and slaked lime.
In the purification of sugar.
As a drying agent.
35. List the uses of Calcium hydroxide
In the preparation of mortar, a building material.
In white wash due to its disinfectant nature.
In glass making, in tanning industry,in the preparation of bleaching powder and for the
purification of sugar.
36. When carbon di oxide is passed through lime water, it turns milky.Explain why?
This is due to the formation of Calcium carbonate.
CaO + CO2 CaCO3
37. What is Slaking of lime or Slaked lime.
When water is added to quicklime, it breaks the lumps of lime. This processis called Slaking of
lime.
CaO + H2O Ca(OH)2
38. Give the action of milk of lime ( calcium carbonate ) with chlorine and
give its uses ?
When milk of lime reacts of chlorine it gives Hypo chlorite.
It is used in Bleaching Power.
2 Ca(OH)2 + 2 Cl2 CaCl2 + CaO2Cl2 + 2H2O
CaO2Cl2-Bleaching Power.
39. Which salt is called Desert Rose ?
Gypsum crystals occur like petals of a flower, called as Desert Rose 40. Write a note on dead burnt plaster?
Gypsum is heated above 393 K, no water of crystallisation is left and anhydrous calcium
sulphate,CaSO4 is formed.
CaSO4 .2 H2O
����� ����
CaSO4 +2 H2O
41. Explain why Ca(OH)2 is used in white washing
In white wash due to its disinfectant nature
42. Among the alkaline earth metals BeO is insoluble in water but other oxides are soluble Why?
Due to high lattice energy
6. GASEOUS STATE
1. State Boyle’s law
At constant temperature the volume of a gas is inversely proportional to thepressure
V α
1
P
2. State Charles law
At constant pressure the volume of a gas is directly proportional to thetemperature
V α T
3. Define Gay-Lussac’s law
At constant volume the pressure of a gas is directly proportional to thetemperature.
P α T
4. Define Avogadro hypothesis
Equal volume of all gases under same temperature and pressure containsequal number of
molecules
V α n
5. Dalton law of Partial pressure
The total pressure of a gaseous mixture is equal to the sum of the partial pressure of the gases
present in the mixture. PTotal= P1 + P2 + P3 ….
6. Define Grahams law of Diffusion or Effusion
The Rate of diffusion (or) effusion of a gas is inversely proportional to the square root of the
molar mass
Rate of Diffusion (r) α
1
M
7. Give the difference between diffusion and Effusion
Diffusion Effusion
The movement of the gasmolecules through
another gasfrom high concentration to
lowconcentration is called asDiffusion.
The movement of the gasmolecules through a
small holefrom high concentration to
lowconcentration is called as Effusion
It refers to the ability of the gases to mix with
each other
It is a ability of a gas to travel through a small
pin - hole
Eg:- spreading of gas from Incense sticks. Eg:- body sprey and deflating ballons.
8. Derive the ideal gas equation
Boyles law V α
1
P
Charles law V α T
Avogadro law V α n
V α
nT
P
(or)
V =
nRT
P
PV = nRT
It is used to prepare Plaster of Paris.
28. List the uses of Strontium
Sr-90 is used in Cancer Treatment
It is used in Dating Rocks
It is used as Radioactive tracer in determining the source of ancient coins.
Sr-87 / Sr-86 ratio is used in Criminal Forensic science.
29. List the uses of Barium
It is used in Metallurgy and Radiology
It is used as De-Oxidizer in refining Copper
Ba-133 is used in the calibration of Gamma rays detectors in nuclear chemistry
It is used to remove Oxygen in Television and Electronic tubes.
30. List the uses of Radium
It is used as self luminous paint in watches, clocks, Dials and Aircraft switches.
31. Explain the Amphoteric nature of Beryllium Hydroxide.
Beryllium Hydroxide reacts both with Acid and Base.
Be(OH)2 + 2HCl BeCl2 + 2H2O
Be(OH)2 + 2NaOH Na2BeO2 + 2H2O
32. Magnesium nitrate crystal is hexa hydrate, whereas barium nitrate crystal is anhydrous–
why?
Size of Magnesium is smaller than Barium
decreasing tendency to form hydrates with increasing size.
33. What is Quick lime give the preparation
Cao - Calcium Oxide is called as Quick lime.
It is prepare by heating lime stone at 1070K.
CaCO3 ⇋ CaO +CO2
34. Mention the uses of Calcium oxide
To manufacture cement, mortar and glass.
In the manufacture of sodium carbonate and slaked lime.
In the purification of sugar.
As a drying agent.
35. List the uses of Calcium hydroxide
In the preparation of mortar, a building material.
In white wash due to its disinfectant nature.
In glass making, in tanning industry,in the preparation of bleaching powder and for the
purification of sugar.
36. When carbon di oxide is passed through lime water, it turns milky.Explain why?
This is due to the formation of Calcium carbonate.
CaO + CO2 CaCO3
37. What is Slaking of lime or Slaked lime.
When water is added to quicklime, it breaks the lumps of lime. This processis called Slaking of
lime.
CaO + H2O Ca(OH)2
38. Give the action of milk of lime ( calcium carbonate ) with chlorine and
give its uses ?
When milk of lime reacts of chlorine it gives Hypo chlorite.
It is used in Bleaching Power.
2 Ca(OH)2 + 2 Cl2 CaCl2 + CaO2Cl2 + 2H2O
CaO2Cl2-Bleaching Power.
39. Which salt is called Desert Rose ?
Gypsum crystals occur like petals of a flower, called as Desert Rose 40. Write a note on dead burnt plaster?
Gypsum is heated above 393 K, no water of crystallisation is left and anhydrous calcium
sulphate,CaSO4 is formed.
CaSO4 .2 H2O
����� ����
CaSO4 +2 H2O
41. Explain why Ca(OH)2 is used in white washing
In white wash due to its disinfectant nature
42. Among the alkaline earth metals BeO is insoluble in water but other oxides are soluble Why?
Due to high lattice energy
6. GASEOUS STATE
1. State Boyle’s law
At constant temperature the volume of a gas is inversely proportional to thepressure
V α
1
P
2. State Charles law
At constant pressure the volume of a gas is directly proportional to thetemperature
V α T
3. Define Gay-Lussac’s law
At constant volume the pressure of a gas is directly proportional to thetemperature.
P α T
4. Define Avogadro hypothesis
Equal volume of all gases under same temperature and pressure containsequal number of
molecules
V α n
5. Dalton law of Partial pressure
The total pressure of a gaseous mixture is equal to the sum of the partial pressure of the gases
present in the mixture. PTotal= P1 + P2 + P3 ….
6. Define Grahams law of Diffusion or Effusion
The Rate of diffusion (or) effusion of a gas is inversely proportional to the square root of the
molar mass
Rate of Diffusion (r) α
1
M
7. Give the difference between diffusion and Effusion
Diffusion Effusion
The movement of the gasmolecules through
another gasfrom high concentration to
lowconcentration is called asDiffusion.
The movement of the gasmolecules through a
small holefrom high concentration to
lowconcentration is called as Effusion
It refers to the ability of the gases to mix with
each other
It is a ability of a gas to travel through a small
pin - hole
Eg:- spreading of gas from Incense sticks. Eg:- body sprey and deflating ballons.
8. Derive the ideal gas equation
Boyles law V α
1
P
Charles law V α T
Avogadro law V α n
V α
nT
P
(or)
V =
nRT
P
PV = nRT
9. What is Compressibility factor Z
The deviation of real gases fromidealbehaviour is measured in terms ofa ratio of PV to nRT.
This is termed ascompressibility factor. Mathematically
Compressibility factor Z =
��
���
10. What are ideal gases?
Ideal gases are gases that obey the ideal gas equation, PV = nRT.
P=Pressure ; V=Volume ; T=Temperature ; R=Gas Constant
11. What are Real gases?
Real gases do not obey the ideal gas equation, PV = nRT
12. In what way real gases differ from ideal gases
Ideal gas assumes that the individual gas molecules occupy negligible volume when
compared to the total volume of the gas and there is no attractive force between the gas
molecules.
For ideal gases the compressibility factor Z = 1
For real gases the compressibility factor Z > 1 or Z < 1
13. Derive Van der Walls equation.
The correction terms in the ideal gas equation to account for (i) volume of the gaseous molecules
themselves in V and (ii) the intermolecular forces of attraction in pressure, P, the equation of state for
the real gas is arrived, Vanderwaal's deduced the equation of state of real gases.
i) Volume correction of real gas :-
The volume of a gas is the free space in the container in which molecules move about.
Volume V of an ideal gas is the same as the volume of the container.
The volume of a real gas is, therefore, ideal volume minus the volume occupied by the gas
molecules themselves.(V–nb).
The corrected volume of the real gas is = (V-nb)
ii) Pressure Correction :-
The pressure of the gas is directly proportional to forces of bombardment of the molecules on
the walls of the container.
A molecule near the wall of the container which is about to strike is surrounded unequally by
other gaseous molecules.
The molecule near the wall experiences attractive forces only such that it will strike the wall
with a lower force which will exert a lower pressure than if such attractive forces are not
operating on it.
The corrected pressure = (P + P' ) where P' is the pressure correction factor.
Van der Waals found out the forces of attraction experienced by a molecule near the wall are
directly proportional to the square of the density of the gas.
. P' α ρ
2
ρ- density of the gas
ρ =
�
??????
n – Number of moles
P
1
α
�
2
??????
2
V - Volume
P
1
= a
�
2
??????
2
a – Van der Waals constant
Pressure Correction= P + P
1
= (P+
a�
2
??????
2
)
Replacing the corrected volume and pressure in the ideal gas equationPV=nRT we get the
van der Waals equation of state for real gases as below.
(P+
a�
2
??????
2
) (V–nb) = nRT (a &bVan der Waals constants) 14. Deduce the relationship between critical constants and Vanderwaal's constants.
The van der Waals equation for n moles is
(P+
a�
2
??????
2
) (V–nb) = nRT …. (1)
For 1 mole
(P+
a
??????
2
) (V–b) = RT ...... (2)
From the equation we can derive the values of critical constants Pc, Vcand Tc in terms
of a and b, the van der Waals constants, On expanding the above equation.
PV – Pb +
a
V
−
ab
??????
2
– RT = 0 …. (3)
Multiply equation (3) by
V
2
P
V
2
P
X (PV – Pb +
a
V
−
ab
V
– RT ) = 0 …. (4)
V
3
– bV
2
+
aV
P
–
ab
P
–
RTV
2
P
= 0 …. (5)
V
3
– b +
RT
P
V
2
+
a
p
v –
ab
p
= 0 …. (6)
At the critical point all these three solutions of V are equal to the critical volume VC. The
pressure and temperature becomes Pc and Tc respectively.
V
3
– b +
RTC
PC
V
2
+
a
PC
v –
ab
PC
= 0 …. (7)
V = Vc
(V–Vc) = 0
(V–Vc)
3
= 0
V
3
–3Vc V
2
+ 3Vc
2
V–Vc
3
= 0 …. (8)
As equation (7) is identical with equation (8), we can equate the coefficients of V2, V and
constant terms in (7) and (8).
– 3Vc V
2
= – [b +
RT
c
P
c
]V
2
3Vc= [b +
RT
c
P
c
] …. (9)
3Vc
2
V =[
a
P
c
]V
3Vc
2
=[
a
P
c
] …. (10)
-
ab
P
c
= –Vc
3
Vc
3
=
ab
P
c
…. (11)
Divide equation (11) by equation (10)
V
c
3
3V
c
2 =
ab
Pc
/
a
Pc
The deviation of real gases fromidealbehaviour is measured in terms ofa ratio of PV to nRT.
This is termed ascompressibility factor. Mathematically
Compressibility factor Z =
��
���
10. What are ideal gases?
Ideal gases are gases that obey the ideal gas equation, PV = nRT.
P=Pressure ; V=Volume ; T=Temperature ; R=Gas Constant
11. What are Real gases?
Real gases do not obey the ideal gas equation, PV = nRT
12. In what way real gases differ from ideal gases
Ideal gas assumes that the individual gas molecules occupy negligible volume when
compared to the total volume of the gas and there is no attractive force between the gas
molecules.
For ideal gases the compressibility factor Z = 1
For real gases the compressibility factor Z > 1 or Z < 1
13. Derive Van der Walls equation.
The correction terms in the ideal gas equation to account for (i) volume of the gaseous molecules
themselves in V and (ii) the intermolecular forces of attraction in pressure, P, the equation of state for
the real gas is arrived, Vanderwaal's deduced the equation of state of real gases.
i) Volume correction of real gas :-
The volume of a gas is the free space in the container in which molecules move about.
Volume V of an ideal gas is the same as the volume of the container.
The volume of a real gas is, therefore, ideal volume minus the volume occupied by the gas
molecules themselves.(V–nb).
The corrected volume of the real gas is = (V-nb)
ii) Pressure Correction :-
The pressure of the gas is directly proportional to forces of bombardment of the molecules on
the walls of the container.
A molecule near the wall of the container which is about to strike is surrounded unequally by
other gaseous molecules.
The molecule near the wall experiences attractive forces only such that it will strike the wall
with a lower force which will exert a lower pressure than if such attractive forces are not
operating on it.
The corrected pressure = (P + P' ) where P' is the pressure correction factor.
Van der Waals found out the forces of attraction experienced by a molecule near the wall are
directly proportional to the square of the density of the gas.
. P' α ρ
2
ρ- density of the gas
ρ =
�
??????
n – Number of moles
P
1
α
�
2
??????
2
V - Volume
P
1
= a
�
2
??????
2
a – Van der Waals constant
Pressure Correction= P + P
1
= (P+
a�
2
??????
2
)
Replacing the corrected volume and pressure in the ideal gas equationPV=nRT we get the
van der Waals equation of state for real gases as below.
(P+
a�
2
??????
2
) (V–nb) = nRT (a &bVan der Waals constants) 14. Deduce the relationship between critical constants and Vanderwaal's constants.
The van der Waals equation for n moles is
(P+
a�
2
??????
2
) (V–nb) = nRT …. (1)
For 1 mole
(P+
a
??????
2
) (V–b) = RT ...... (2)
From the equation we can derive the values of critical constants Pc, Vcand Tc in terms
of a and b, the van der Waals constants, On expanding the above equation.
PV – Pb +
a
V
−
ab
??????
2
– RT = 0 …. (3)
Multiply equation (3) by
V
2
P
V
2
P
X (PV – Pb +
a
V
−
ab
V
– RT ) = 0 …. (4)
V
3
– bV
2
+
aV
P
–
ab
P
–
RTV
2
P
= 0 …. (5)
V
3
– b +
RT
P
V
2
+
a
p
v –
ab
p
= 0 …. (6)
At the critical point all these three solutions of V are equal to the critical volume VC. The
pressure and temperature becomes Pc and Tc respectively.
V
3
– b +
RTC
PC
V
2
+
a
PC
v –
ab
PC
= 0 …. (7)
V = Vc
(V–Vc) = 0
(V–Vc)
3
= 0
V
3
–3Vc V
2
+ 3Vc
2
V–Vc
3
= 0 …. (8)
As equation (7) is identical with equation (8), we can equate the coefficients of V2, V and
constant terms in (7) and (8).
– 3Vc V
2
= – [b +
RT
c
P
c
]V
2
3Vc= [b +
RT
c
P
c
] …. (9)
3Vc
2
V =[
a
P
c
]V
3Vc
2
=[
a
P
c
] …. (10)
-
ab
P
c
= –Vc
3
Vc
3
=
ab
P
c
…. (11)
Divide equation (11) by equation (10)
V
c
3
3V
c
2 =
ab
Pc
/
a
Pc
??????c
3
= b
Vc = 3b …. (12)
when equation (12) is substituted in (10)
3Vc
2
=[
a
P
c
]
3 x (3b)
2
=
a
P
c
27b
2
=
a
Pc
Pc =
a
27b
2
…. (13)
substituting the values of Vcand Pc in equation (9),
3Vc= [b +
RT
c
P
c
]
(3Vc– b)=
RT
c
P
c
Tc = (3Vc– b) x
P
c
R
= ( 3 x 3b –b) x
a
27Rb
2
= 8b x
a
27Rb
2
Tc =
8a
27Rb
…. (14)
15. Can a Van der Waals gas with a = 0 be liquefied? Explain
a = 0 for a Van der Waals gas i.e for a real gas. Van der Waals constan a = 0. It cannot be
liquefied.
There is a very less interaction between the molecules of gas.
16. Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting
this area stick there permanently. Is the pressure greater or less than on the ordinary
area of walls?
The pressure on the sticky wall is greater than on the ordinary area of walls
17. Explain the following observations
a) Aerated water bottles are kept under water during summer
During summer, the temperature increases, so the solubility of the gas inwater decreases.
So the pressure inside the Aerated water bottles increases and explodes.
b) Liquid ammonia bottles is cooled before opening.
At room temperature the vapour pressure of liquid ammonia is very high.
Ifthe bottle is opened suddenly the pressure decreases and the volumeincreases.
This cause the breakage of the bottle.
c) The tyres of Automobile is inflated slightly lesser pressure in summer.
The Pressure is directly proportional to the temperature.
During summer,the temperature increases and the Pressure of air inside the tubeincreases.
So the tube will burst.
d) The size of weather balloons becomes larger and larger as it ascends up into
altitude.
When the balloons ascents, the atmospheric pressure decreases,
So the volume of the balloons increases according to Boyles Law.
18. a) Why does gases don’t settle at the bottom of the container.
Gases have less density and the molecules can freely move.
All gases molecules have high inter molecular force of attractionthe gases molecules are
constantly bombarding with one another andon the wall on the container.
So they don‟t settle down.
b) Why there is no gas in the atmosphere? And why moon has noatmosphere.?
Hydrogen gas is light in weight. So it escapes high from the earth‟satmosphere.
The moon is having very low gravity and low escape velocity
The gases have higher thermal velocity.
c) Why aerosol cans carry warning of heating ?
On heating the gas inside the aerosol cans expands and the pressureinside the cans increases,
leading to an explosion..
19. Suggest why there is no hydrogen (H2) in our atmosphere. Why does the moon have no
atmosphere?
Hydrogen is the lightest element thus when produced in free state.
Moon has no atmosphere because the value of acceleration due to gravity on the surface of
the moon is small.
20. Which of the following gases would you expect to deviate from ideal behaviour under
conditions of low temperature F2, Cl2, or Br2? Explain.
Bromine deviates from the ideal gas maximum
Bromine deviates from ideal behavior because it has largest atomic radii compared to others.
So Van der Waals forces are stronger in Bromine.
21. Aerosol cans carry clear warning of heating of the can. Why?
This is because the pressure will build up to so much that the can will burst.
It is due to two reason.
1, The gas pressure increases
2, More of the liquefied propellant turns into a gas.
22. When the driver of an automobile applies brake, the passengers are pushed toward the
front of the car but a helium balloon is pushed toward back of the car. Upon forward
acceleration the passengers are pushed toward the front of the car. Why?
Upon forward acceleration, the passengers are pushed toward the front of the car, becaused
the body in motion tends to stay in motion until acted upon by an outside force.
Helium balloon is going to move opposite to this pseudo gravitational force.
23. Would it be easier to drink water with a straw on the top of Mount Everest?
It is difficult to drink water with a straw on the top of mount Everest. This is because at higher
altitudes the pressure decreases and hence temperature also decreases.
24. Why do astronauts have to wear protective suits when they are on the surface of moon?
In space there is no air to breath and no air pressure if we do not Wear, Our body will continue to push out
and blow up like a balloon. It would look cool, but we will be dead, so the astronauts in space must wear a
protective suits.
25. When ammonia combines with HCl, NH4Cl is formed as white dense fumes. Why do
more fumes appear near HCl?
Molecular weightof Ammonia is 17 and Molecular weight of HCl is 36.45
The Molecular weight of Ammonia is less than HCl.
The rate of diffusion is inversely proportional to the molecular weight of the gas. Lower the
molecular weight, faster is the diffusion.
Hence ammonia diffuses faster than HCl and thus appear as dense whiter fumes near HCl
3
= b
Vc = 3b …. (12)
when equation (12) is substituted in (10)
3Vc
2
=[
a
P
c
]
3 x (3b)
2
=
a
P
c
27b
2
=
a
Pc
Pc =
a
27b
2
…. (13)
substituting the values of Vcand Pc in equation (9),
3Vc= [b +
RT
c
P
c
]
(3Vc– b)=
RT
c
P
c
Tc = (3Vc– b) x
P
c
R
= ( 3 x 3b –b) x
a
27Rb
2
= 8b x
a
27Rb
2
Tc =
8a
27Rb
…. (14)
15. Can a Van der Waals gas with a = 0 be liquefied? Explain
a = 0 for a Van der Waals gas i.e for a real gas. Van der Waals constan a = 0. It cannot be
liquefied.
There is a very less interaction between the molecules of gas.
16. Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting
this area stick there permanently. Is the pressure greater or less than on the ordinary
area of walls?
The pressure on the sticky wall is greater than on the ordinary area of walls
17. Explain the following observations
a) Aerated water bottles are kept under water during summer
During summer, the temperature increases, so the solubility of the gas inwater decreases.
So the pressure inside the Aerated water bottles increases and explodes.
b) Liquid ammonia bottles is cooled before opening.
At room temperature the vapour pressure of liquid ammonia is very high.
Ifthe bottle is opened suddenly the pressure decreases and the volumeincreases.
This cause the breakage of the bottle.
c) The tyres of Automobile is inflated slightly lesser pressure in summer.
The Pressure is directly proportional to the temperature.
During summer,the temperature increases and the Pressure of air inside the tubeincreases.
So the tube will burst.
d) The size of weather balloons becomes larger and larger as it ascends up into
altitude.
When the balloons ascents, the atmospheric pressure decreases,
So the volume of the balloons increases according to Boyles Law.
18. a) Why does gases don’t settle at the bottom of the container.
Gases have less density and the molecules can freely move.
All gases molecules have high inter molecular force of attractionthe gases molecules are
constantly bombarding with one another andon the wall on the container.
So they don‟t settle down.
b) Why there is no gas in the atmosphere? And why moon has noatmosphere.?
Hydrogen gas is light in weight. So it escapes high from the earth‟satmosphere.
The moon is having very low gravity and low escape velocity
The gases have higher thermal velocity.
c) Why aerosol cans carry warning of heating ?
On heating the gas inside the aerosol cans expands and the pressureinside the cans increases,
leading to an explosion..
19. Suggest why there is no hydrogen (H2) in our atmosphere. Why does the moon have no
atmosphere?
Hydrogen is the lightest element thus when produced in free state.
Moon has no atmosphere because the value of acceleration due to gravity on the surface of
the moon is small.
20. Which of the following gases would you expect to deviate from ideal behaviour under
conditions of low temperature F2, Cl2, or Br2? Explain.
Bromine deviates from the ideal gas maximum
Bromine deviates from ideal behavior because it has largest atomic radii compared to others.
So Van der Waals forces are stronger in Bromine.
21. Aerosol cans carry clear warning of heating of the can. Why?
This is because the pressure will build up to so much that the can will burst.
It is due to two reason.
1, The gas pressure increases
2, More of the liquefied propellant turns into a gas.
22. When the driver of an automobile applies brake, the passengers are pushed toward the
front of the car but a helium balloon is pushed toward back of the car. Upon forward
acceleration the passengers are pushed toward the front of the car. Why?
Upon forward acceleration, the passengers are pushed toward the front of the car, becaused
the body in motion tends to stay in motion until acted upon by an outside force.
Helium balloon is going to move opposite to this pseudo gravitational force.
23. Would it be easier to drink water with a straw on the top of Mount Everest?
It is difficult to drink water with a straw on the top of mount Everest. This is because at higher
altitudes the pressure decreases and hence temperature also decreases.
24. Why do astronauts have to wear protective suits when they are on the surface of moon?
In space there is no air to breath and no air pressure if we do not Wear, Our body will continue to push out
and blow up like a balloon. It would look cool, but we will be dead, so the astronauts in space must wear a
protective suits.
25. When ammonia combines with HCl, NH4Cl is formed as white dense fumes. Why do
more fumes appear near HCl?
Molecular weightof Ammonia is 17 and Molecular weight of HCl is 36.45
The Molecular weight of Ammonia is less than HCl.
The rate of diffusion is inversely proportional to the molecular weight of the gas. Lower the
molecular weight, faster is the diffusion.
Hence ammonia diffuses faster than HCl and thus appear as dense whiter fumes near HCl
26. What is Boyle’s temperature ? What happens to real gases above and below the Boyle’s
temperature ?
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called
Boyle temperature or Boyle point.
Above the Boyle point, for real gases, Z > 1, ie., the real gases show positive deviation.
Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with
the increase in pressure.
27. Name the different methods of liquefaction of gases
1) Linde‟s method 2) Claude‟s process 3) Adiabatic process
28. What is Inversion temperature ?
Temperature below which a gas obeys Joule-Thomson effect is called inversion temperature (Ti).
29. Define Joule-Thomson effect
The phenomenon of lowering of temperature when a gas is made to expand adiabatically from a
region of high pressure into a region of low pressure is known as Joule-Thomson effect.
7.THERMODYNAMICS
1. Define system and explain its types?
The system is the part of universe which is under thermodynamic consideration.
Types:-
Types Explanation
Open system A System which can exchange both matter and energy with its surrounding
Closed system A system which can exchange only energy but not matter with its
surroundings.
Isolated system A system which can exchange neither matter nor energy with its surroundings
2. State the first law of thermodynamics
"Energy can neither be created nor destroyed, but may be converted from one form to another".
3. State the various statements of second law of thermodynamics.
Entropy statement:-
Entropy is a measure of the molecular disorder (randomness) of a system.
The entropy of an isolated system increases during a spontaneous process.
Entropy is state function
Kelvin-Planck statement:-
“It is impossible to construct an engine which operated in a complete cycle will absorb heat
from a single body and convert it completely to work without leaving some changes in the working
system”.
Clausius statement:-
It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing
some work.
Efficiency :-
Efficiency =
work performed
heat absorbed
4. State the Third law of thermodynamics
The entropy of a perfectly crystalline material at absolute zero is zero
5. State the Zeroth law of thermodynamics
When two systems are separately in equilibrium with a third system, they are in equilibrium
with each other.
6. Explain Extensive and Intensive properties with two examples
Extensive properties :- The property that depends on the mass or the size of the system is
called an extensive property.
Eg: Volume, Number of moles,Mass
Intensive properties :- The properties which do not depend upon the quantity of matter
present in the system
Eg: Boiling point, Melting point. 7. What are state and path functions? Give two examples.
State Function Path Function
A state function is a thermodynamic property of a
system, which has a specific value for a given state
and does not depend on the path (or manner) by
which the particular state is reached.
A path function is a thermodynamic property of the
system whose value depends on the path by which
the system changes from its initial to final states.
Example : Pressure (P), Volume (V),
Temperature(T), Internal energy (U),
Enthalpy (H), free energy (G) etc.
Example: Work (w), Heat (q).
8. Define the following terms:
a. isothermal process b. adiabatic process
c. isobaric process d. isochoric process
a. isothermal process : - The thermodynamic process in which temperature remains
constantthroughout are called isothermal process.
b. adiabatic process :- The thermodynamic process in which heat exchange between system and
surrounding is not possible.
c. isobaric process :- The thermodynamic process occurring at constant pressure.
d. isochoric process :- The thermodynamic process occurring at constant volume.
9. What are Reversible and Irreversible process?
Reversible process :- The process in which the system and surrounding can be restored to the
initial state from the final state without producing any changes in thethermodynamic properties of the
universe.
Irreversible process:-The process in which the system and surrounding cannot be restored to the
initial state from the final state.
10. Define Heat.
Heat is regarded as the energy in transit across the boundary separating a system from its
surrounding.
Heat is a path function.
The SI unit of heat is joule (J)
11. Write the sign conventions for heat and work
If heat is absorbed by the system +q
If heat is evolved by the system +q
If work is done by the system -w
If work is done on the system +W
12. Define Work
Work is defined as the force (F)multiplied by the displacement -w=F.x,
Work is measured in Joules,
i.e the SI unit of work is Joule
13. Define Enthalpy
Enthalpy is a thermodynamic property of a system.
Enthalpy H is defined as the sum of the internal energy and pressure volume work.
H = U + PV
14. Define Standard heat of formation
The change in enthalpy that takes place when one mole of a compound is formed from its
elements, present in their standard states (298 K and 1 bar pressure).
15. Define Gibb’s free energy.
G is expressed as G = H - TS,
G Gibb‟s free energy ; H Enthalpy ; T Temperature ; S Entropy
Free energy change of a process is given by the relation ΔG= ΔH – TΔS
Gibbs free energy (G) is an extensive property and it is a single valued state function.
16. List the characteristics of Gibbs free energy.
Free energy is defined as G = H – TS
(where G – free energy, H – enthalpy and S – entropy).
temperature ?
The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called
Boyle temperature or Boyle point.
Above the Boyle point, for real gases, Z > 1, ie., the real gases show positive deviation.
Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with
the increase in pressure.
27. Name the different methods of liquefaction of gases
1) Linde‟s method 2) Claude‟s process 3) Adiabatic process
28. What is Inversion temperature ?
Temperature below which a gas obeys Joule-Thomson effect is called inversion temperature (Ti).
29. Define Joule-Thomson effect
The phenomenon of lowering of temperature when a gas is made to expand adiabatically from a
region of high pressure into a region of low pressure is known as Joule-Thomson effect.
7.THERMODYNAMICS
1. Define system and explain its types?
The system is the part of universe which is under thermodynamic consideration.
Types:-
Types Explanation
Open system A System which can exchange both matter and energy with its surrounding
Closed system A system which can exchange only energy but not matter with its
surroundings.
Isolated system A system which can exchange neither matter nor energy with its surroundings
2. State the first law of thermodynamics
"Energy can neither be created nor destroyed, but may be converted from one form to another".
3. State the various statements of second law of thermodynamics.
Entropy statement:-
Entropy is a measure of the molecular disorder (randomness) of a system.
The entropy of an isolated system increases during a spontaneous process.
Entropy is state function
Kelvin-Planck statement:-
“It is impossible to construct an engine which operated in a complete cycle will absorb heat
from a single body and convert it completely to work without leaving some changes in the working
system”.
Clausius statement:-
It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing
some work.
Efficiency :-
Efficiency =
work performed
heat absorbed
4. State the Third law of thermodynamics
The entropy of a perfectly crystalline material at absolute zero is zero
5. State the Zeroth law of thermodynamics
When two systems are separately in equilibrium with a third system, they are in equilibrium
with each other.
6. Explain Extensive and Intensive properties with two examples
Extensive properties :- The property that depends on the mass or the size of the system is
called an extensive property.
Eg: Volume, Number of moles,Mass
Intensive properties :- The properties which do not depend upon the quantity of matter
present in the system
Eg: Boiling point, Melting point. 7. What are state and path functions? Give two examples.
State Function Path Function
A state function is a thermodynamic property of a
system, which has a specific value for a given state
and does not depend on the path (or manner) by
which the particular state is reached.
A path function is a thermodynamic property of the
system whose value depends on the path by which
the system changes from its initial to final states.
Example : Pressure (P), Volume (V),
Temperature(T), Internal energy (U),
Enthalpy (H), free energy (G) etc.
Example: Work (w), Heat (q).
8. Define the following terms:
a. isothermal process b. adiabatic process
c. isobaric process d. isochoric process
a. isothermal process : - The thermodynamic process in which temperature remains
constantthroughout are called isothermal process.
b. adiabatic process :- The thermodynamic process in which heat exchange between system and
surrounding is not possible.
c. isobaric process :- The thermodynamic process occurring at constant pressure.
d. isochoric process :- The thermodynamic process occurring at constant volume.
9. What are Reversible and Irreversible process?
Reversible process :- The process in which the system and surrounding can be restored to the
initial state from the final state without producing any changes in thethermodynamic properties of the
universe.
Irreversible process:-The process in which the system and surrounding cannot be restored to the
initial state from the final state.
10. Define Heat.
Heat is regarded as the energy in transit across the boundary separating a system from its
surrounding.
Heat is a path function.
The SI unit of heat is joule (J)
11. Write the sign conventions for heat and work
If heat is absorbed by the system +q
If heat is evolved by the system +q
If work is done by the system -w
If work is done on the system +W
12. Define Work
Work is defined as the force (F)multiplied by the displacement -w=F.x,
Work is measured in Joules,
i.e the SI unit of work is Joule
13. Define Enthalpy
Enthalpy is a thermodynamic property of a system.
Enthalpy H is defined as the sum of the internal energy and pressure volume work.
H = U + PV
14. Define Standard heat of formation
The change in enthalpy that takes place when one mole of a compound is formed from its
elements, present in their standard states (298 K and 1 bar pressure).
15. Define Gibb’s free energy.
G is expressed as G = H - TS,
G Gibb‟s free energy ; H Enthalpy ; T Temperature ; S Entropy
Free energy change of a process is given by the relation ΔG= ΔH – TΔS
Gibbs free energy (G) is an extensive property and it is a single valued state function.
16. List the characteristics of Gibbs free energy.
Free energy is defined as G = H – TS
(where G – free energy, H – enthalpy and S – entropy).
G is a state function.
G – Extensive property, ΔG – intensive property.
G has a single value for the thermodynamic state of the system.
Process Spontaneous Non - spontaneous Equilibrium
ΔG – ve +ve zero
17. The equilibrium constant of a reaction is 10, what will be the sign of ΔG? Will this
reaction be spontaneous?
Van‟tHoff equation. G = - 2.303RTlogK
Equilibrium constant K = 10 G = - 2.303RTlog10 = - 2.303RT
G is –ve ; so the reaction is Spontaneous.
18. Define enthalpy of combustion.
The enthalpy change when one mole of any substance is completely burnt in excess of oxygen.
Ex:-The heat of combustion of Methane is – 87.78 kJ mol
-1
19. Define molar heat capacity. Give its unit.
“The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.
The SI unit of molar heat capacity is JK
−1
mol
-1
20. Define Specific heat capacity
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a
specified temperature
21. Define the calorific value of food. What is the unit of calorific value?
The calorific value is defined as the amount of heat produced in calories (or joules) when one
gram of the substance is completely burnt.
The SI unit of calorific value is J kg
−1
.
It is usually expressed in cal g
-1.
22. DefineHeat of neutralization
The change in enthalpy when one gram equivalent of an acid is completely neutralised by one
gram equivalent of a base or vice versa in dilute solution.
The heat of neutralisation of a strong acid and strong base is constant (H = 57.32 KJ)
23. The heat of neutralisation of a strong acid and strong base is constant – Why?
The reason for this can be explained on the basis of Arrhenius theory of acidsand bases which
states that strong acids and strong bases completely ioniseinaqueous solution to produce H+ and OH-
ions respectively.
H
+
+ OH
H2O H = 57.32 KJ
24. What is lattice energy?
It is defined as the amount of energy required to completely remove the constituent ions from
its crystal lattice to an infinite distance. It is also referred as lattice enthalpy.
25. Define Hess's law of constant heat summation.
It states that “the enthalpy change of a reaction either at constant volume or constant pressure
is the same whether it takes place in a single or multiple steps”.
26. What are spontaneous reactions? What are the conditions for the spontaneity of a
process?
Spontaneous reaction:- A reaction that occurs under the given set of conditions without any
externaldriving force is called spontaneous reaction.
Conditions for the spontaneity : Δ� = −��, Δ� = + ��, Δ� = −��
27. Identify the state and path functions out of the following: a)Enthalpy b)Entropy c) Heat
d) Temperature e) Work f) Free energy.
a) Enthalpy state function
b) Entropy state function
c) Heat path function
d) Temperature state function
e) Work path function f) Free energy state function
28. Predict the feasibility of a reaction wheni) both ΔH and ΔS positive
ii) both ΔH and ΔS negativeiii) ΔH decreases but ΔS increases
ΔH ΔS ΔG= ΔH−TΔS Description Example
i, +ve +ve
+ (at low T)
non-spontaneous at low
temperature
spontaneous at high
temperature
Melting of a solid
− (at high T)
ii, -ve -ve
− (at low T) spontaneous at low temperature
Adsorption of gases + (at high T) non-spontaneous at high
temperature
iii
,
-ve +ve − (at all T) Spontaneous at all temperature 2O3 (g)3O2 (g)
29. Define Internal energy
Internal energy of a system is equal to the energy possessed by all its constituents namely
atoms, ions and molecules.
30. List the Characteristics of internal energy
The internal energy of a system is an extensive property
The internal energy of a system is astate function.
The change in internal energy of a system is expressed as U= Uf– Ui
In a cyclic process, there is no internal energy change. U(cyclic)= 0
ΔU = Uf– Ui = −ve (Uf< Ui) ΔU would be negative.
ΔU = Uf– Ui = +ve (Uf> Ui) ΔU would be positive.
31. Derive the relation between ΔH and ΔU for an ideal gas. Explain each term involved in
the equation.
When the system at constant pressure undergoes changes from an initial state with H1, U1 and
V1 to a final state with H2, U2 and V2 the change in enthalpy ΔH, can be calculated as follows:
H = U + PV
In the initial state
H1 = U1 + PV1 …….(1)
In the final state
H2 = U2 + PV2 …….(2)
change in enthalpy is (2) - (1)
(H2 -H1) = (U1 – U2) + P(V2-V1)
H = U + PV…. (3)
Consider a closed system of gases which are chemically reacting to form gaseous products at constant
temperature and pressure with V1and V2as the total volumes of the reactant and product gases
respectively, and n1and n2as the number of moles of gaseous reactants and products, then,
For reactants (initial state) :
PV1 = n1RT ….. (4)
For products (final state) :
PV2 = n2RT ….. (5)
(5) – (4)
P(V2-V1) = (n2-n1) RT
PV = n(g)RT ….. (6)
Substituting (6) in (3)
H = U + n(g)RT …. (7)
H- Change in Enthalpy ;U – Change in Internal Energy;
n(g) – Change in number of Moles ; R- Gas constant ; T- Temperature.
G – Extensive property, ΔG – intensive property.
G has a single value for the thermodynamic state of the system.
Process Spontaneous Non - spontaneous Equilibrium
ΔG – ve +ve zero
17. The equilibrium constant of a reaction is 10, what will be the sign of ΔG? Will this
reaction be spontaneous?
Van‟tHoff equation. G = - 2.303RTlogK
Equilibrium constant K = 10 G = - 2.303RTlog10 = - 2.303RT
G is –ve ; so the reaction is Spontaneous.
18. Define enthalpy of combustion.
The enthalpy change when one mole of any substance is completely burnt in excess of oxygen.
Ex:-The heat of combustion of Methane is – 87.78 kJ mol
-1
19. Define molar heat capacity. Give its unit.
“The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.
The SI unit of molar heat capacity is JK
−1
mol
-1
20. Define Specific heat capacity
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a
specified temperature
21. Define the calorific value of food. What is the unit of calorific value?
The calorific value is defined as the amount of heat produced in calories (or joules) when one
gram of the substance is completely burnt.
The SI unit of calorific value is J kg
−1
.
It is usually expressed in cal g
-1.
22. DefineHeat of neutralization
The change in enthalpy when one gram equivalent of an acid is completely neutralised by one
gram equivalent of a base or vice versa in dilute solution.
The heat of neutralisation of a strong acid and strong base is constant (H = 57.32 KJ)
23. The heat of neutralisation of a strong acid and strong base is constant – Why?
The reason for this can be explained on the basis of Arrhenius theory of acidsand bases which
states that strong acids and strong bases completely ioniseinaqueous solution to produce H+ and OH-
ions respectively.
H
+
+ OH
H2O H = 57.32 KJ
24. What is lattice energy?
It is defined as the amount of energy required to completely remove the constituent ions from
its crystal lattice to an infinite distance. It is also referred as lattice enthalpy.
25. Define Hess's law of constant heat summation.
It states that “the enthalpy change of a reaction either at constant volume or constant pressure
is the same whether it takes place in a single or multiple steps”.
26. What are spontaneous reactions? What are the conditions for the spontaneity of a
process?
Spontaneous reaction:- A reaction that occurs under the given set of conditions without any
externaldriving force is called spontaneous reaction.
Conditions for the spontaneity : Δ� = −��, Δ� = + ��, Δ� = −��
27. Identify the state and path functions out of the following: a)Enthalpy b)Entropy c) Heat
d) Temperature e) Work f) Free energy.
a) Enthalpy state function
b) Entropy state function
c) Heat path function
d) Temperature state function
e) Work path function f) Free energy state function
28. Predict the feasibility of a reaction wheni) both ΔH and ΔS positive
ii) both ΔH and ΔS negativeiii) ΔH decreases but ΔS increases
ΔH ΔS ΔG= ΔH−TΔS Description Example
i, +ve +ve
+ (at low T)
non-spontaneous at low
temperature
spontaneous at high
temperature
Melting of a solid
− (at high T)
ii, -ve -ve
− (at low T) spontaneous at low temperature
Adsorption of gases + (at high T) non-spontaneous at high
temperature
iii
,
-ve +ve − (at all T) Spontaneous at all temperature 2O3 (g)3O2 (g)
29. Define Internal energy
Internal energy of a system is equal to the energy possessed by all its constituents namely
atoms, ions and molecules.
30. List the Characteristics of internal energy
The internal energy of a system is an extensive property
The internal energy of a system is astate function.
The change in internal energy of a system is expressed as U= Uf– Ui
In a cyclic process, there is no internal energy change. U(cyclic)= 0
ΔU = Uf– Ui = −ve (Uf< Ui) ΔU would be negative.
ΔU = Uf– Ui = +ve (Uf> Ui) ΔU would be positive.
31. Derive the relation between ΔH and ΔU for an ideal gas. Explain each term involved in
the equation.
When the system at constant pressure undergoes changes from an initial state with H1, U1 and
V1 to a final state with H2, U2 and V2 the change in enthalpy ΔH, can be calculated as follows:
H = U + PV
In the initial state
H1 = U1 + PV1 …….(1)
In the final state
H2 = U2 + PV2 …….(2)
change in enthalpy is (2) - (1)
(H2 -H1) = (U1 – U2) + P(V2-V1)
H = U + PV…. (3)
Consider a closed system of gases which are chemically reacting to form gaseous products at constant
temperature and pressure with V1and V2as the total volumes of the reactant and product gases
respectively, and n1and n2as the number of moles of gaseous reactants and products, then,
For reactants (initial state) :
PV1 = n1RT ….. (4)
For products (final state) :
PV2 = n2RT ….. (5)
(5) – (4)
P(V2-V1) = (n2-n1) RT
PV = n(g)RT ….. (6)
Substituting (6) in (3)
H = U + n(g)RT …. (7)
H- Change in Enthalpy ;U – Change in Internal Energy;
n(g) – Change in number of Moles ; R- Gas constant ; T- Temperature.
32. Write the Applications of bomb calorimeter:
It is used to determine the amount of heat released in combustion reaction.
It is used to determine the calorific value of food.
It is used in many industries such as metabolic study, food processing, explosive testing etc.
33. How heat absorbed at constant volume is measured using
bomb calorimeter with a neat diagram.
For chemical reactions, heat evolved at constant volume, is
measured in a bomb calorimeter.
The inner vessel (the bomb) and its cover are made of
strong steel.
The coveris fitted tightly to the vessel by means of metal lid
and screws.
A weighed amount of the substance is taken in a platinum
cup and pressurized with excess oxygen.
The bomb is immersed in water,in the inner volume of the
calorimeter.
The reaction is started by striking the substance through
electrical heating.
Heat evolved during the reaction is absorbed by the
calorimeter as well as the water in which the bomb is
immersed.
The change in temperature is measured using a Beckman
thermometer. Since the bomb is sealed its volume does not
change and hence the heat measurements is equal to the
heat of combustion at a constant volume (U
c
0
).
The amount of heat produced in the reaction U
c
0
is equal
to the sum of the heat abosrbed by the calorimeter and water.
Heat absorbed by the calorimeter :-
q1 = k T
k - calorimeter constant
k = mcCc
mc- mass of the calorimeter ;Cc- heat capacity of calorimeter.
Heat absorbed by the water :-
q2 = mwCwT
mw- molar mass of water
Cw- molar heat capacity of water (75.29 JK
-1
mol
-1
)
U
c
0
= q1 + q2
U
c
0
= kT + mwCwT
= (k +mwCw)T
Calorimeter constant can be determined by burning a known mass of standard sample (benzoic
acid) for which the heat of combustion is known (-3227 kJmol
-1
) The enthalpy of combustion
at constant pressure of the substance is calculated from the equation .
??????
�
�
(Pressure) = �
�
�
(volume) + RTn(g)
34. Write down the Born-Haber cycle for the formation of CaCl2
ΔH1 = heat of sublimation of Ca(S)
ΔH2 = dissociation energy of Cl2(g)
ΔH3 = ionisation energy of Ca(S)
ΔH4 = Electron affinity of Cl (g)
ΔH5 = the lattice enthalpy for the formation
of CaCl2(S)
ΔHf = enthalpy change for the formation of
CaCl2(S) directly form elements
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5
ΔH5 = ΔHf- (ΔH1 + ΔH2 + ΔH3 + ΔH4)
35.
.
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride
crystal.
ΔH1 = Heat of sublimation of Na(S)
ΔH2 = Ionisation energy of Na(S)
ΔH3 = Dissociation energy of Cl2(g)
ΔH4 = Electron affinity of Cl (g)
U = The lattice Energy of NaCl(S)
ΔHf = Enthalpy change for the formation of NaCl(S) directly
form its elements
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + U
U = ΔHf - (ΔH1 + ΔH2 + ΔH3 + ΔH4)
8. PHYSICAL AND CHEMICAL EQUILIBRIUM
1. Why chemical equilibrium is considered as dynamic equilibrium (or) What is Dynamic
Equilibrium
At equilibrium the forward and the backward reactions will proceed atthe same rate.So no
macroscopic changes is observed.
2. Write a note on Solid-liquid equilibrium
Equilibrium exists only at a particular temperature and pressure. The temperature at which
the solid and liquid phases of a substance are at equilibrium.
Eg : - H2O(s) ⇌ H2O(l)
3. Write a note on Liquid - Vapour equilibrium
Equilibrium exists between the liquid phase and the vapour phase of a substance Eg: - at 373 K
and1 atm pressure
Eg: - H2O(l) ⇌ H2O(g)
4. What are boiling point (or) condensation point
The temperature at which the liquid and vapour phases are at equilibrium
5. Write a note on Solid in liquid equilibrium
A dynamic equilibrium is established between the solute molecules in the solid phase and
in the solution phase..Eg: - Sugar
(Solid) ⇌ Sugar
(Solution)
6. Write a note on Gas in liquids equilibrium
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas
molecules in the gaseous state and those dissolved in the liquid.
Eg: - CO2
(g) ⇌ CO2
(Solution)
7. What is Boiling point (or) Condensation point.
The temperature at which the liquid and vapour phases are at equilibrium is called the
boiling point and condensation point of the liquid.
8. Define Homogeneous Equilibrium
When the reactant and the product are in the same phase it is called as
Homogeneous_Equilibrium
Eg:- H2(g) + I2(g) ⇌ 2HI(g)
It is used to determine the amount of heat released in combustion reaction.
It is used to determine the calorific value of food.
It is used in many industries such as metabolic study, food processing, explosive testing etc.
33. How heat absorbed at constant volume is measured using
bomb calorimeter with a neat diagram.
For chemical reactions, heat evolved at constant volume, is
measured in a bomb calorimeter.
The inner vessel (the bomb) and its cover are made of
strong steel.
The coveris fitted tightly to the vessel by means of metal lid
and screws.
A weighed amount of the substance is taken in a platinum
cup and pressurized with excess oxygen.
The bomb is immersed in water,in the inner volume of the
calorimeter.
The reaction is started by striking the substance through
electrical heating.
Heat evolved during the reaction is absorbed by the
calorimeter as well as the water in which the bomb is
immersed.
The change in temperature is measured using a Beckman
thermometer. Since the bomb is sealed its volume does not
change and hence the heat measurements is equal to the
heat of combustion at a constant volume (U
c
0
).
The amount of heat produced in the reaction U
c
0
is equal
to the sum of the heat abosrbed by the calorimeter and water.
Heat absorbed by the calorimeter :-
q1 = k T
k - calorimeter constant
k = mcCc
mc- mass of the calorimeter ;Cc- heat capacity of calorimeter.
Heat absorbed by the water :-
q2 = mwCwT
mw- molar mass of water
Cw- molar heat capacity of water (75.29 JK
-1
mol
-1
)
U
c
0
= q1 + q2
U
c
0
= kT + mwCwT
= (k +mwCw)T
Calorimeter constant can be determined by burning a known mass of standard sample (benzoic
acid) for which the heat of combustion is known (-3227 kJmol
-1
) The enthalpy of combustion
at constant pressure of the substance is calculated from the equation .
??????
�
�
(Pressure) = �
�
�
(volume) + RTn(g)
34. Write down the Born-Haber cycle for the formation of CaCl2
ΔH1 = heat of sublimation of Ca(S)
ΔH2 = dissociation energy of Cl2(g)
ΔH3 = ionisation energy of Ca(S)
ΔH4 = Electron affinity of Cl (g)
ΔH5 = the lattice enthalpy for the formation
of CaCl2(S)
ΔHf = enthalpy change for the formation of
CaCl2(S) directly form elements
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5
ΔH5 = ΔHf- (ΔH1 + ΔH2 + ΔH3 + ΔH4)
35.
.
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride
crystal.
ΔH1 = Heat of sublimation of Na(S)
ΔH2 = Ionisation energy of Na(S)
ΔH3 = Dissociation energy of Cl2(g)
ΔH4 = Electron affinity of Cl (g)
U = The lattice Energy of NaCl(S)
ΔHf = Enthalpy change for the formation of NaCl(S) directly
form its elements
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + U
U = ΔHf - (ΔH1 + ΔH2 + ΔH3 + ΔH4)
8. PHYSICAL AND CHEMICAL EQUILIBRIUM
1. Why chemical equilibrium is considered as dynamic equilibrium (or) What is Dynamic
Equilibrium
At equilibrium the forward and the backward reactions will proceed atthe same rate.So no
macroscopic changes is observed.
2. Write a note on Solid-liquid equilibrium
Equilibrium exists only at a particular temperature and pressure. The temperature at which
the solid and liquid phases of a substance are at equilibrium.
Eg : - H2O(s) ⇌ H2O(l)
3. Write a note on Liquid - Vapour equilibrium
Equilibrium exists between the liquid phase and the vapour phase of a substance Eg: - at 373 K
and1 atm pressure
Eg: - H2O(l) ⇌ H2O(g)
4. What are boiling point (or) condensation point
The temperature at which the liquid and vapour phases are at equilibrium
5. Write a note on Solid in liquid equilibrium
A dynamic equilibrium is established between the solute molecules in the solid phase and
in the solution phase..Eg: - Sugar
(Solid) ⇌ Sugar
(Solution)
6. Write a note on Gas in liquids equilibrium
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas
molecules in the gaseous state and those dissolved in the liquid.
Eg: - CO2
(g) ⇌ CO2
(Solution)
7. What is Boiling point (or) Condensation point.
The temperature at which the liquid and vapour phases are at equilibrium is called the
boiling point and condensation point of the liquid.
8. Define Homogeneous Equilibrium
When the reactant and the product are in the same phase it is called as
Homogeneous_Equilibrium
Eg:- H2(g) + I2(g) ⇌ 2HI(g)
9. Define Heterogeneous Equilibrium
When the reactant and the product are in the different phase it is called as
Heterogeneous_Equilibrium
Eg:- CaCO3(s) ⇌ CaO(s) + CO2(g)
10. Define Law of Mass action
At a given temperature, the rate of a chemical reaction is directlyproportional to the product of
the active masses of the reactants.
Rate [Reactant]
x
11. List the application of Equilibrium constant
It is used to find the direction of the reaction taking place
It is used to find the extend of the reaction
It is used to find the equilibrium concentration of the reactant and the product.
12. Define Reaction Quotient ‘Q’
Reaction Quotient is defined as the ratio between the product of the active
masses of the products and the reactants raised to a Stoichiometriccoefficientunder non equilibrium
conditions
xA(g) + yB(g) mC(g) + nD(g)
Q =
[�]
�
[�]
�
[�]
�
[�]
�
13. Explain how the equilibrium constant Kc is used to find out the direction of the reaction
If Q = Kc, Equilibrium reaction
If Q > Kc, the reaction proceeds in the Reverse direction.
If Q < Kc, the reaction proceeds in the Forward direction
14. Explain the effect of concentration on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which
equilibrium shifts
Concentration
Addition of reactants (increase in reactant
concentration)
Forward reaction
Removal of products (decrease in product
concentration)
Addition of products (increase in product
concentration)
Reverse reaction
Removal of reactants (decrease in reactant
concentration)
15. Explain the effect of pressure on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Pressure
Increase of pressure (Decrease in
volume)
Reaction that favours fewer moles of
the gaseous molecules
Decrease of pressure (Increase in
volume)
Reaction that favours more moles of
the gaseous molecules
16. Explain the effect of Temperature on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Temperature
Increase (High T) Towards endothermic reaction
decrease (Low T) Towards endothermic reaction
17. Explain the effect of Catalyst on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Catalyst Addition of catalyst No effect
18. Explain the Effect of inert gas on a equilibrium reaction
Addition of inert gas at constant volume - No effect
19. Derive the relation between Kp and Kc.
Let us consider a reaction
xA(g) + yB(g)⇌mC(g) + nD(g)
Kc =
[�]
�
[�]
�
[�]
�
[�]
�
------------- 1
Kp =
�
�
�
��
�
�
�
�
�
��
�
�------------- 2
From Ideal gas equation,
PV = nRT
P=
�
�
RT------------------- 3
Pi = Ci RT
C
i
v
= Ci
Substituting the partial pressure terms in eqn(3)
�
�
�
= [A]
x
(RT)
x
�
�
�
= [B]
y
(RT)
y
�
�
�
= [C]
m
(RT)
m
�
�
�
= [D]
n
(RT)
n
Substitute the above pressure terms in eqn( 2 )
Kp =
[C]
m
[D]
n
[A]
x
[B]
y
X
(RT)
m
×(RT)
n
(RT)
x
×(RT)
y
Kp = Kc RT
m+n − (x+y)
n(g) = (m+n) – (x+y)
Kp = Kc RT
n(g)
20. Give a balanced chemical equation for the equilibrium reaction for which the
equilibrium constant is given by expression Kc =
[�??????�]
�
[��]
�
[��]
�
[ ??????��]
�
4 NH3 (g) + 7O2 (g) ⇌ 4NO(g)+ 6H2O(g)
21. Define le-Chatlier Braun principle
If a system at equilibrium is subjected to a disturbance, then the systemwill move in the
direction to nullify the effect of the disturbance
22. For a given reaction the equilibrium constant has a constant value. If the value of Q also
constant ?. explain.
As the reaction proceeds, there is a change in the concentration of the reactant and the products.
The Q value also changes until the equilibrium is reached.
At equilibrium the Q value is equal to Kc.
Once the equilibrium is reached there is no change in the Q value.
23. What the relation between Kp and Kc.Give one example for Kp is equal to Kc
Kp = Kc ��
�(�)
H2(g) + I2 (g) ⇌ 2HI(g)
Δn(g) = np – nr= 2 – 2 = 0
Hence Kp = Kc
When the reactant and the product are in the different phase it is called as
Heterogeneous_Equilibrium
Eg:- CaCO3(s) ⇌ CaO(s) + CO2(g)
10. Define Law of Mass action
At a given temperature, the rate of a chemical reaction is directlyproportional to the product of
the active masses of the reactants.
Rate [Reactant]
x
11. List the application of Equilibrium constant
It is used to find the direction of the reaction taking place
It is used to find the extend of the reaction
It is used to find the equilibrium concentration of the reactant and the product.
12. Define Reaction Quotient ‘Q’
Reaction Quotient is defined as the ratio between the product of the active
masses of the products and the reactants raised to a Stoichiometriccoefficientunder non equilibrium
conditions
xA(g) + yB(g) mC(g) + nD(g)
Q =
[�]
�
[�]
�
[�]
�
[�]
�
13. Explain how the equilibrium constant Kc is used to find out the direction of the reaction
If Q = Kc, Equilibrium reaction
If Q > Kc, the reaction proceeds in the Reverse direction.
If Q < Kc, the reaction proceeds in the Forward direction
14. Explain the effect of concentration on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which
equilibrium shifts
Concentration
Addition of reactants (increase in reactant
concentration)
Forward reaction
Removal of products (decrease in product
concentration)
Addition of products (increase in product
concentration)
Reverse reaction
Removal of reactants (decrease in reactant
concentration)
15. Explain the effect of pressure on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Pressure
Increase of pressure (Decrease in
volume)
Reaction that favours fewer moles of
the gaseous molecules
Decrease of pressure (Increase in
volume)
Reaction that favours more moles of
the gaseous molecules
16. Explain the effect of Temperature on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Temperature
Increase (High T) Towards endothermic reaction
decrease (Low T) Towards endothermic reaction
17. Explain the effect of Catalyst on a Equilibrium reaction- using Le-Chatlier Braun
principle.
Condition Stress Direction in which equilibrium shifts
Catalyst Addition of catalyst No effect
18. Explain the Effect of inert gas on a equilibrium reaction
Addition of inert gas at constant volume - No effect
19. Derive the relation between Kp and Kc.
Let us consider a reaction
xA(g) + yB(g)⇌mC(g) + nD(g)
Kc =
[�]
�
[�]
�
[�]
�
[�]
�
------------- 1
Kp =
�
�
�
��
�
�
�
�
�
��
�
�------------- 2
From Ideal gas equation,
PV = nRT
P=
�
�
RT------------------- 3
Pi = Ci RT
C
i
v
= Ci
Substituting the partial pressure terms in eqn(3)
�
�
�
= [A]
x
(RT)
x
�
�
�
= [B]
y
(RT)
y
�
�
�
= [C]
m
(RT)
m
�
�
�
= [D]
n
(RT)
n
Substitute the above pressure terms in eqn( 2 )
Kp =
[C]
m
[D]
n
[A]
x
[B]
y
X
(RT)
m
×(RT)
n
(RT)
x
×(RT)
y
Kp = Kc RT
m+n − (x+y)
n(g) = (m+n) – (x+y)
Kp = Kc RT
n(g)
20. Give a balanced chemical equation for the equilibrium reaction for which the
equilibrium constant is given by expression Kc =
[�??????�]
�
[��]
�
[��]
�
[ ??????��]
�
4 NH3 (g) + 7O2 (g) ⇌ 4NO(g)+ 6H2O(g)
21. Define le-Chatlier Braun principle
If a system at equilibrium is subjected to a disturbance, then the systemwill move in the
direction to nullify the effect of the disturbance
22. For a given reaction the equilibrium constant has a constant value. If the value of Q also
constant ?. explain.
As the reaction proceeds, there is a change in the concentration of the reactant and the products.
The Q value also changes until the equilibrium is reached.
At equilibrium the Q value is equal to Kc.
Once the equilibrium is reached there is no change in the Q value.
23. What the relation between Kp and Kc.Give one example for Kp is equal to Kc
Kp = Kc ��
�(�)
H2(g) + I2 (g) ⇌ 2HI(g)
Δn(g) = np – nr= 2 – 2 = 0
Hence Kp = Kc
24. Explain how will you predict the direction of a equilibrium reaction
Value of KC Prediction
Kc < 10
-3
The reaction proceeds in the Reverse direction
10
-3
< Kc < 10
3
Neither forward nor reverse reaction predominates
Kc > 10
-3
The reaction proceeds in the Forward direction
Kc is large The reaction reaches equilibrium with high product concentration`
Kc is small The reaction reaches equilibrium with low product concentration
25. Derive the Van’t Hoff equation
Van‟t Hoff equation gives the effect of temperature of the equilibrium Constant.(K)
ΔG° = –RTln K ……………. (1)
We know that
ΔG° = ΔH° – TΔS° …………… (2)
Substituting (1) in equation (2)
–RTln K= ΔH° – TΔS°
Rearranging
ln K=
1
–RT
(ΔH° – TΔS°)
ln K=
−ΔH°
RT
+
−ΔS°
R
…………… (3)
Differentiating equation (3) with respect to temperature,
d(lnK)
dT
=
ΔH°
RT
2
…………… (4)
Rearranging
d(lnK)=
ΔH°
R
.
dT
T
2
Equation (4) is known as differential form of van‟t Hoff equation.
On integrating the equation 4, between T1and T2with their respective equilibrium constants K1and K2.
Equation (5) is known as integrated form of van‟t Hoff equation. 26. Derive the Kp and Kc for the following equilibrium reaction. H2 (g)+ I2 (g)⇌ 2HI(g)
H2 (g)+ I2 (g)⇌ 2HI(g)
H2 I2 HI
Initial number of moles a b 0
Number of moles reacted x x 0
Number of moles remaining (a-x) (b-x) 2x
Molar concentration (a−x)
??????
(b−x)
??????
2x
??????
Applying law of mass action
Kc =
[HI]
2
[ H
2] [ I
2]
----------(1)
Substitute the above Molar concentration terms in eqn(1)
=
2�
V
2
(a−x)
V
(b−x)
V
=
4x
2
a−x (b−x)
Kc =
��
�
�−� (�−�)
Kp = Kc RT
n(g)
n(g) = 2- (1+1) = 0
For this reaction n(g) = 0 Kp = Kc
So Kp = Kc =
��
�
�−� (�−�)
27. Derive the Kp and Kc for the following equilibrium reactionPCl5(g)⇌ PCl3(g) + Cl2(g)
PCl5 PCl3 Cl2
Initial number of moles a 0 0
Number of moles reacted x 0 0
Number of moles remaining (a-x) x x
Molar concentration (a−x)
??????
x
??????
x
??????
Value of KC Prediction
Kc < 10
-3
The reaction proceeds in the Reverse direction
10
-3
< Kc < 10
3
Neither forward nor reverse reaction predominates
Kc > 10
-3
The reaction proceeds in the Forward direction
Kc is large The reaction reaches equilibrium with high product concentration`
Kc is small The reaction reaches equilibrium with low product concentration
25. Derive the Van’t Hoff equation
Van‟t Hoff equation gives the effect of temperature of the equilibrium Constant.(K)
ΔG° = –RTln K ……………. (1)
We know that
ΔG° = ΔH° – TΔS° …………… (2)
Substituting (1) in equation (2)
–RTln K= ΔH° – TΔS°
Rearranging
ln K=
1
–RT
(ΔH° – TΔS°)
ln K=
−ΔH°
RT
+
−ΔS°
R
…………… (3)
Differentiating equation (3) with respect to temperature,
d(lnK)
dT
=
ΔH°
RT
2
…………… (4)
Rearranging
d(lnK)=
ΔH°
R
.
dT
T
2
Equation (4) is known as differential form of van‟t Hoff equation.
On integrating the equation 4, between T1and T2with their respective equilibrium constants K1and K2.
Equation (5) is known as integrated form of van‟t Hoff equation. 26. Derive the Kp and Kc for the following equilibrium reaction. H2 (g)+ I2 (g)⇌ 2HI(g)
H2 (g)+ I2 (g)⇌ 2HI(g)
H2 I2 HI
Initial number of moles a b 0
Number of moles reacted x x 0
Number of moles remaining (a-x) (b-x) 2x
Molar concentration (a−x)
??????
(b−x)
??????
2x
??????
Applying law of mass action
Kc =
[HI]
2
[ H
2] [ I
2]
----------(1)
Substitute the above Molar concentration terms in eqn(1)
=
2�
V
2
(a−x)
V
(b−x)
V
=
4x
2
a−x (b−x)
Kc =
��
�
�−� (�−�)
Kp = Kc RT
n(g)
n(g) = 2- (1+1) = 0
For this reaction n(g) = 0 Kp = Kc
So Kp = Kc =
��
�
�−� (�−�)
27. Derive the Kp and Kc for the following equilibrium reactionPCl5(g)⇌ PCl3(g) + Cl2(g)
PCl5 PCl3 Cl2
Initial number of moles a 0 0
Number of moles reacted x 0 0
Number of moles remaining (a-x) x x
Molar concentration (a−x)
??????
x
??????
x
??????
Applying law of mass action
Kc =
[ PCl
3] [ Cl
2]
[PCl5]
----------(1)
Substitute the above Molar concentration terms in eqn(1)
Kc =
x
V
×
x
V
(a−x)
V
Kc =
�
�
�(�−�)
Kp = Kc RT
n(g)
n(g) = (2-1) = 1
For this reaction n(g) = 1
Kp = Kc(RT)
PV = nRT
RT =
PV
n
Where n is the total number of moles at equilibrium.
n = (a-x) + x + x = (a+x)
RT =
�??????
(a+x)
Kp =
x
2
V(a−x)
(RT)
Kp =
x
2
V a−x
X
�??????
a+x
Kp =
�
�
�−�
X
�
�+�
28. Derive the Kp and Kc for the following equilibrium reactionN2(g) + 3H2(g)⇌2NH3(g)
N2 H2 NH3
Initial number of moles a b 0
Number of moles reacted x x 0
Number of moles remaining (a-x) (b-3x) 2x
Molar concentration (a−x)
??????
(b−3x)
??????
2x
??????
Applying law of mass action
Kc =
[ NH
3]
2
[ N
2][ H
2]
3
---------(1)
Substitute the above Molar concentration terms in eqn(1)
Kc =
(
2�
??????
)
2
(a−x)
??????
(
(b−3x)
??????
)
3
Kc =
��
�
�
�
�−� (�−��)
�
Kp = Kc ��
�(??????)
n(g) = (24) = 2
For this reaction n(g) =2
Kp = Kc(RT)
2
Kp =
Kc
(RT)
2
PV = nRT
RT =
PV
n
Where n is the total number of moles at equilibrium.
n = 2x + (a-x) + (b 3x) = (a+b2x)
RT =
�??????
(a+b−2x)
Kp =
4x
2
V
2
a−x (b−3x)
3
x (
�??????
(a+b−2x)
)
2
= 4x
2
V
2
a−x (b−3x)
3
x (
(a+b−2x)
�??????
)
2
Kp =
��
�
(�+�−��)
�
�−� (�−��)
�
�
�
9 . SOLUTIONS
1. Define Molality and its unit
Molality =
Number of moles of solute
Mass of the solvent (Kg)
unit : Mole / Kg
2. Define Normality and its unit
Equivalent Weight =
Number of gram equivalent of solute
volume of solution (L)
Kc =
[ PCl
3] [ Cl
2]
[PCl5]
----------(1)
Substitute the above Molar concentration terms in eqn(1)
Kc =
x
V
×
x
V
(a−x)
V
Kc =
�
�
�(�−�)
Kp = Kc RT
n(g)
n(g) = (2-1) = 1
For this reaction n(g) = 1
Kp = Kc(RT)
PV = nRT
RT =
PV
n
Where n is the total number of moles at equilibrium.
n = (a-x) + x + x = (a+x)
RT =
�??????
(a+x)
Kp =
x
2
V(a−x)
(RT)
Kp =
x
2
V a−x
X
�??????
a+x
Kp =
�
�
�−�
X
�
�+�
28. Derive the Kp and Kc for the following equilibrium reactionN2(g) + 3H2(g)⇌2NH3(g)
N2 H2 NH3
Initial number of moles a b 0
Number of moles reacted x x 0
Number of moles remaining (a-x) (b-3x) 2x
Molar concentration (a−x)
??????
(b−3x)
??????
2x
??????
Applying law of mass action
Kc =
[ NH
3]
2
[ N
2][ H
2]
3
---------(1)
Substitute the above Molar concentration terms in eqn(1)
Kc =
(
2�
??????
)
2
(a−x)
??????
(
(b−3x)
??????
)
3
Kc =
��
�
�
�
�−� (�−��)
�
Kp = Kc ��
�(??????)
n(g) = (24) = 2
For this reaction n(g) =2
Kp = Kc(RT)
2
Kp =
Kc
(RT)
2
PV = nRT
RT =
PV
n
Where n is the total number of moles at equilibrium.
n = 2x + (a-x) + (b 3x) = (a+b2x)
RT =
�??????
(a+b−2x)
Kp =
4x
2
V
2
a−x (b−3x)
3
x (
�??????
(a+b−2x)
)
2
= 4x
2
V
2
a−x (b−3x)
3
x (
(a+b−2x)
�??????
)
2
Kp =
��
�
(�+�−��)
�
�−� (�−��)
�
�
�
9 . SOLUTIONS
1. Define Molality and its unit
Molality =
Number of moles of solute
Mass of the solvent (Kg)
unit : Mole / Kg
2. Define Normality and its unit
Equivalent Weight =
Number of gram equivalent of solute
volume of solution (L)
Normality =
mass of solute
Equivalent weight of solute X volume of solution (L)
unit : N
3. Define Molarity and its unit
Molarity =
Number of moles of solute
volume of solution (L)
unit : Mole / Lit
4. Define Formality and its unit
Formality =
Number of Formula weight of solute
volume of solution (L)
unit :F
5. Define Mass percentage
Mass percentage (%W/W) =
Mass of the Solute (in g)
Mass of Solution (in g)
x 100
6. Define Volume percentage
Volume percentage (%V/V) =
Volume of the Solute (in mL)
Volume of Solution (in mL)
x 100
7. Define Mass / Volume percentage
Mass / Volume percentage (%W/V) =
Mass of the Solute (in g)
Volume of Solution (in mL)
x 100
8. Define Mole fraction of Solvent and its unit
Mole fraction of solvent =
Number of moles of Solvent
Total number of moles of solute and solvent
Unit : No unit
9. Define Mole fraction of Solute and its unit
Mole fraction of solute =
Number of moles of Solute
Total number of moles of solute and solvent
Unit : No unit
10. Define Parts per Million
Parts per Million (ppm) =
Number of parts of the Component
Total number of all components
x 10
6
=
Mass of the Solute (in g)
Mass of Solution (in g)
x 10
6
ppm
11. Write a notes on Standard solution
A standard solution or a stock solution is a solution whose Normality is accurately known
12. What are the Advantages of using standard solutions?
The error due to weighing the solute can be minimised by using concentrated stock solution
that requires large quantity of solute.
We can prepare working standards of different concentrations by diluting the stock
solution, which is more efficient since consistency is maintained.
Some of the concentrated solutions are more stable and are less likely to support microbial
growth than working standards used in the experiments.
13. Define vapour pressure of liquid
The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid
at the given temperature.
14. What is relative lowering of vapour pressure?
Relative lowering of vapour pressure is the ratio of the lowering of vapour pressure of the
solution to the vapour pressure of the pure solvent. Relative lowering of vapour pressure =
vapour pressure of the pure solvent−vapour pressure of the solution
vapour pressure of the pure solvent
15. Define Henry's law
“The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly
proportional to the mole fraction(x) of the gaseous solute in the solution at low concentrations”.
Psolute α Xsolute in solution
Psolute = KHXsolute in solution
16. Mention the limitations of Henry’s law
Henry‟s law is applicable at moderate temperature and pressure only.
Only the less soluble gases obeys Henry‟s law
The gases reacting with the solvent do not obey Henry‟s law.
17 Why ammonia and HCl does not obey Henry’s law.
The gases obeying Henry‟s law should not associate or dissociate while dissolving in the solvent.
18. Explain the Vapour pressure of Binary solution of Liquid in Liquid.According to
Raoult’s law
Let us consider the liquid solution containing toluene (solute) in benzene (solvent).
Vapour pressure of Solvent = Vapour pressure of pure solvent x Mole fraction of solvent
P(Solvent) = P
0
(Pure Solvent) X(Solvent)
Vapour pressure of Solute = Vapour pressure of pure Solute x Mole fraction of Solute
P(Solute) = P
0
(Pure Solute) X(Solute)
Total pressure of solution = Vapour pressure of Solvent + Vapour pressure of Solute
P(solution)= P
0
(Pure Solvent) X(Solvent) + P
0
(Pure Solute) X(Solute)
19. Derive an expression for the lowering of vapour pressure when a nonvolatile solute is
dissolved in a solvent
When a nonvolatile solute is dissolved in a pure solvent, the vapour pressure of the pure
solvent will decrease. In such solutions, the vapour pressure of the solution will depend only on
the solvent molecules as the solute is nonvolatile.
The vapour pressure of the solution is determined by the number of molecules of the
solvent present in the surface at any time and is proportional to the mole fraction of the solvent.
P(Solution) X(Solvent)
P(Solution)= K X(Solvent)
K = P
0
(Solvent) (X(Solvent) = 1)
P(Solution) = P
0
(Solvent) X(Solvent)
P Solution
P
Solvent
0 = X(Solvent)
1 -
P
Solution
P
Solvent
0 = (1 – X(Solvent))
P
Solvent
0
−P
Solution
P
Solvent
0 = X(Solute)
Relative lowering of vapour pressure = X(Solute)
The relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute
is equal to the mole fraction of the solute at a given temperature.
20. What is Ideal Solutions and give one example ?
Ideal Solutions:The solute as well as the solvent obeys the Raoult’s law over the entire range
of concentration. Eg: Benzene &Toluene
mass of solute
Equivalent weight of solute X volume of solution (L)
unit : N
3. Define Molarity and its unit
Molarity =
Number of moles of solute
volume of solution (L)
unit : Mole / Lit
4. Define Formality and its unit
Formality =
Number of Formula weight of solute
volume of solution (L)
unit :F
5. Define Mass percentage
Mass percentage (%W/W) =
Mass of the Solute (in g)
Mass of Solution (in g)
x 100
6. Define Volume percentage
Volume percentage (%V/V) =
Volume of the Solute (in mL)
Volume of Solution (in mL)
x 100
7. Define Mass / Volume percentage
Mass / Volume percentage (%W/V) =
Mass of the Solute (in g)
Volume of Solution (in mL)
x 100
8. Define Mole fraction of Solvent and its unit
Mole fraction of solvent =
Number of moles of Solvent
Total number of moles of solute and solvent
Unit : No unit
9. Define Mole fraction of Solute and its unit
Mole fraction of solute =
Number of moles of Solute
Total number of moles of solute and solvent
Unit : No unit
10. Define Parts per Million
Parts per Million (ppm) =
Number of parts of the Component
Total number of all components
x 10
6
=
Mass of the Solute (in g)
Mass of Solution (in g)
x 10
6
ppm
11. Write a notes on Standard solution
A standard solution or a stock solution is a solution whose Normality is accurately known
12. What are the Advantages of using standard solutions?
The error due to weighing the solute can be minimised by using concentrated stock solution
that requires large quantity of solute.
We can prepare working standards of different concentrations by diluting the stock
solution, which is more efficient since consistency is maintained.
Some of the concentrated solutions are more stable and are less likely to support microbial
growth than working standards used in the experiments.
13. Define vapour pressure of liquid
The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid
at the given temperature.
14. What is relative lowering of vapour pressure?
Relative lowering of vapour pressure is the ratio of the lowering of vapour pressure of the
solution to the vapour pressure of the pure solvent. Relative lowering of vapour pressure =
vapour pressure of the pure solvent−vapour pressure of the solution
vapour pressure of the pure solvent
15. Define Henry's law
“The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly
proportional to the mole fraction(x) of the gaseous solute in the solution at low concentrations”.
Psolute α Xsolute in solution
Psolute = KHXsolute in solution
16. Mention the limitations of Henry’s law
Henry‟s law is applicable at moderate temperature and pressure only.
Only the less soluble gases obeys Henry‟s law
The gases reacting with the solvent do not obey Henry‟s law.
17 Why ammonia and HCl does not obey Henry’s law.
The gases obeying Henry‟s law should not associate or dissociate while dissolving in the solvent.
18. Explain the Vapour pressure of Binary solution of Liquid in Liquid.According to
Raoult’s law
Let us consider the liquid solution containing toluene (solute) in benzene (solvent).
Vapour pressure of Solvent = Vapour pressure of pure solvent x Mole fraction of solvent
P(Solvent) = P
0
(Pure Solvent) X(Solvent)
Vapour pressure of Solute = Vapour pressure of pure Solute x Mole fraction of Solute
P(Solute) = P
0
(Pure Solute) X(Solute)
Total pressure of solution = Vapour pressure of Solvent + Vapour pressure of Solute
P(solution)= P
0
(Pure Solvent) X(Solvent) + P
0
(Pure Solute) X(Solute)
19. Derive an expression for the lowering of vapour pressure when a nonvolatile solute is
dissolved in a solvent
When a nonvolatile solute is dissolved in a pure solvent, the vapour pressure of the pure
solvent will decrease. In such solutions, the vapour pressure of the solution will depend only on
the solvent molecules as the solute is nonvolatile.
The vapour pressure of the solution is determined by the number of molecules of the
solvent present in the surface at any time and is proportional to the mole fraction of the solvent.
P(Solution) X(Solvent)
P(Solution)= K X(Solvent)
K = P
0
(Solvent) (X(Solvent) = 1)
P(Solution) = P
0
(Solvent) X(Solvent)
P Solution
P
Solvent
0 = X(Solvent)
1 -
P
Solution
P
Solvent
0 = (1 – X(Solvent))
P
Solvent
0
−P
Solution
P
Solvent
0 = X(Solute)
Relative lowering of vapour pressure = X(Solute)
The relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute
is equal to the mole fraction of the solute at a given temperature.
20. What is Ideal Solutions and give one example ?
Ideal Solutions:The solute as well as the solvent obeys the Raoult’s law over the entire range
of concentration. Eg: Benzene &Toluene
21. What is Non-ideal solutionsand give one example ?
Non-ideal solutions:The solutions which do not obey Raoult’s law over the entire range of
concentration,Eg:Ethyl alcohol and Water
22. What are the conditions for ideal solutions?
There is no change in the volume on mixing the two components (solute & solvents).
(ΔVmixing= 0)
There is no exchange of heat when the solute is dissolved in solvent (ΔHmixing= 0).
Escaping tendency of the solute and the solvent present in it should be same as in pure liquids.
23. What are the factors responsible for the deviation from Raoult’s law ?
Solute-Solvent interactions:- If the interaction between the Solute-Solute and the Solvent-
Solventmolecules is dissimilar then there will be deviation from Raoult‟s law
Dissociation of the solute:- When the solute is dissociated, it gives ions. If the Ions will interact with
thesolvent, then there will be deviation from Raoult‟s law
Association of the solute:- When the solute molecules associates, then there will be deviation from
Raoult‟s law
Pressure:-At high pressure, the intermolecular force of attraction increases and therewill be deviation
from Raoult‟s law.
Temperature:- At high temperature, the intermolecular force of attraction decreases andthere will be
deviation from Raoult‟s law.
Concentration:-At low concentration the Solvent-Solute interaction decreases.At high concentration
the Solvent-Solute interaction increases.
24. Definecryoscopic constant (or) molal freezing point depression constant.
It is defined as the elevation of boiling point of one molal solution
25. Define Molal elevation constant (or) Ebullioscopic constant
The Depression in the Freezing point of one molal solution is called asCryoscopic constant.
26. What is Osmosis ?
Which is a spontaneous process by which the solvent molecules pass through a semi
permeable membrane from a solution of lower concentration to a solution of higher concentration.
27 Explain Reverse osmosis
It can be defined as a process in which a solvent passes through a semi permeable
membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater
than the osmotic pressure.
Use: Purification of water
28. What are the Significances of osmotic pressure over other colligative properties
The osmotic pressure can be measured at room temperature enables to determine the
molecular mass of biomolecules which are unstable at higher temperatures.
29. Explain the term Isotonic solutions
Two solutions having same osmotic pressure at a given temperature are called isotonic
solutions.
30. State van’t Hoff factor
It is defined as the ratio of the actual molar mass to the abnormal (calculated) molar mass
of the solute.
van‟t Hoff factor(i) =
Normal ( actual) molar mass
Observed (abnormal) molar mass
31. Explain the effect of pressure on the solubility.
According to Le-Chatelier principle, the increase in pressure will shift the equilibrium in
the direction which will reduce the pressure. Therefore, more number of gaseous molecules
dissolves in the solvent and the solubility increases. 32. Write equation and explain the determination of Molar mass of solute from relative
lowering of vapour pressure
MB–Molecular weight of solute, �
??????
�
– Vapour pressure of pure solvent,
WB - Weight of solute MA – Molecular weight of pure solvent,
p – Relative lowering of vapour pressure, WA – Weight of pure solvent.
33. Write equation and explain the determination of molar mass of solute from elevation of
boiling point
M2- Molecular weight of solute, Kb- Ebullioscopicconstant,
W2 - Weight of solute, M1 - Molecular weight of pure solvent,
Tb – Elevation of Bioling point, W1 - Weight of pure solvent.
34. Write equation and explain the determination of molar mass of solute from depression
in freezing point
M2- Molecular weight of solute, Kf- Cryoscopic constant,
W2 - Weight of solute, M1 - Molecular weight of pure solvent,
Tf – Depression in freezing point, W1 - Weight of pure solvent.
35. Write equation and explain the determination of molar mass of solute from osmotic
pressure
M2- Molecular weight of solute, W2 - Weight of solute,
V – Volume of solution, – Osmotic pressure
R – Gas constantT – Room Temperature.
36. Define Gaseous solution with an example
State of solution Solute Solvent Examples
Gaseous solution
Gas Gas Air (A mixture of nitrogen,
oxygen and other gases)
Liquid Gas Humid oxygen (Oxygen
containing water)
Solid Gas Camphor in nitrogen gas
37. Define Liquid solution with an example
State of solution Solute Solvent Examples
Liquid solution
Gas Liquid CO2 dissolved in water
(carbonated water)
Liquid Liquid Ethanol dissolved in water
Solid Liquid Salt water
Non-ideal solutions:The solutions which do not obey Raoult’s law over the entire range of
concentration,Eg:Ethyl alcohol and Water
22. What are the conditions for ideal solutions?
There is no change in the volume on mixing the two components (solute & solvents).
(ΔVmixing= 0)
There is no exchange of heat when the solute is dissolved in solvent (ΔHmixing= 0).
Escaping tendency of the solute and the solvent present in it should be same as in pure liquids.
23. What are the factors responsible for the deviation from Raoult’s law ?
Solute-Solvent interactions:- If the interaction between the Solute-Solute and the Solvent-
Solventmolecules is dissimilar then there will be deviation from Raoult‟s law
Dissociation of the solute:- When the solute is dissociated, it gives ions. If the Ions will interact with
thesolvent, then there will be deviation from Raoult‟s law
Association of the solute:- When the solute molecules associates, then there will be deviation from
Raoult‟s law
Pressure:-At high pressure, the intermolecular force of attraction increases and therewill be deviation
from Raoult‟s law.
Temperature:- At high temperature, the intermolecular force of attraction decreases andthere will be
deviation from Raoult‟s law.
Concentration:-At low concentration the Solvent-Solute interaction decreases.At high concentration
the Solvent-Solute interaction increases.
24. Definecryoscopic constant (or) molal freezing point depression constant.
It is defined as the elevation of boiling point of one molal solution
25. Define Molal elevation constant (or) Ebullioscopic constant
The Depression in the Freezing point of one molal solution is called asCryoscopic constant.
26. What is Osmosis ?
Which is a spontaneous process by which the solvent molecules pass through a semi
permeable membrane from a solution of lower concentration to a solution of higher concentration.
27 Explain Reverse osmosis
It can be defined as a process in which a solvent passes through a semi permeable
membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater
than the osmotic pressure.
Use: Purification of water
28. What are the Significances of osmotic pressure over other colligative properties
The osmotic pressure can be measured at room temperature enables to determine the
molecular mass of biomolecules which are unstable at higher temperatures.
29. Explain the term Isotonic solutions
Two solutions having same osmotic pressure at a given temperature are called isotonic
solutions.
30. State van’t Hoff factor
It is defined as the ratio of the actual molar mass to the abnormal (calculated) molar mass
of the solute.
van‟t Hoff factor(i) =
Normal ( actual) molar mass
Observed (abnormal) molar mass
31. Explain the effect of pressure on the solubility.
According to Le-Chatelier principle, the increase in pressure will shift the equilibrium in
the direction which will reduce the pressure. Therefore, more number of gaseous molecules
dissolves in the solvent and the solubility increases. 32. Write equation and explain the determination of Molar mass of solute from relative
lowering of vapour pressure
MB–Molecular weight of solute, �
??????
�
– Vapour pressure of pure solvent,
WB - Weight of solute MA – Molecular weight of pure solvent,
p – Relative lowering of vapour pressure, WA – Weight of pure solvent.
33. Write equation and explain the determination of molar mass of solute from elevation of
boiling point
M2- Molecular weight of solute, Kb- Ebullioscopicconstant,
W2 - Weight of solute, M1 - Molecular weight of pure solvent,
Tb – Elevation of Bioling point, W1 - Weight of pure solvent.
34. Write equation and explain the determination of molar mass of solute from depression
in freezing point
M2- Molecular weight of solute, Kf- Cryoscopic constant,
W2 - Weight of solute, M1 - Molecular weight of pure solvent,
Tf – Depression in freezing point, W1 - Weight of pure solvent.
35. Write equation and explain the determination of molar mass of solute from osmotic
pressure
M2- Molecular weight of solute, W2 - Weight of solute,
V – Volume of solution, – Osmotic pressure
R – Gas constantT – Room Temperature.
36. Define Gaseous solution with an example
State of solution Solute Solvent Examples
Gaseous solution
Gas Gas Air (A mixture of nitrogen,
oxygen and other gases)
Liquid Gas Humid oxygen (Oxygen
containing water)
Solid Gas Camphor in nitrogen gas
37. Define Liquid solution with an example
State of solution Solute Solvent Examples
Liquid solution
Gas Liquid CO2 dissolved in water
(carbonated water)
Liquid Liquid Ethanol dissolved in water
Solid Liquid Salt water
38. Define Solid solution with an example
State of solution Solute Solvent Examples
Solid solution
Gas Solid Solution of H2 in palladium
Liquid Solid Amalgam of potassium (used for
dental filling)
Solid Solid Gold alloy (of copper used in
making Jewelery)
39. Explain the term osmotic pressure
“The pressure that must be applied to the solution to stop the influx of the solvent (to stop
osmosis) through the semipermeable membrane”
40. State and explain Raoult’s Law.
This law states that “in the case of a solution of volatile liquids, the partial vapour pressure
of each component (A & B) of the solution is directly proportional to its mole fraction”.
PAα XA
PA= k XA
whenXA= 1, k = P
0
A
PA= P
0
AXA
Similarly, for component „B‟
PB= P
0
BXB
41.
Write a notes on Colligative properties
Depends only on the number of solute particles (ions/molecules) present in the
solution.The properties, namely, relative lowering of vapour pressure, elevation of boiling point,
depression in freezing point and osmotic pressure.
10.CHEMICAL BONDING
1. Define Bond Order
Bond order is the number of Covalent Bonds between two atoms.
Bond order =
N
b− N
a
2
Nb =Total number of Bonding Electrons ; Na = Total number of Anti Bonding Electrons
2. Define Hybridization
Hybridisation is the process of mixing of atomic orbitals of the same atom with
comparable energy to form equal number of new equivalent orbitals with same energy.
3. Explain the formation of () sigma bond
When two atomic orbitals overlap along the axis linearly it forms Sigma bond.
4. Explain the formation of bond
When two atomic orbitals overlap only side ways it forms pi bond
The unhybridised 2Pz orbital of both carbon atoms can overlap only sideways as they are not in the
molecular axis.
5. Which bond is stronger σ or π? Why?
σ -Bond the bonding occurs by the linear overlap of the orbitals, where as in the
π- bond parallel overlapping occurs.
Electron density is maximum between the nuclei of bonded atoms in
σ -bond as compound to the π –bond
Sigma (σ ) bond is stronger bond
6. Define bond energy.
The minimum amount of energy required to break one mole of a particular bond in
molecules in their gaseous state.The unit of bond energy is KJ.mol
-1
7. What is Polar Covalent bond? explain with example.
Covalent bond formed between atoms having different electronegativities, The atom with higher electronegativity will have greater tendency to attract the shared
pair of electrons more towards itself than the other atom.
As a result the cloud of shared electron pair gets distorted.
Example: H
+
– F
–
The electronegativities of hydrogen and fluorine on Pauling's scale are 2.1and 4 respectively.
8. What is dipolment?
The polarity of a covalent bond can be measured in terms of dipole moment
which is defined as μ = q × 2d
Where „μ’ is the dipole moment,
„q’ is the charge
„d’ is the distance between the two charges.
The dipole moment is a vector and the direction of the dipole moment vector points from the
negative charge to positive charge.
9. Linear form of carbondioxide molecule has two polar bonds. yet the molecule has Zero
dipolement why?
The linear form of carbon dioxide has zero dipole moment,
Even though it has two polar bonds. In CO2,
The dipole moments of two polar bonds (CO) are equal in magnitude but have opposite
direction.
Hence, the net dipole moment of the CO2is, μ = μ1 + μ2 = μ1 + (-μ1) = 0
10. Hydrogen gas is diatomic where as Inert gases are monoatomic – explain on the basis of
MO theory.
Bond order of Hydrogen molecule is ONE,
Where as Bond order of Innert gases are ZERO. So it does not form molecule.
11. Define Bond length and how to calculate it?
The distance between the two nucleus of the two covalently bonded atoms is called as Bond
length
It can be determined by spectroscopic, x-ray diffraction and electron-diffraction
techniques.
12. Define Bond angle
This directional nature creates a fixed angle between two covalent bonds in a molecule and this
angle is termed as bond angle.
It can be determined by spectroscopic methods.
13. Define Hybridisation.
Hybridisation is the process of mixing of atomic orbitals of the same atom with comparable
energy to form equal number of new equivalent orbitals with same energy.
Hybridisation of Methane (CH4) is sp
3
14. What is resonance?
When a compound is represented by more than one structure, whichdiffers only in the
position of bonding and lone pair of electrons arecalled as Resonance .
Eg :-
State of solution Solute Solvent Examples
Solid solution
Gas Solid Solution of H2 in palladium
Liquid Solid Amalgam of potassium (used for
dental filling)
Solid Solid Gold alloy (of copper used in
making Jewelery)
39. Explain the term osmotic pressure
“The pressure that must be applied to the solution to stop the influx of the solvent (to stop
osmosis) through the semipermeable membrane”
40. State and explain Raoult’s Law.
This law states that “in the case of a solution of volatile liquids, the partial vapour pressure
of each component (A & B) of the solution is directly proportional to its mole fraction”.
PAα XA
PA= k XA
whenXA= 1, k = P
0
A
PA= P
0
AXA
Similarly, for component „B‟
PB= P
0
BXB
41.
Write a notes on Colligative properties
Depends only on the number of solute particles (ions/molecules) present in the
solution.The properties, namely, relative lowering of vapour pressure, elevation of boiling point,
depression in freezing point and osmotic pressure.
10.CHEMICAL BONDING
1. Define Bond Order
Bond order is the number of Covalent Bonds between two atoms.
Bond order =
N
b− N
a
2
Nb =Total number of Bonding Electrons ; Na = Total number of Anti Bonding Electrons
2. Define Hybridization
Hybridisation is the process of mixing of atomic orbitals of the same atom with
comparable energy to form equal number of new equivalent orbitals with same energy.
3. Explain the formation of () sigma bond
When two atomic orbitals overlap along the axis linearly it forms Sigma bond.
4. Explain the formation of bond
When two atomic orbitals overlap only side ways it forms pi bond
The unhybridised 2Pz orbital of both carbon atoms can overlap only sideways as they are not in the
molecular axis.
5. Which bond is stronger σ or π? Why?
σ -Bond the bonding occurs by the linear overlap of the orbitals, where as in the
π- bond parallel overlapping occurs.
Electron density is maximum between the nuclei of bonded atoms in
σ -bond as compound to the π –bond
Sigma (σ ) bond is stronger bond
6. Define bond energy.
The minimum amount of energy required to break one mole of a particular bond in
molecules in their gaseous state.The unit of bond energy is KJ.mol
-1
7. What is Polar Covalent bond? explain with example.
Covalent bond formed between atoms having different electronegativities, The atom with higher electronegativity will have greater tendency to attract the shared
pair of electrons more towards itself than the other atom.
As a result the cloud of shared electron pair gets distorted.
Example: H
+
– F
–
The electronegativities of hydrogen and fluorine on Pauling's scale are 2.1and 4 respectively.
8. What is dipolment?
The polarity of a covalent bond can be measured in terms of dipole moment
which is defined as μ = q × 2d
Where „μ’ is the dipole moment,
„q’ is the charge
„d’ is the distance between the two charges.
The dipole moment is a vector and the direction of the dipole moment vector points from the
negative charge to positive charge.
9. Linear form of carbondioxide molecule has two polar bonds. yet the molecule has Zero
dipolement why?
The linear form of carbon dioxide has zero dipole moment,
Even though it has two polar bonds. In CO2,
The dipole moments of two polar bonds (CO) are equal in magnitude but have opposite
direction.
Hence, the net dipole moment of the CO2is, μ = μ1 + μ2 = μ1 + (-μ1) = 0
10. Hydrogen gas is diatomic where as Inert gases are monoatomic – explain on the basis of
MO theory.
Bond order of Hydrogen molecule is ONE,
Where as Bond order of Innert gases are ZERO. So it does not form molecule.
11. Define Bond length and how to calculate it?
The distance between the two nucleus of the two covalently bonded atoms is called as Bond
length
It can be determined by spectroscopic, x-ray diffraction and electron-diffraction
techniques.
12. Define Bond angle
This directional nature creates a fixed angle between two covalent bonds in a molecule and this
angle is termed as bond angle.
It can be determined by spectroscopic methods.
13. Define Hybridisation.
Hybridisation is the process of mixing of atomic orbitals of the same atom with comparable
energy to form equal number of new equivalent orbitals with same energy.
Hybridisation of Methane (CH4) is sp
3
14. What is resonance?
When a compound is represented by more than one structure, whichdiffers only in the
position of bonding and lone pair of electrons arecalled as Resonance .
Eg :-
15. Describe Fajan’s rule
Fajan found that the ionic compounds show the covalent characters under the following
conditions.
Both ions (cation and anion) have higher charges.
Smaller cation and larger anion.
Cations having ns
2
np
6
nd
10
configuration show greater covalent character than cations having
ns
2
np
6
configuration.
16. Calculate the formal charge on carbon and oxygen for the following structure
Formal charge on Carbon = Nv - N
l+
N
b
2
= 4 – 0 +
8
2
= 0
Formal charge on Oxygen = Nv - N
l+
N
b
2
= 6 – 4 +
4
2
= 0
17. Define ionic bond. Give example.
The bond formed by transfer of electron(s) from one atom to another atom is called ionic
bond. Example: In NaCl, one electron is transferred from sodium atom to chlorine atom.
18. Define covalent bond. Give example.
The bond formed between two atoms by mutual sharing of electrons is called covalent
bond. Example: CH4.
19. Discuss the formation of Hydrogen molecule using MO Theory
Electronic configuration of Hydrogen :- 1s
1
Atomic orbitalAtomic orbital
of Hydrogen of Hydrogen
20. Discuss the formation of N2 molecule using MO Theory
Electronic configuration of Nitrogen:-
1s
2
2s
2
2px
1
2py
1
2pz
1
21. Draw the M.O diagram for oxygen molecule calculate its bond order and show that O2 is
paramagnetic.
Electronic configuration of Oxygen:-
1s
2
2s
2
2px
2
2py
1
2pz
1
22. Draw MO diagram of CO and calculate its bond order
23. In CH4, NH3 and H2O, the central atom undergoes sp
3
hybridisation - yet their bond
angles are different. why?
CH4
Methane
Molecule has no lone pair electron
Bond angle : 109.5
0
Shape :
Tetrahedral
NH3
Ammonia
Molecule has one lone pair electron .
So repulsive interaction takes place. Bond angle is reduced
to 107
0
Shape :
Pyramidal
H2O
Water
Molecule has Two lone pair electrons .
So repulsive interaction takes place more. Bond angle is reduced
to 104.5
0
Shape :
V Shape
24. What type of hybridisations are possible in the following geometeries?
a) octahedral b) tetrahedral c) square planer.?
a) octahedral d
2
sp
3
(or) sp
3
d
2
b) tetrahedral sp
3
c) square planer dsp
2
Fajan found that the ionic compounds show the covalent characters under the following
conditions.
Both ions (cation and anion) have higher charges.
Smaller cation and larger anion.
Cations having ns
2
np
6
nd
10
configuration show greater covalent character than cations having
ns
2
np
6
configuration.
16. Calculate the formal charge on carbon and oxygen for the following structure
Formal charge on Carbon = Nv - N
l+
N
b
2
= 4 – 0 +
8
2
= 0
Formal charge on Oxygen = Nv - N
l+
N
b
2
= 6 – 4 +
4
2
= 0
17. Define ionic bond. Give example.
The bond formed by transfer of electron(s) from one atom to another atom is called ionic
bond. Example: In NaCl, one electron is transferred from sodium atom to chlorine atom.
18. Define covalent bond. Give example.
The bond formed between two atoms by mutual sharing of electrons is called covalent
bond. Example: CH4.
19. Discuss the formation of Hydrogen molecule using MO Theory
Electronic configuration of Hydrogen :- 1s
1
Atomic orbitalAtomic orbital
of Hydrogen of Hydrogen
20. Discuss the formation of N2 molecule using MO Theory
Electronic configuration of Nitrogen:-
1s
2
2s
2
2px
1
2py
1
2pz
1
21. Draw the M.O diagram for oxygen molecule calculate its bond order and show that O2 is
paramagnetic.
Electronic configuration of Oxygen:-
1s
2
2s
2
2px
2
2py
1
2pz
1
22. Draw MO diagram of CO and calculate its bond order
23. In CH4, NH3 and H2O, the central atom undergoes sp
3
hybridisation - yet their bond
angles are different. why?
CH4
Methane
Molecule has no lone pair electron
Bond angle : 109.5
0
Shape :
Tetrahedral
NH3
Ammonia
Molecule has one lone pair electron .
So repulsive interaction takes place. Bond angle is reduced
to 107
0
Shape :
Pyramidal
H2O
Water
Molecule has Two lone pair electrons .
So repulsive interaction takes place more. Bond angle is reduced
to 104.5
0
Shape :
V Shape
24. What type of hybridisations are possible in the following geometeries?
a) octahedral b) tetrahedral c) square planer.?
a) octahedral d
2
sp
3
(or) sp
3
d
2
b) tetrahedral sp
3
c) square planer dsp
2
25. Considering x- axis as molecular axis, which out of the following will form a sigma bond.
i) 1s and 2py ii) 2Px and 2Px iii) 2px and 2pz iv) 1s and 2pz
Orbital Bond Reason
i) 1s and 2py - Bond overlap along the axis
ii) 2px and 2px - Bond overlap side ways
iii) 2px and 2pz - Bond overlap side ways
iv) 1s and 2pz - Bond overlap along the axis
26. Draw the Lewis structures for the following species.
i)
NO3
-
iii)
HNO3
ii)
SO4
2-
iv)
O3
27. CO2 and H2O both are triatomic molecule but their dipole moment values are different.
Why?
Dipole moment of the CO2 is, μ = μ1 + μ2 = μ1 + (-μ1) = 0
Incase of water net dipole moment is the vector sum of μ1+ μ2 as shown
28. Which one of the following has highest bond order? N2, N2
+
or N2
–
SL.No Molecule Bond order
1 N2 3
2 N2
+
2.5
3 N2
–
2.5
N2 Molecule has highest bond order 29. Explain Sp
2
hybridisation in BF3
Consider boron trifluoride molecule. The valence shell electronic configuration of boron atom
is [He]
2
2s
2
2P
1
1s 2s 2p
Electronic configuration of boron in Ground state ↿⇂ ↿⇂ ↿
1s 2s 2p
Electronic configuration of boron in Excited state ↿⇂ ↿ ↿ ↿
sp
2
Hybridisation
30. Explain the bond formation in BeCl2 and MgCl2.
1s 2s 2p
Electronic configuration of Berliumin Ground state ↿⇂ ↿⇂
1s 2s 2p
Electronic configuration of Berliumin Excited state ↿⇂ ↿ ↿
sp Hybridisation
BeCl2::- Each of the sp hybridized orbitals linearly overlap with pz orbital of the chlorine to form
a covalent bond between Be and Cl as shown in the Figure.
MgCl2:In MgCl2, Mg will lose its electrons and Chlorine will accept those electrons to form MgCl2.
Hence ionic boding is present in MgCl2 .
Mg Mg
2+
+ 2e
-
(2s
2
2p
6
3s
2
) (2s
2
2p
6
)
2Cl + 2e
-
2Cl
-
(2s
2
2p
6
3s
2
3p
5
) (2s
2
2p
6
3s
2
3p
6
)
Mg
2+
+ 2Cl
-
MgCl2
31. Explain resonance with reference to carbonate ion?
When we write Lewis structures for a molecule, more than one valid Lewis structures are
possible in certain cases.
For example resonance structure of carbonate ion [CO3]
2-
.
The skeletal structure of carbonate ion (Th e oxygen atoms are denoted as OA, OB& OC
i) 1s and 2py ii) 2Px and 2Px iii) 2px and 2pz iv) 1s and 2pz
Orbital Bond Reason
i) 1s and 2py - Bond overlap along the axis
ii) 2px and 2px - Bond overlap side ways
iii) 2px and 2pz - Bond overlap side ways
iv) 1s and 2pz - Bond overlap along the axis
26. Draw the Lewis structures for the following species.
i)
NO3
-
iii)
HNO3
ii)
SO4
2-
iv)
O3
27. CO2 and H2O both are triatomic molecule but their dipole moment values are different.
Why?
Dipole moment of the CO2 is, μ = μ1 + μ2 = μ1 + (-μ1) = 0
Incase of water net dipole moment is the vector sum of μ1+ μ2 as shown
28. Which one of the following has highest bond order? N2, N2
+
or N2
–
SL.No Molecule Bond order
1 N2 3
2 N2
+
2.5
3 N2
–
2.5
N2 Molecule has highest bond order 29. Explain Sp
2
hybridisation in BF3
Consider boron trifluoride molecule. The valence shell electronic configuration of boron atom
is [He]
2
2s
2
2P
1
1s 2s 2p
Electronic configuration of boron in Ground state ↿⇂ ↿⇂ ↿
1s 2s 2p
Electronic configuration of boron in Excited state ↿⇂ ↿ ↿ ↿
sp
2
Hybridisation
30. Explain the bond formation in BeCl2 and MgCl2.
1s 2s 2p
Electronic configuration of Berliumin Ground state ↿⇂ ↿⇂
1s 2s 2p
Electronic configuration of Berliumin Excited state ↿⇂ ↿ ↿
sp Hybridisation
BeCl2::- Each of the sp hybridized orbitals linearly overlap with pz orbital of the chlorine to form
a covalent bond between Be and Cl as shown in the Figure.
MgCl2:In MgCl2, Mg will lose its electrons and Chlorine will accept those electrons to form MgCl2.
Hence ionic boding is present in MgCl2 .
Mg Mg
2+
+ 2e
-
(2s
2
2p
6
3s
2
) (2s
2
2p
6
)
2Cl + 2e
-
2Cl
-
(2s
2
2p
6
3s
2
3p
5
) (2s
2
2p
6
3s
2
3p
6
)
Mg
2+
+ 2Cl
-
MgCl2
31. Explain resonance with reference to carbonate ion?
When we write Lewis structures for a molecule, more than one valid Lewis structures are
possible in certain cases.
For example resonance structure of carbonate ion [CO3]
2-
.
The skeletal structure of carbonate ion (Th e oxygen atoms are denoted as OA, OB& OC
Total number of valence electrons =
[1 x 4(carbon)] + [3 x 6 (oxygen)] + [2 (charge)] = 24 electrons.
Distribution of these valence electrons gives us the following structure.
Complete the octet for carbon by moving a lone pair from one of the oxygens (OA) and write the
charge of the ion (2-) on the upper right side as shown in the figure.
In this case, we can draw two additional Lewis structures by moving the lone pairs from the
other two oxygens (OB and OC) thus creating three similar structures as shown below in which
the relative position of the atoms are same.
They only diff er in the position of bonding and lone pair of electrons.
Such structures are called resonance structures (canonical structures) and this phenomenon is
called resonance.
32. Explain the bond formation in ethylene and acetylene.
Bonding in ethylene:
The molecular formula of ethylene is C2H4. The valency of carbon is 4. Th e electronic
configuration of valence shell of carbon in ground state is 1s
2
2s
2
2px
1
2py
1
2pz
0
.
1s 2s 2p
Electronic configuration of Carbonin Ground state ↿⇂ ↿⇂ ↿ ↿
1s 2s 2p
Electronic configuration of Carbonin Excited state ↿⇂ ↿ ↿ ↿ ↿
sp
2
Hybridisation
Hybridised state ↿ ↿ ↿ ↿
sp
2
sp
2
sp
2
2pz
Formation of sigma bond: One of the sp
2
hybridised orbitals of each carbon lying on the
molecular axis (x-axis) linearly overlaps with each other resulting in the formation a C-C sigma
bond. Other two sp
2
hybridised orbitals of both carbons linearly overlap with the four 1s orbitals
of four hydrogen atoms leading to the formation of two C-H sigma bonds on each carbon.
Formation of pi bond: The unhybridised 2Pz orbital of both carbon atoms can overlap only
sideways as they are not in the molecular axis. This lateral overlap results in the formation a pi
bond between the two carbon atoms as shown in the figure.
Bonding in acetylene:
The molecular formula of acetylene is C2H2. The electronic configuration of valence shell of
carbon in ground state is 1s
2
2s
2
2px
1
2py
1
2pz
0
.
1s 2s 2p
Electronic configuration of Carbonin Ground state ↿⇂ ↿⇂ ↿ ↿
1s 2s 2p
Electronic configuration of Carbonin Excited state ↿⇂ ↿ ↿ ↿ ↿
spHybridisation
Hybridised state ↿ ↿ ↿ ↿
sp sp 2py 2pz
Formation of sigma bond:
One of the two sphybridised orbitals of each carbon linearly overlaps with each other
resulting in the formation a C-C sigma bond. The other sphybridised orbital of both carbons
linearly overlap with the two 1s orbitals of two hydrogen atoms leading to the formation of one C-
H sigma bonds on each carbon.
Formation of pi bond:
The unhybridised 2Py and 2Pz orbitals of each carbon overlap sideways. This lateral
overlap results in the formation of two pi bonds ( Py-Py and Pz-Pz) between the two carbon atoms
33. Explain VSEPR theory.
The shape of the molecules depends on the number of valence shell electron pair around the
central atom.
There are two types of electron pairs namely bond pairs (bp) and lone pairs (lp).
The bond pair(bp) of electrons are those shared between two atoms,
while the lone pairs (lp) are the valence electron pairs that are not involved in bonding.
Each pair of valence electrons around the central atom repels each other and hence, they are
located as far away as possible in three dimensional space to minimize the repulsion between
them.
The repulsive interaction between the different types of electron pairs is in the following order.
lp – lp > lp – bp > bp – bp
[1 x 4(carbon)] + [3 x 6 (oxygen)] + [2 (charge)] = 24 electrons.
Distribution of these valence electrons gives us the following structure.
Complete the octet for carbon by moving a lone pair from one of the oxygens (OA) and write the
charge of the ion (2-) on the upper right side as shown in the figure.
In this case, we can draw two additional Lewis structures by moving the lone pairs from the
other two oxygens (OB and OC) thus creating three similar structures as shown below in which
the relative position of the atoms are same.
They only diff er in the position of bonding and lone pair of electrons.
Such structures are called resonance structures (canonical structures) and this phenomenon is
called resonance.
32. Explain the bond formation in ethylene and acetylene.
Bonding in ethylene:
The molecular formula of ethylene is C2H4. The valency of carbon is 4. Th e electronic
configuration of valence shell of carbon in ground state is 1s
2
2s
2
2px
1
2py
1
2pz
0
.
1s 2s 2p
Electronic configuration of Carbonin Ground state ↿⇂ ↿⇂ ↿ ↿
1s 2s 2p
Electronic configuration of Carbonin Excited state ↿⇂ ↿ ↿ ↿ ↿
sp
2
Hybridisation
Hybridised state ↿ ↿ ↿ ↿
sp
2
sp
2
sp
2
2pz
Formation of sigma bond: One of the sp
2
hybridised orbitals of each carbon lying on the
molecular axis (x-axis) linearly overlaps with each other resulting in the formation a C-C sigma
bond. Other two sp
2
hybridised orbitals of both carbons linearly overlap with the four 1s orbitals
of four hydrogen atoms leading to the formation of two C-H sigma bonds on each carbon.
Formation of pi bond: The unhybridised 2Pz orbital of both carbon atoms can overlap only
sideways as they are not in the molecular axis. This lateral overlap results in the formation a pi
bond between the two carbon atoms as shown in the figure.
Bonding in acetylene:
The molecular formula of acetylene is C2H2. The electronic configuration of valence shell of
carbon in ground state is 1s
2
2s
2
2px
1
2py
1
2pz
0
.
1s 2s 2p
Electronic configuration of Carbonin Ground state ↿⇂ ↿⇂ ↿ ↿
1s 2s 2p
Electronic configuration of Carbonin Excited state ↿⇂ ↿ ↿ ↿ ↿
spHybridisation
Hybridised state ↿ ↿ ↿ ↿
sp sp 2py 2pz
Formation of sigma bond:
One of the two sphybridised orbitals of each carbon linearly overlaps with each other
resulting in the formation a C-C sigma bond. The other sphybridised orbital of both carbons
linearly overlap with the two 1s orbitals of two hydrogen atoms leading to the formation of one C-
H sigma bonds on each carbon.
Formation of pi bond:
The unhybridised 2Py and 2Pz orbitals of each carbon overlap sideways. This lateral
overlap results in the formation of two pi bonds ( Py-Py and Pz-Pz) between the two carbon atoms
33. Explain VSEPR theory.
The shape of the molecules depends on the number of valence shell electron pair around the
central atom.
There are two types of electron pairs namely bond pairs (bp) and lone pairs (lp).
The bond pair(bp) of electrons are those shared between two atoms,
while the lone pairs (lp) are the valence electron pairs that are not involved in bonding.
Each pair of valence electrons around the central atom repels each other and hence, they are
located as far away as possible in three dimensional space to minimize the repulsion between
them.
The repulsive interaction between the different types of electron pairs is in the following order.
lp – lp > lp – bp > bp – bp
34. Explain VSEPR theory. Applying this theory to predict the shapes of IF7, and SF6
Shape of IF7 :-
Number of lone pairs is zero
It belongs in AB7 type
It consists of an Iodine atom at the center and seven
Fluorine atoms around it arranged in a regular pentagonal
bipyramidal in bond pair.
Shape of SF6 :-
Number of lone pairs is zero
It belongs in AB6type
It consists of Sulphur atom at the center and six Fluorine
atoms around it arranged in a regular octahedral in bond
pair
35. Explain the covalent character in ionic bond.
The partial covalent character in ionic compounds can be explained on the basis of a
phenomenon called polarisation.
We know that in an ionic compound, there is an electrostatic attractive force between the cation
and anion.
The positively charged cation attracts the valence electrons of anion while repelling the nucleus.
This causes a distortion in the electron cloud of the anion and its electron density drift s towards
the cation, which results in some sharing of the valence electrons between these ions.
Thus, a partial covalent character is developed between them. This phenomenon is called
polarization.
36. Explain Molecular orbital theory :
In a molecule, electrons are present in new orbitals called molecular orbitals.
Molecular orbitals are formed by combination of atomic orbitals of equal energies (in case of
homonuclear molecules) or of comparable energies (in case of heteronuclear molecules).
The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing
combination.
Two atomic orbitals can combine to form two molecular orbitals. One of these two molecular
orbitals one has a lower energy and the other has a higher energy. The molecular orbital with
lower energy is called bonding molecular orbital and the other with higher energy is called anti
bondingmolecular orbital.
The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
The bonding molecular orbitals are represented by (sigma), (pi), (delta) and the antibonding
molecular orbitals are represented by *.
The molecular orbitals are filled in the increasing order of their energies, starting with orbital of
least energy. (Aufbau principle).
A molecular orbital can accommodate only two electrons and these two electrons must have
opposite spins. (Paul’s exclusion principle).
While filling molecular orbitals of equal energy, pairing of electrons does not take place until all
such molecular orbitals are singly filled with electrons having parallel spins. (Hund’s rule).
Bond order =
N
b − N
a
2
Nb = Total number of electrons present in the bonding molecular orbitals
Na = Total number of electrons present in the antibonding molecular orbitals
11. FUNDAMENTALS OF ORGANIC CHEMISTRY
1. Give the general characteristics of organic compounds?
They are covalent compounds insoluble in water and readily soluble in organic solvent such
as benzene, toluene, ether, chloroform etc...
Many of the organic compounds are inflammable (except CCl4).Organic compounds are
characterised by functional groups. They exhibit isomerism
2. Write a note on homologous series.
A series of organic compounds each containing a characteric functional group and the
successive members differ from each other in molecular formula by a - CH2- group is called
homologous series.
Eg.Alkanes: Methane (CH4), Ethane (C2H6), etc..Alcohols: Methanol (CH3OH), Ethanol
(C2H5OH) etc..
Compounds of the homologous series are represented by a general formula
Alkanes CnH2n+2, Alkenes CnH2n, Alkynes CnH2n-2
It can be prepared by general methods.
They show regular gradation in physical properties but have almost similar chemical property
3. What is meant by a functional group? Identify the functional group in the following
compounds.
(a) acetaldehyde (b) oxalic acid (c) di methyl ether (d) methylamine
The chemical properties of all the members of a homologous series are characterised by a group
called the functional group
Compound Functional group
(a) acetaldehyde -CHO Aldehyde
(b) oxalic acid -COOH Carboxylic acid
(c) di methyl ether -O- Ether
(d) methylamine -NH2 amine
4. Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol (b) Aliphatic ketones. (c) Aliphatic amines.
Compound General molecular formula
(a) Aliphatic monohydric alcohol
R-OH
Ex. CH3-OH, CH3-CH2-OH
(b) Aliphatic ketones
R-CO-R
Ex. CH3-CO-CH3
(c) Aliphatic amines
R-NH2
Ex. CH3-NH2, CH3-CH2-NH2
Shape of IF7 :-
Number of lone pairs is zero
It belongs in AB7 type
It consists of an Iodine atom at the center and seven
Fluorine atoms around it arranged in a regular pentagonal
bipyramidal in bond pair.
Shape of SF6 :-
Number of lone pairs is zero
It belongs in AB6type
It consists of Sulphur atom at the center and six Fluorine
atoms around it arranged in a regular octahedral in bond
pair
35. Explain the covalent character in ionic bond.
The partial covalent character in ionic compounds can be explained on the basis of a
phenomenon called polarisation.
We know that in an ionic compound, there is an electrostatic attractive force between the cation
and anion.
The positively charged cation attracts the valence electrons of anion while repelling the nucleus.
This causes a distortion in the electron cloud of the anion and its electron density drift s towards
the cation, which results in some sharing of the valence electrons between these ions.
Thus, a partial covalent character is developed between them. This phenomenon is called
polarization.
36. Explain Molecular orbital theory :
In a molecule, electrons are present in new orbitals called molecular orbitals.
Molecular orbitals are formed by combination of atomic orbitals of equal energies (in case of
homonuclear molecules) or of comparable energies (in case of heteronuclear molecules).
The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing
combination.
Two atomic orbitals can combine to form two molecular orbitals. One of these two molecular
orbitals one has a lower energy and the other has a higher energy. The molecular orbital with
lower energy is called bonding molecular orbital and the other with higher energy is called anti
bondingmolecular orbital.
The shapes of molecular orbitals depend upon the shapes of combining atomic orbitals.
The bonding molecular orbitals are represented by (sigma), (pi), (delta) and the antibonding
molecular orbitals are represented by *.
The molecular orbitals are filled in the increasing order of their energies, starting with orbital of
least energy. (Aufbau principle).
A molecular orbital can accommodate only two electrons and these two electrons must have
opposite spins. (Paul’s exclusion principle).
While filling molecular orbitals of equal energy, pairing of electrons does not take place until all
such molecular orbitals are singly filled with electrons having parallel spins. (Hund’s rule).
Bond order =
N
b − N
a
2
Nb = Total number of electrons present in the bonding molecular orbitals
Na = Total number of electrons present in the antibonding molecular orbitals
11. FUNDAMENTALS OF ORGANIC CHEMISTRY
1. Give the general characteristics of organic compounds?
They are covalent compounds insoluble in water and readily soluble in organic solvent such
as benzene, toluene, ether, chloroform etc...
Many of the organic compounds are inflammable (except CCl4).Organic compounds are
characterised by functional groups. They exhibit isomerism
2. Write a note on homologous series.
A series of organic compounds each containing a characteric functional group and the
successive members differ from each other in molecular formula by a - CH2- group is called
homologous series.
Eg.Alkanes: Methane (CH4), Ethane (C2H6), etc..Alcohols: Methanol (CH3OH), Ethanol
(C2H5OH) etc..
Compounds of the homologous series are represented by a general formula
Alkanes CnH2n+2, Alkenes CnH2n, Alkynes CnH2n-2
It can be prepared by general methods.
They show regular gradation in physical properties but have almost similar chemical property
3. What is meant by a functional group? Identify the functional group in the following
compounds.
(a) acetaldehyde (b) oxalic acid (c) di methyl ether (d) methylamine
The chemical properties of all the members of a homologous series are characterised by a group
called the functional group
Compound Functional group
(a) acetaldehyde -CHO Aldehyde
(b) oxalic acid -COOH Carboxylic acid
(c) di methyl ether -O- Ether
(d) methylamine -NH2 amine
4. Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol (b) Aliphatic ketones. (c) Aliphatic amines.
Compound General molecular formula
(a) Aliphatic monohydric alcohol
R-OH
Ex. CH3-OH, CH3-CH2-OH
(b) Aliphatic ketones
R-CO-R
Ex. CH3-CO-CH3
(c) Aliphatic amines
R-NH2
Ex. CH3-NH2, CH3-CH2-NH2
5. Describe the classification of organic compounds based on their structure
6. Write the molecular formula of the first six members of homologous series of nitro
alkanes.
Molecular formula NAME
CH3-NO2 Nitro methane
CH3-CH2-NO2 Nitro ehtane
CH3-CH2-CH2-NO2 1-Nitro propane
CH3-CH2-CH2-CH2-NO2 1-Nitro butane
CH3-CH2-CH2-CH2-CH2-NO2 1-Nitro pentane
CH3-CH2-CH2-CH2-CH2-CH2-NO2 1-Nitro hexane
7. Write the molecular and possible structural formula of the first four members of
homologous series of carboxylic acids.
Molecular formula NAME
H-COOH Methanoic acid
CH3- COOH Ethanoic acid
CH3- CH2- COOH Propanaoic acid
CH3- CH2-CH2- COOH Butanoic acid
CH3- CH(COOH )CH3 2-Methyl propanoic acid
8. Give the IUPAC names of the following compounds.
i,(CH3)2CH-CH2-CH(CH3)-CH(CH3)2
6 5 4 3 2 1
CH3 – CH–CH2– CH – CH – CH3
CH3 CH3 CH3 2,3,5-Trimethylhexane
ii, CH3 – CH – CH – CH3
CH3 Br
4 3 2 1
CH3 – CH – CH - CH3
CH3 Br 2–Bromo–3–Methylbutane
iii, CH3– O – CH3
CH3 – O – CH3 Methoxymethane
iv, CH3 – CH2 –CH – CHO
OH
4 3 2 1
CH3 – CH2 –CH – CHO 2-Hydroxybutanal
OH
v, CH2= CH – CH = CH2
1 2 3 4
CH2= CH – CH = CH2 Buta-1,3-diene (or) 1,3-butadiene
vi, CH3 – C C – CH – CH3
Cl
1 2 3 4 5
CH3 – C C – CH – CH3
Cl 4-Chloropent-2-yne (or) 4-Chloro-2-pentyne
vii,
4 3 2 1
CH3 – CH = CH – CH2 – Br 1-Bromobut-2-ene (or) 1-Bromo-2-butene
6. Write the molecular formula of the first six members of homologous series of nitro
alkanes.
Molecular formula NAME
CH3-NO2 Nitro methane
CH3-CH2-NO2 Nitro ehtane
CH3-CH2-CH2-NO2 1-Nitro propane
CH3-CH2-CH2-CH2-NO2 1-Nitro butane
CH3-CH2-CH2-CH2-CH2-NO2 1-Nitro pentane
CH3-CH2-CH2-CH2-CH2-CH2-NO2 1-Nitro hexane
7. Write the molecular and possible structural formula of the first four members of
homologous series of carboxylic acids.
Molecular formula NAME
H-COOH Methanoic acid
CH3- COOH Ethanoic acid
CH3- CH2- COOH Propanaoic acid
CH3- CH2-CH2- COOH Butanoic acid
CH3- CH(COOH )CH3 2-Methyl propanoic acid
8. Give the IUPAC names of the following compounds.
i,(CH3)2CH-CH2-CH(CH3)-CH(CH3)2
6 5 4 3 2 1
CH3 – CH–CH2– CH – CH – CH3
CH3 CH3 CH3 2,3,5-Trimethylhexane
ii, CH3 – CH – CH – CH3
CH3 Br
4 3 2 1
CH3 – CH – CH - CH3
CH3 Br 2–Bromo–3–Methylbutane
iii, CH3– O – CH3
CH3 – O – CH3 Methoxymethane
iv, CH3 – CH2 –CH – CHO
OH
4 3 2 1
CH3 – CH2 –CH – CHO 2-Hydroxybutanal
OH
v, CH2= CH – CH = CH2
1 2 3 4
CH2= CH – CH = CH2 Buta-1,3-diene (or) 1,3-butadiene
vi, CH3 – C C – CH – CH3
Cl
1 2 3 4 5
CH3 – C C – CH – CH3
Cl 4-Chloropent-2-yne (or) 4-Chloro-2-pentyne
vii,
4 3 2 1
CH3 – CH = CH – CH2 – Br 1-Bromobut-2-ene (or) 1-Bromo-2-butene
viii, O O
HO
1 2 3 4 5 6
HOOC – CH2 – CH2 – CH2 - CO – CH3 5-Oxohexanoic acid
ix,
1CH2 CH3
2 CH CH2
CH3 – CH2 – CH2 – CH – CH – CH2 – CH3 4-Ethyl-3-prpopyl-1-hexene
3 4 5 6
x,
CH3 CH3
CH3 – C – CH = C – CH3 2,4,4-Trimethyl-2-pentene
5 4
3 2 1
CH3
xi,
NH2 CH3
C6H5 – CH – CH – CH3 2,- Methyl-1-phenylpropane-1-amine
1 2 3
xii,
1CN
5 4 3 2
CH3 – C – CH2 – C – CH3 2,2 –Dimethyl-4-Oxopentanenitryl
O CH3
xiii,
1
CH3
CH3 – CH2 – O – CH – CH3 Ethoxy-2-propane (or) 2-Ethoxypropane
2 3
xiv, O2N CH3
F
O2N 3 CH3
2 4
1 5 1 –Fluro-4-methyl-2-nitrobenzene
F 6
xv, Br
OHC.
CH3 Br
OHC – CH – CH – CH3 3-Bromo–2-Methylbutanal
1 2 3 4
xvi,
C6H5 – CO – CH3 Acetophenone
9.
Give the structure for the following compound.
(i) 3- ethyl - 2 methyl -1-pentene
1 2 3 4 5
CH2 = C – CH– CH2– CH3
CH3 CH2 – CH3
HO
1 2 3 4 5 6
HOOC – CH2 – CH2 – CH2 - CO – CH3 5-Oxohexanoic acid
ix,
1CH2 CH3
2 CH CH2
CH3 – CH2 – CH2 – CH – CH – CH2 – CH3 4-Ethyl-3-prpopyl-1-hexene
3 4 5 6
x,
CH3 CH3
CH3 – C – CH = C – CH3 2,4,4-Trimethyl-2-pentene
5 4
3 2 1
CH3
xi,
NH2 CH3
C6H5 – CH – CH – CH3 2,- Methyl-1-phenylpropane-1-amine
1 2 3
xii,
1CN
5 4 3 2
CH3 – C – CH2 – C – CH3 2,2 –Dimethyl-4-Oxopentanenitryl
O CH3
xiii,
1
CH3
CH3 – CH2 – O – CH – CH3 Ethoxy-2-propane (or) 2-Ethoxypropane
2 3
xiv, O2N CH3
F
O2N 3 CH3
2 4
1 5 1 –Fluro-4-methyl-2-nitrobenzene
F 6
xv, Br
OHC.
CH3 Br
OHC – CH – CH – CH3 3-Bromo–2-Methylbutanal
1 2 3 4
xvi,
C6H5 – CO – CH3 Acetophenone
9.
Give the structure for the following compound.
(i) 3- ethyl - 2 methyl -1-pentene
1 2 3 4 5
CH2 = C – CH– CH2– CH3
CH3 CH2 – CH3
(ii) 1,3,5- Trimethylcyclohex - 1 –ene
CH3
C
1
6CH2 2CH
5CH 3CH
CH3 CH3
4CH2
(iii) tertiary butyl iodide
CH3
CH3 – C – I I
CH3
(iv) 3 –Chlorobutanal
4 3 2 1 O
CH3– CH – CH2– CHO
Cl Cl
(v) 3 - Chlorobutanol
4 3 2 1 OH
CH3– CH – CH2– CH2–OH
Cl Cl
(vi) 2 - Chloro - 2- methyl propane
3 2 1
CH3– C – CH3
Cl CH3 Cl
(vii) 2,2-dimethyl-1-chloropropane
CH3 Cl
CH3– C – CH2 – Cl
CH3
(viii) 3 - methylbut -1- ene
CH3 – CH – CH = CH2
CH3
(ix) Butan - 2, 2 - diol
OH OH
CH3 – C – CH2 – CH3
OH OH
(x) Octane - 1,3- diene
1 2 3 4 5 6 7 8
CH2 = CH – CH = CH–CH2– CH2– CH2– CH3
(xi) 1,5- Dimethylcyclohexane
CH3
CH
1
6CH2 2CH2
5CH 3 CH2
CH3
4CH2
(xii) 2-Chlorobut - 3 –ene
1 2 3 4
CH3 – CH – CH = CH2
Cl Cl
CH3
C
1
6CH2 2CH
5CH 3CH
CH3 CH3
4CH2
(iii) tertiary butyl iodide
CH3
CH3 – C – I I
CH3
(iv) 3 –Chlorobutanal
4 3 2 1 O
CH3– CH – CH2– CHO
Cl Cl
(v) 3 - Chlorobutanol
4 3 2 1 OH
CH3– CH – CH2– CH2–OH
Cl Cl
(vi) 2 - Chloro - 2- methyl propane
3 2 1
CH3– C – CH3
Cl CH3 Cl
(vii) 2,2-dimethyl-1-chloropropane
CH3 Cl
CH3– C – CH2 – Cl
CH3
(viii) 3 - methylbut -1- ene
CH3 – CH – CH = CH2
CH3
(ix) Butan - 2, 2 - diol
OH OH
CH3 – C – CH2 – CH3
OH OH
(x) Octane - 1,3- diene
1 2 3 4 5 6 7 8
CH2 = CH – CH = CH–CH2– CH2– CH2– CH3
(xi) 1,5- Dimethylcyclohexane
CH3
CH
1
6CH2 2CH2
5CH 3 CH2
CH3
4CH2
(xii) 2-Chlorobut - 3 –ene
1 2 3 4
CH3 – CH – CH = CH2
Cl Cl
(xiii) 2 - methylbutan- 3 - ol
1 2 3 4 OH
CH3– CH – CH – CH3
OH CH3
(xiv) acetaldehyde
CH3 – CHO
O
10. Describe the reactions involved in the detection of nitrogen in an organic compound by
Lassaigne method.
If nitrogen is present it gets converted to sodium cyanide
which reacts with freshly prepared ferrous sulphate and ferric ion followed by conc. HCl and gives
a Prussian blue color or green color or precipitate.
It confirms the presence of nitrogen.
Na +C + N NaCN(sodium fusion extract)
FeSO4 +2NaOH Na2SO4 +Fe(OH)2
6NaCN +Fe(OH)2 2NaOH + Na4[Fe(CN)6]
Na4[Fe(CN)6] +4 FeCl3 12NaCl + Fe4[Fe(CN)6]3
Prussian blue color precipitate.
11. Give the principle involved in the estimation of halogen in an organic compound by
carius method.
A known mass of the organic compound is heated with fuming HNO3 and AgNO3.
C,H&S get oxidized to CO2, H2O & SO2 and halogen combines with AgNO3 to form a
precipitate of silver halide.
The ppt of AgX is filtered, washed, dried and weighed.
From the mass of AgX and the mass of the organic compound taken,percentage of halogens
are calculated.
X
??????����� ??????��
� +����
�
AgX
12. Give a brief description of the principles of
i) Fractional distillation
This is one method to purify and separate liquids present in the mixture having their boiling point
close to each other.
The process of separation of the components in a liquid mixture at their respective boilingpoints in
the form of vapours and the subsequent condensation of those vapours is called fractional
distillation.
ii) Column Chromatography
This is the simplest chromatographic method carried out in long glass column having a stop
cock near the lower end.
This method involves separation of a mixture over a column of adsorbent (Stationery phase)
packed in a column.
In the column a plug of cotton or glass wool is placed at the lower end of the column to
support the adsorbent powder.
The tube is uniformly packed with suitable absorbent constitute the stationary phase.
(Activated aluminum oxides (alumina), Magnesium oxide, starch are also used as absorbents).
13. Explain paper chromatography
This method involves continues differential portioning of components of a mixture between
stationary and mobile phase.
In paper chromatography, a special quality paper known as chromatography paper is used. This
paper act as a stationary phase.
Paper chromatography is a technique that involves placing a small dot orline of sample solution
onto a strip of chromatography paper.
The paper is placed in a jar containing a shallow layer of solvent and sealed. As the solvent rises
through the paper, it meets the sample mixture,
which starts to travel up the paper with the solvent.
14. Explain varions types of constitutional isomerism (structural isomerism) in organic
compounds
Chain Isomerism:
Chain isomers possess the same molecular formula, but different number of carbon inparent–chains
(ie. Straight or branched)
Ex:-
CH3 - CH2 - CH2 - CH3 CH3 - CH - CH3
n Butane Isobutane
CH3
Position Isomerism:
Compounds having same molecular formula but the position of functional group,multiple bond or
branches along the same chain length of carbon atoms varies.
Ex:-
CH3 - CH2 - CH2 - OH CH3 - CH - CH3
1-Propanol 2-Propanol
OH
Functional Group Isomerism:
When isomers have the same molecular formula but different functionalgroups,
Ex:-
CH3 - CH2 - OH CH3 - O - CH3
Ethylalcohol Dimethylether
15. Describe optical isomerism with suitable example.
Compounds having same physical and chemical property but differ only in the rotation of plane of
the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Eg:- d- Lacticacid &l-Lacticacid
16. Briefly explain geometrical isomerism in alkene by considering 2- butene as an example.
The cis isomer is one in which two similar groups are on the same side of the double bond.
The trans isomers is that in which the two similar groups are on the opposite side of the double
bond,
Hence this type of isomerism is often called cis-trans isomerism.
H CH3 H CH3
C C
|| ||
C C
H CH3 H3C H
Cis 2-butene Trans 2- butane
1 2 3 4 OH
CH3– CH – CH – CH3
OH CH3
(xiv) acetaldehyde
CH3 – CHO
O
10. Describe the reactions involved in the detection of nitrogen in an organic compound by
Lassaigne method.
If nitrogen is present it gets converted to sodium cyanide
which reacts with freshly prepared ferrous sulphate and ferric ion followed by conc. HCl and gives
a Prussian blue color or green color or precipitate.
It confirms the presence of nitrogen.
Na +C + N NaCN(sodium fusion extract)
FeSO4 +2NaOH Na2SO4 +Fe(OH)2
6NaCN +Fe(OH)2 2NaOH + Na4[Fe(CN)6]
Na4[Fe(CN)6] +4 FeCl3 12NaCl + Fe4[Fe(CN)6]3
Prussian blue color precipitate.
11. Give the principle involved in the estimation of halogen in an organic compound by
carius method.
A known mass of the organic compound is heated with fuming HNO3 and AgNO3.
C,H&S get oxidized to CO2, H2O & SO2 and halogen combines with AgNO3 to form a
precipitate of silver halide.
The ppt of AgX is filtered, washed, dried and weighed.
From the mass of AgX and the mass of the organic compound taken,percentage of halogens
are calculated.
X
??????����� ??????��
� +����
�
AgX
12. Give a brief description of the principles of
i) Fractional distillation
This is one method to purify and separate liquids present in the mixture having their boiling point
close to each other.
The process of separation of the components in a liquid mixture at their respective boilingpoints in
the form of vapours and the subsequent condensation of those vapours is called fractional
distillation.
ii) Column Chromatography
This is the simplest chromatographic method carried out in long glass column having a stop
cock near the lower end.
This method involves separation of a mixture over a column of adsorbent (Stationery phase)
packed in a column.
In the column a plug of cotton or glass wool is placed at the lower end of the column to
support the adsorbent powder.
The tube is uniformly packed with suitable absorbent constitute the stationary phase.
(Activated aluminum oxides (alumina), Magnesium oxide, starch are also used as absorbents).
13. Explain paper chromatography
This method involves continues differential portioning of components of a mixture between
stationary and mobile phase.
In paper chromatography, a special quality paper known as chromatography paper is used. This
paper act as a stationary phase.
Paper chromatography is a technique that involves placing a small dot orline of sample solution
onto a strip of chromatography paper.
The paper is placed in a jar containing a shallow layer of solvent and sealed. As the solvent rises
through the paper, it meets the sample mixture,
which starts to travel up the paper with the solvent.
14. Explain varions types of constitutional isomerism (structural isomerism) in organic
compounds
Chain Isomerism:
Chain isomers possess the same molecular formula, but different number of carbon inparent–chains
(ie. Straight or branched)
Ex:-
CH3 - CH2 - CH2 - CH3 CH3 - CH - CH3
n Butane Isobutane
CH3
Position Isomerism:
Compounds having same molecular formula but the position of functional group,multiple bond or
branches along the same chain length of carbon atoms varies.
Ex:-
CH3 - CH2 - CH2 - OH CH3 - CH - CH3
1-Propanol 2-Propanol
OH
Functional Group Isomerism:
When isomers have the same molecular formula but different functionalgroups,
Ex:-
CH3 - CH2 - OH CH3 - O - CH3
Ethylalcohol Dimethylether
15. Describe optical isomerism with suitable example.
Compounds having same physical and chemical property but differ only in the rotation of plane of
the polarized light are known as optical isomers and the phenomenon is known as optical isomerism.
Eg:- d- Lacticacid &l-Lacticacid
16. Briefly explain geometrical isomerism in alkene by considering 2- butene as an example.
The cis isomer is one in which two similar groups are on the same side of the double bond.
The trans isomers is that in which the two similar groups are on the opposite side of the double
bond,
Hence this type of isomerism is often called cis-trans isomerism.
H CH3 H CH3
C C
|| ||
C C
H CH3 H3C H
Cis 2-butene Trans 2- butane
12 . BASIC CONCEPT OF ORGANIC REACTIONS
1. Write short notes on
(a) Resonance
Certain organic compounds can be represented by more than one structure and they differ
only in the position of bonding and lone pair of electrons. Such structures are called resonance
structures and this phenomenon is called resonance.
(b) Hyperconjucation
The delocalization of bond is called as hyper conjucation
hyperconjucation is a permanent effect
2. What are electrophiles and nucleophiles? Give suitable examples for each.
Electrophiles:
Electron deficiency species
Positive changed ions
All Lewis acids act as electrophiles
Eg. Cl
-
, CN
-
, Br
-
, I
-
, NH3
Nucleophiles :
Electron rich species having a lone pair of electron
Negatively charged ions
All Lewis bases act as nucleophiles
Eg. NO2
+
, AlCl3 , BF3
3. Show the heterolysis of covalent bond by using curved arrow notation and complete the
following equations. Identify the nucleophile is each case.
(i) CH3 - Br + KOH →
(ii) CH3 - OCH3 + HI →
(i) CH3 - Br + KOH CH3 - OH + KBr
Nucleophile OH
Step – 1 CH3 - Br CH3
+
+ Br
-
Step – 2 CH3
+
+OH
-
CH3– OH
(ii) CH3– O - CH3 + HI CH3 – OH + CH3–I
Nucleophile I
Step – 1 CH3– O - CH3 CH3
+
+
-
O–CH3
Step – 2 CH3
+
+ I
-
CH3 – I
4. Explain Inductive effect with suitable example.
Inductive effect is defined as the change in the polarisation of a covalent bond due to the
presence of adjacent bonds, atoms or groups in the molecule. This is a permanent phenomenon.
5. Explain electromeric effect.
It is defined as the complete transfer of shared pair of electrons of multiple bonds to one
of the atoms in presence of an attaching reagent.
O O
|| |
CN
+ – C – – C –
|
CN. 6. Give examples of the following types of organic reactions
(i) β–elimination
A chemical reaction in which atoms or groups are eliminated from adjacent atoms resulting in
a new bond.
alcoholic KOH
CH3 – CH = CH2 + H2O + Br
ii) Electrophilic substitution.
7. Write a note on Homolytic Cleavage
The process in which a covalent bond breaks symmetrically in such way that each of the
bonded atoms retains one electron.
CH
3 – CH
3
.
CH
3 +
.
CH
3
8. Write a note on Heterolytic Cleavage
The process in which a covalent bond breaks unsymmetrically such that one of the bonded
atoms retains the bond pair of electrons.
9. What is Positive Mesomericeffect ? Give example.
Positive resonance effect occurs, when the electrons move away from substituent
attached to the conjugated system.
The electron releasing substituent are attached to the conjugated system.
These electron releasing groups are usually denoted as +R or +M groups.
Eg :- OH, SH, OR, SR, NH2
10. What is Negative Mesomeric effect? Give example.
Negative resonance effect occurs, when the electrons move towards the substituent
attached to the conjugated system.
The electron withdrawing substituents are attached to the conjugated system.
These electron withdrawing groups are usually denoted as –R or –M groups.
Eg :- NO2, COOH, > C = O, C N
11. What is Nucleophilic substitution and give one example ?
Nucleophilic substitution reactions are chemical reactios in which an nucleophile (electron
rich chemical spicies (or) negative charge ion) replaces a function group in a compound.
CH3 – Br + OH
-
(dil) CH3 – OH + Br
-
12. What is Electophilic substitution and give one example ?
Electrophilic substitution reactions are chemical reactios in which an electrophile
(positive charge ion) replaces a function group in a compound.
Eg :- Nitration of Benzene
C6H5 – H + NO2
+
C6H5 – NO2 + H
+
13. What is Free radical substitution and give one example ?
Free radicals are atoms or groups of atoms that have a single unpaired electron a free
aradical substitution reaction in one involving these radicals.
CH3 – H + Cl CH3 + HCl
14. Give the example for Electophilic Addition reaction.
CH2 = CH2 + Br – Br CH2Br – CH2Br
Ethylene Ethylenebromide
1. Write short notes on
(a) Resonance
Certain organic compounds can be represented by more than one structure and they differ
only in the position of bonding and lone pair of electrons. Such structures are called resonance
structures and this phenomenon is called resonance.
(b) Hyperconjucation
The delocalization of bond is called as hyper conjucation
hyperconjucation is a permanent effect
2. What are electrophiles and nucleophiles? Give suitable examples for each.
Electrophiles:
Electron deficiency species
Positive changed ions
All Lewis acids act as electrophiles
Eg. Cl
-
, CN
-
, Br
-
, I
-
, NH3
Nucleophiles :
Electron rich species having a lone pair of electron
Negatively charged ions
All Lewis bases act as nucleophiles
Eg. NO2
+
, AlCl3 , BF3
3. Show the heterolysis of covalent bond by using curved arrow notation and complete the
following equations. Identify the nucleophile is each case.
(i) CH3 - Br + KOH →
(ii) CH3 - OCH3 + HI →
(i) CH3 - Br + KOH CH3 - OH + KBr
Nucleophile OH
Step – 1 CH3 - Br CH3
+
+ Br
-
Step – 2 CH3
+
+OH
-
CH3– OH
(ii) CH3– O - CH3 + HI CH3 – OH + CH3–I
Nucleophile I
Step – 1 CH3– O - CH3 CH3
+
+
-
O–CH3
Step – 2 CH3
+
+ I
-
CH3 – I
4. Explain Inductive effect with suitable example.
Inductive effect is defined as the change in the polarisation of a covalent bond due to the
presence of adjacent bonds, atoms or groups in the molecule. This is a permanent phenomenon.
5. Explain electromeric effect.
It is defined as the complete transfer of shared pair of electrons of multiple bonds to one
of the atoms in presence of an attaching reagent.
O O
|| |
CN
+ – C – – C –
|
CN. 6. Give examples of the following types of organic reactions
(i) β–elimination
A chemical reaction in which atoms or groups are eliminated from adjacent atoms resulting in
a new bond.
alcoholic KOH
CH3 – CH = CH2 + H2O + Br
ii) Electrophilic substitution.
7. Write a note on Homolytic Cleavage
The process in which a covalent bond breaks symmetrically in such way that each of the
bonded atoms retains one electron.
CH
3 – CH
3
.
CH
3 +
.
CH
3
8. Write a note on Heterolytic Cleavage
The process in which a covalent bond breaks unsymmetrically such that one of the bonded
atoms retains the bond pair of electrons.
9. What is Positive Mesomericeffect ? Give example.
Positive resonance effect occurs, when the electrons move away from substituent
attached to the conjugated system.
The electron releasing substituent are attached to the conjugated system.
These electron releasing groups are usually denoted as +R or +M groups.
Eg :- OH, SH, OR, SR, NH2
10. What is Negative Mesomeric effect? Give example.
Negative resonance effect occurs, when the electrons move towards the substituent
attached to the conjugated system.
The electron withdrawing substituents are attached to the conjugated system.
These electron withdrawing groups are usually denoted as –R or –M groups.
Eg :- NO2, COOH, > C = O, C N
11. What is Nucleophilic substitution and give one example ?
Nucleophilic substitution reactions are chemical reactios in which an nucleophile (electron
rich chemical spicies (or) negative charge ion) replaces a function group in a compound.
CH3 – Br + OH
-
(dil) CH3 – OH + Br
-
12. What is Electophilic substitution and give one example ?
Electrophilic substitution reactions are chemical reactios in which an electrophile
(positive charge ion) replaces a function group in a compound.
Eg :- Nitration of Benzene
C6H5 – H + NO2
+
C6H5 – NO2 + H
+
13. What is Free radical substitution and give one example ?
Free radicals are atoms or groups of atoms that have a single unpaired electron a free
aradical substitution reaction in one involving these radicals.
CH3 – H + Cl CH3 + HCl
14. Give the example for Electophilic Addition reaction.
CH2 = CH2 + Br – Br CH2Br – CH2Br
Ethylene Ethylenebromide
15. Give the example for Nucleophilic Addition reaction.
OH
CH3 – CHO + HCN
�??????
−
CH3 – CH - CN
Acetaldehyde Acetaldehyde cyanohydrine
16. Give the example for Free radical Addition reaction..
CH2 = CH2 + HBr
Benzoyl chloride
CH3 – CH2 – Br
Ethylene Ethylbromide
13 . HYDROCARBONS
1. Give IUPAC names for the following compounds
1) CH3 – CH = CH – CH = CH – C ≡ C – CH3 Oct – 6 –yne –2,4–diene
C2H5 CH3
| |
2) CH3– C – C – C ≡ C – CH3
| |
CH3 H
5- Ethyl-4,5-dimethylhex-2-yne
3) (CH3)3 C – C ≡ C – CH (CH3)2 2,2’5-Trimethylhex-3-yne
4, CH3
|
CH3– C – C ≡ C – CH2 – CH3
Ethyl isopropyl acetylene
2-Methylhex-3-yne
5) CH ≡ C – C ≡ C – C ≡ CH Hex-1,3,5-triyne
2. Identify the compound A, B, C and D in the following series of reactions
CH3– CH2–Br
���.��??????
A
??????�
�/????????????�
�
B
���??????�
D
�
�/��−??????��
C
CH3– CH2–Br
alc.KOH
CH2 = CH2 (A)
CH2 = CH2
��
2/���
4
CH2Cl – CH2Cl (B)
CH2Cl – CH2Cl
NaNH2
CH2 = CHCl (D)
CH2 = CH2
�
3/Zn−H2O
2 H – CHO (C)
(A) CH2 = CH2 Ethylene
(B) CH2Cl – CH2Cl Ethylenechloride
(C) H–CHO Formaldehyde
(D) CH2 = CHCl Vinylchloride
3.
Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
All the activating groups are „ortho - para‟ directors.
Eg. –OH, –NH2, –CH3 etc.
In Phenol the (-) charge residue is present on ortho and para position of ring structure.
It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene.
The electron density at ortho and parapositions increases as compared to the meta position.
Therefore phenolic group activates the benzene ring for electrophilic attack at „ortho‟ and „para
positions and hence –OH group is an ortho-para director and activator.
4. How is propyne prepared from an alkyenedihalide ?
Br Br
| |
CH3 – CH – CH2
� ���??????
�
CH3 – C ≡ CH + 2NaBr + 2NH3
Propyne
5. An alkylhalide with molecular formula C6H13Br on dehydro halogenation gave two
isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and
Y gave four compounds CH3COCH3, CH3CHO, CH3CH2CHO and (CH3)2 CHCHO. Find
the alkylhalide.
CH3 Br CH3
| | |
CH3 – CH – CH – CH2 – CH3
���.��??????
CH3 – C = CH – CH2 – CH3 (X)
2-methylpent -2- ene
CH3
|
+ CH3 – CH– C = CH – CH3 (Y)
4-methylpent -2- ene
CH3 CH3
| |
CH3 – C = CH – CH2 – CH3
��
CH3 – C – O – CH – CH2 – CH3
2-methylpent -2- ene | |
O –– O
Zn / H2O
CH3 – CO – CH3 + CH3 – CH2 – CHO
Acetone Propanaldehyde
CH3 CH3
| |
CH3– CH - CH = CH – CH3
��
CH3– CH – CH – O – CH – CH3
4-methylpent -2- ene | |
O –– O
Zn / H2O
CH3
|
CH3 – CH – CHO + CH3 – CHO
Isobutanaldehyde Acetaldehyde
6. Describe the mechanism of Nitration of benzene.
Reagent Conc HNO3 + ConcH2SO4
Electrophiles -NO2
+
OH
CH3 – CHO + HCN
�??????
−
CH3 – CH - CN
Acetaldehyde Acetaldehyde cyanohydrine
16. Give the example for Free radical Addition reaction..
CH2 = CH2 + HBr
Benzoyl chloride
CH3 – CH2 – Br
Ethylene Ethylbromide
13 . HYDROCARBONS
1. Give IUPAC names for the following compounds
1) CH3 – CH = CH – CH = CH – C ≡ C – CH3 Oct – 6 –yne –2,4–diene
C2H5 CH3
| |
2) CH3– C – C – C ≡ C – CH3
| |
CH3 H
5- Ethyl-4,5-dimethylhex-2-yne
3) (CH3)3 C – C ≡ C – CH (CH3)2 2,2’5-Trimethylhex-3-yne
4, CH3
|
CH3– C – C ≡ C – CH2 – CH3
Ethyl isopropyl acetylene
2-Methylhex-3-yne
5) CH ≡ C – C ≡ C – C ≡ CH Hex-1,3,5-triyne
2. Identify the compound A, B, C and D in the following series of reactions
CH3– CH2–Br
���.��??????
A
??????�
�/????????????�
�
B
���??????�
D
�
�/��−??????��
C
CH3– CH2–Br
alc.KOH
CH2 = CH2 (A)
CH2 = CH2
��
2/���
4
CH2Cl – CH2Cl (B)
CH2Cl – CH2Cl
NaNH2
CH2 = CHCl (D)
CH2 = CH2
�
3/Zn−H2O
2 H – CHO (C)
(A) CH2 = CH2 Ethylene
(B) CH2Cl – CH2Cl Ethylenechloride
(C) H–CHO Formaldehyde
(D) CH2 = CHCl Vinylchloride
3.
Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
All the activating groups are „ortho - para‟ directors.
Eg. –OH, –NH2, –CH3 etc.
In Phenol the (-) charge residue is present on ortho and para position of ring structure.
It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene.
The electron density at ortho and parapositions increases as compared to the meta position.
Therefore phenolic group activates the benzene ring for electrophilic attack at „ortho‟ and „para
positions and hence –OH group is an ortho-para director and activator.
4. How is propyne prepared from an alkyenedihalide ?
Br Br
| |
CH3 – CH – CH2
� ���??????
�
CH3 – C ≡ CH + 2NaBr + 2NH3
Propyne
5. An alkylhalide with molecular formula C6H13Br on dehydro halogenation gave two
isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and
Y gave four compounds CH3COCH3, CH3CHO, CH3CH2CHO and (CH3)2 CHCHO. Find
the alkylhalide.
CH3 Br CH3
| | |
CH3 – CH – CH – CH2 – CH3
���.��??????
CH3 – C = CH – CH2 – CH3 (X)
2-methylpent -2- ene
CH3
|
+ CH3 – CH– C = CH – CH3 (Y)
4-methylpent -2- ene
CH3 CH3
| |
CH3 – C = CH – CH2 – CH3
��
CH3 – C – O – CH – CH2 – CH3
2-methylpent -2- ene | |
O –– O
Zn / H2O
CH3 – CO – CH3 + CH3 – CH2 – CHO
Acetone Propanaldehyde
CH3 CH3
| |
CH3– CH - CH = CH – CH3
��
CH3– CH – CH – O – CH – CH3
4-methylpent -2- ene | |
O –– O
Zn / H2O
CH3
|
CH3 – CH – CHO + CH3 – CHO
Isobutanaldehyde Acetaldehyde
6. Describe the mechanism of Nitration of benzene.
Reagent Conc HNO3 + ConcH2SO4
Electrophiles -NO2
+
Total reaction
Nitrobenzene
Mechanism HNO3 + H2SO4 NO2
+
+ HSO4
-
+ H2O
7. How does Huckel rule help to decide the aromatic character of a compound
The molecule must be co-planar
Complete delocalization of π electron in the ring
Presence of (4n+2) π electrons in the ring where n is an integer (n=0,1,2….)
8. Suggest the route for the preparation of the following from benzene.
1) 3 – chloro nitrobenzene
NO2
|
+ HNO3
���.??????
���
� /����
Nitrobenzene
Cl
2
/ FeCl
3
Cl 3 – chloro nitrobenzene
2) 4 – chlorotoluene
CH3
|
+ CH3Cl
��������� ����
�
Toluene
Cl
2 / FeCl
3
4 – chlorotoluene
Cl
3) Bromo benzene
Br
Br
2
/ FeCl
3
����� �������
4) m - dinitro benzene NO2
|
+ HNO3
���.??????
���
� /����
Nitrobenzene
HNO
3
NO2 m - dinitro benzene
9. What happens when isobutylene is treated with acidified potassium permanganate ?
CH3 CH3
| |
CH3 – CH = CH2
��������� ????????????��4
CH3 – C = O
Isobutylene Acetone
10 Suggest a simple chemical test to distinguish propane and propene.
TEST PROPANE
(CH3-CH2-CH3)
PROPENE
(CH3-CH=CH2)
1,With bromine water No colour change Colourless
2, Addition Reaction No reaction Takes place
3, With Ozone No reaction Aldehyde Form
11
.
How will you convert ethyl chloride in to
i) Ethane
CH3 – CH2 – Cl
Pt / H2
CH3 – CH3
Ethyl chloride Ethane
ii) n – Butane
CH3 – CH2 – Cl + 2Na + CH3 – CH2 – Cl
Ether
CH3 – CH2 – CH2 – CH3 + 2NaCl
Ethyl chloride n – Butane
12
.
Describe the conformers of n - butane.
The two tetrahedral groups can rotate about the carbon – carbon bond axis yielding several
arrangements called conformers. The extreme conformations are staggered and eclipsed
conformation.
Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is
maximum repulsion between them and it is the least stable conformer.
Eclipsed conformations
Anti or staggered form
In this conformation, the distance between the two methyl groups is maximum and so there
is minimum repulsion between them. And it is the most stable conformer.
Nitrobenzene
Mechanism HNO3 + H2SO4 NO2
+
+ HSO4
-
+ H2O
7. How does Huckel rule help to decide the aromatic character of a compound
The molecule must be co-planar
Complete delocalization of π electron in the ring
Presence of (4n+2) π electrons in the ring where n is an integer (n=0,1,2….)
8. Suggest the route for the preparation of the following from benzene.
1) 3 – chloro nitrobenzene
NO2
|
+ HNO3
���.??????
���
� /����
Nitrobenzene
Cl
2
/ FeCl
3
Cl 3 – chloro nitrobenzene
2) 4 – chlorotoluene
CH3
|
+ CH3Cl
��������� ����
�
Toluene
Cl
2 / FeCl
3
4 – chlorotoluene
Cl
3) Bromo benzene
Br
Br
2
/ FeCl
3
����� �������
4) m - dinitro benzene NO2
|
+ HNO3
���.??????
���
� /����
Nitrobenzene
HNO
3
NO2 m - dinitro benzene
9. What happens when isobutylene is treated with acidified potassium permanganate ?
CH3 CH3
| |
CH3 – CH = CH2
��������� ????????????��4
CH3 – C = O
Isobutylene Acetone
10 Suggest a simple chemical test to distinguish propane and propene.
TEST PROPANE
(CH3-CH2-CH3)
PROPENE
(CH3-CH=CH2)
1,With bromine water No colour change Colourless
2, Addition Reaction No reaction Takes place
3, With Ozone No reaction Aldehyde Form
11
.
How will you convert ethyl chloride in to
i) Ethane
CH3 – CH2 – Cl
Pt / H2
CH3 – CH3
Ethyl chloride Ethane
ii) n – Butane
CH3 – CH2 – Cl + 2Na + CH3 – CH2 – Cl
Ether
CH3 – CH2 – CH2 – CH3 + 2NaCl
Ethyl chloride n – Butane
12
.
Describe the conformers of n - butane.
The two tetrahedral groups can rotate about the carbon – carbon bond axis yielding several
arrangements called conformers. The extreme conformations are staggered and eclipsed
conformation.
Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is
maximum repulsion between them and it is the least stable conformer.
Eclipsed conformations
Anti or staggered form
In this conformation, the distance between the two methyl groups is maximum and so there
is minimum repulsion between them. And it is the most stable conformer.
Gauche Planar Gauche
13
.
Write the chemical equations for combustion of propane
CH3 – CH2 – CH3+ O2
∆
3CO2 + 5H2O
14
.
Explain Markownikoff's rule with suitable example.
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon
that has more number of hydrogen and halogen add to the car- bon having fewer hydrogen”.
CH
3 – CH = CH
2 + HBr CH
3 – CH – CH
3
Propene |
Br 2- Bromopropane ( Main product)
15
.
Explain Anti-Markovnikoff ’s Rule (Or) Peroxide Effect (Or) KharaschAddition
The addition of HBr to an alkene in the presence of organic peroxide, gives the anti
Markovnikoff’s product. This effect is called peroxide effect (Or) KharaschAddition reaction.
CH
3 – CH = CH
2 + HBr
PEROXIDE
CH
3 – CH
2 – CH
2 – Br
Propene 1- Bromopropane ( Main product)
16
.
What happens when ethylene is passed through cold dilute alkaline potassium
permanganate.
CH2 = CH2 + H2O + (O)
���� ������ ??????��??????���� KMnO4
CH2-OH
Ethylene |
CH2-OH Ethyleneglycol
17
.
Write the structures of folowing alkanes.
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
CH3
|
CH3 CH3 CH2– CH – CH3
| | |
CH3 – CH – CH – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH3
2) 5 – (2 – Ethyl butyl) – 3, 3 – dimethyldecane
CH2 - CH3
|
CH3 CH2 – CH – CH2 – CH3
| |
CH3 – CH2 – C – CH2 – CH – CH2 – CH2 – CH2 – CH2 – CH3
|
CH3
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
CH3 CH3
| |
CH3 CH – CH – CH3
| |
CH3 – CH – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH3
18
.
How will you prepare propane from a sodium salt of fatty acid ?
CH3 – CH2 – CH2 – COONa
soda lime /
CH3 – CH2 – CH3 19
.
How will you distinguish 1 – butyne and 2 – butyne?
Test 1-butyne 2-butyne
1, With
AgNO3 + NH4OH
It gives silver butynide
No reaction
2, With
O3 + H2O/H2O2
It gives formic acid and propanoic acid It gives two molecules Acetic
acid.
20
.
CH3–CH(CH3)–CH(OH)–CH3
??????
+
/∆
(A) major product
HBr
(B) major product
Identify A and B
CH3 OH CH3
| | |
CH3 – CH – CH – CH3
??????
+
/∆
CH3 – CH – CH = CH2 (A)
3-methyl-1-butene
CH3 CH3 Br
| | |
CH3 – CH – CH = CH2
HBr
CH3 – CH – CH – CH3 (B)
3-methyl-1-butene 2-Bromo-3-methylbutane
21
.
Complete the following :
i) 2 – butyne
������� ��������
?
H H
CH3 C ≡ C CH3
Lindlar Catalyst
C = C
H3C CH3 Cis-2-butene
ii) CH2 = CH2
��
?
CH2 = CH2
I2
[CH2I CH2I]
− I2
CH2 = CH2
Unstable Ethelyne
iii) CH2 – CH2
| | + Zn / C2H5OH ?
Br Br
CH2 – CH2
| | + Zn/C2H5OH CH2 = CH2+ ZnBr2
Br Br Ethelyne
iv) CaC2
??????��
?
CaC2 + 2H2O CH ≡ CH + Ca(OH)2
Acetelyne
22
.
Write a notes on Wurtz reaction
When a solution of halo alkanes in dry ether is treated with sodium metalhigher alkanes
are produced
CH
3 Br + 2Na + CH
3-Br
Ether
CH
3 CH
3 + 2NaBr
(2 molecules methylbromide) Ethane
13
.
Write the chemical equations for combustion of propane
CH3 – CH2 – CH3+ O2
∆
3CO2 + 5H2O
14
.
Explain Markownikoff's rule with suitable example.
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon
that has more number of hydrogen and halogen add to the car- bon having fewer hydrogen”.
CH
3 – CH = CH
2 + HBr CH
3 – CH – CH
3
Propene |
Br 2- Bromopropane ( Main product)
15
.
Explain Anti-Markovnikoff ’s Rule (Or) Peroxide Effect (Or) KharaschAddition
The addition of HBr to an alkene in the presence of organic peroxide, gives the anti
Markovnikoff’s product. This effect is called peroxide effect (Or) KharaschAddition reaction.
CH
3 – CH = CH
2 + HBr
PEROXIDE
CH
3 – CH
2 – CH
2 – Br
Propene 1- Bromopropane ( Main product)
16
.
What happens when ethylene is passed through cold dilute alkaline potassium
permanganate.
CH2 = CH2 + H2O + (O)
���� ������ ??????��??????���� KMnO4
CH2-OH
Ethylene |
CH2-OH Ethyleneglycol
17
.
Write the structures of folowing alkanes.
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
CH3
|
CH3 CH3 CH2– CH – CH3
| | |
CH3 – CH – CH – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH3
2) 5 – (2 – Ethyl butyl) – 3, 3 – dimethyldecane
CH2 - CH3
|
CH3 CH2 – CH – CH2 – CH3
| |
CH3 – CH2 – C – CH2 – CH – CH2 – CH2 – CH2 – CH2 – CH3
|
CH3
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
CH3 CH3
| |
CH3 CH – CH – CH3
| |
CH3 – CH – CH2 – CH2 – CH – CH2 – CH2 – CH2 – CH3
18
.
How will you prepare propane from a sodium salt of fatty acid ?
CH3 – CH2 – CH2 – COONa
soda lime /
CH3 – CH2 – CH3 19
.
How will you distinguish 1 – butyne and 2 – butyne?
Test 1-butyne 2-butyne
1, With
AgNO3 + NH4OH
It gives silver butynide
No reaction
2, With
O3 + H2O/H2O2
It gives formic acid and propanoic acid It gives two molecules Acetic
acid.
20
.
CH3–CH(CH3)–CH(OH)–CH3
??????
+
/∆
(A) major product
HBr
(B) major product
Identify A and B
CH3 OH CH3
| | |
CH3 – CH – CH – CH3
??????
+
/∆
CH3 – CH – CH = CH2 (A)
3-methyl-1-butene
CH3 CH3 Br
| | |
CH3 – CH – CH = CH2
HBr
CH3 – CH – CH – CH3 (B)
3-methyl-1-butene 2-Bromo-3-methylbutane
21
.
Complete the following :
i) 2 – butyne
������� ��������
?
H H
CH3 C ≡ C CH3
Lindlar Catalyst
C = C
H3C CH3 Cis-2-butene
ii) CH2 = CH2
��
?
CH2 = CH2
I2
[CH2I CH2I]
− I2
CH2 = CH2
Unstable Ethelyne
iii) CH2 – CH2
| | + Zn / C2H5OH ?
Br Br
CH2 – CH2
| | + Zn/C2H5OH CH2 = CH2+ ZnBr2
Br Br Ethelyne
iv) CaC2
??????��
?
CaC2 + 2H2O CH ≡ CH + Ca(OH)2
Acetelyne
22
.
Write a notes on Wurtz reaction
When a solution of halo alkanes in dry ether is treated with sodium metalhigher alkanes
are produced
CH
3 Br + 2Na + CH
3-Br
Ether
CH
3 CH
3 + 2NaBr
(2 molecules methylbromide) Ethane
23
.
Write a notes on Corey- House Mechanism
An alkyl halide and lithium dialkylcuprate are reacted to give higher alkane.
CH3CH2 Br + (CH3)2LiCu CH3CH2CH3 + CH3CuLiBr
24
.
Explain Birch reduction reaction
25
.
Write a notes on Wurtz – Fittig Reaction
C6H5 – Br + 2Na + CH3 – I
??????����
C6H5 – CH3 + 2NaI
Bromobenzene Methylbromide Tolune
26
.
Write a notes on Friedel Craft’s Alkylation.
C6H5 – Br + CH3 – Cl
??????��??????� ??????�??????��
C6H5 – CH3 + HCl
Bromobenzene Methylchloride Tolune
27 How will you prepare ethene from Kolbe’s method?
CH3COONa + 2H2O
??????�������??????���
CH3 – CH3 + 2CO2 + H2 + 2NaOH
At anode At cathode
28 What happens Methane react with Steam ?
CH4(g) + H2O(g)
�� / �����
CO(g) + 3H2(g)
29
.
Define Isomerisation reaction
Isomerisation is a chemical process by which a compound is transformed into any its isomeric
forms.
CH3 – CH2 – CH2 –CH3
����� / ??????�� / ����
CH3 – CH – CH3
n – Butane Isobutane
CH3
30
.
How will you prepare Alkene from Vicinal dihalide ?
CH2 – Br CH2 – Br CH2
�� / �??????��?????? / ∆
+ ZnBr2
CH2 – Br CH2 – ZnBr CH2
1,2- dibromoethane Ethene (Alkene)
31
.
Explain the preparation of Benzene from coal tar ?
During fractional distillation, coal tar is heated and distills away its volatile compounds
namely benzene, toluene, xylene in the temperature range
NAME OF THE FRACTION
TEMPERATURE
NAME OF THE COMPENENTS
1, Crude light oil 350 – 443 K Benzene, Toluene, Xylenes
2, Middle oil 443 – 503 K Phenol, Naphthalene
3, Heavy oil 503 – 543 K Naphthalene , Cresol
4, Green oil 543 – 633 K Anthracene
5, Pitch 633 K Residue
32
.
Convert Acetelyne to Benzene
Acetylene on passing through a red –hot tube trimerises to give benzene
3CH CH
red –hot tube
C6H6
Acetylene benzene 33
.
How will you prepare ethylene from Kolbe’s method?
CH2 – COOK CH2 – COO
-
??????�������??????���
+ 2K
+
CH2 – COOK CH2 – COO
-
At anode
CH2 – COO
-
CH2
+ CO2 + 2e
-
CH2 – COO
-
CH2
Ethylene (Ethene)
34
.
Write a note on ortho and para directing groups.
Those which increases electron density at „ortho‟ and „para‟ position are known as ortho-
para directors
Eg :- –OH, –NH2, –NHR, –NHCOCH 3.
35
.
Write a note on meta directing groups
Those which increases electron density at „meta‟ position is known as meta director.
Eg :- –NO2, –CN, –CHO, –COOH,
With meta directing groups positive charge is created on the ortho and
para positions, but the meta positions are free of such charges, and hence
these positions are more reactive than the ortho and para position.
36
.
Structure of benzene :
Molecular formula C6H6. This indicates it is highly unsaturated.
As it does not show the properties of alkanes, it should not be a straight chain compound.
X-ray and electron diffraction studies inidicated that all carbon-carbon bonds are of equal length 1.39 Å
which is in between that of a single bond (1.54 Å) and that of a double bond (1.34 Å).
As it gives mono bromo benzene, all six hydrogen atoms are identical. It is possible only if benzene
is cyclic.
Benzene reacts with hydrogen (catalyst Raney Nickel) to give cyclohexane. So there are three
double bonds.
Benzene has resonance structures as follows:
Delocalisation of electrons in benzene
14. HALOALKANES AND HALOARENES
1. Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides.
a, CH3 – CH = CH – Cl vinylhalide
b, C6H5 – CH2 – I benzylichalide
c, CH3 – CH – CH3
Br
alkylhalide
d, CH2 = CH – Cl vinylhalide
2. Why chlorination of methane is not possible in dark?
Chlorination of methane is Free Radical substitution reaction.
Chlorine is notable to convertinto free radicals in the dark.
.
Write a notes on Corey- House Mechanism
An alkyl halide and lithium dialkylcuprate are reacted to give higher alkane.
CH3CH2 Br + (CH3)2LiCu CH3CH2CH3 + CH3CuLiBr
24
.
Explain Birch reduction reaction
25
.
Write a notes on Wurtz – Fittig Reaction
C6H5 – Br + 2Na + CH3 – I
??????����
C6H5 – CH3 + 2NaI
Bromobenzene Methylbromide Tolune
26
.
Write a notes on Friedel Craft’s Alkylation.
C6H5 – Br + CH3 – Cl
??????��??????� ??????�??????��
C6H5 – CH3 + HCl
Bromobenzene Methylchloride Tolune
27 How will you prepare ethene from Kolbe’s method?
CH3COONa + 2H2O
??????�������??????���
CH3 – CH3 + 2CO2 + H2 + 2NaOH
At anode At cathode
28 What happens Methane react with Steam ?
CH4(g) + H2O(g)
�� / �����
CO(g) + 3H2(g)
29
.
Define Isomerisation reaction
Isomerisation is a chemical process by which a compound is transformed into any its isomeric
forms.
CH3 – CH2 – CH2 –CH3
����� / ??????�� / ����
CH3 – CH – CH3
n – Butane Isobutane
CH3
30
.
How will you prepare Alkene from Vicinal dihalide ?
CH2 – Br CH2 – Br CH2
�� / �??????��?????? / ∆
+ ZnBr2
CH2 – Br CH2 – ZnBr CH2
1,2- dibromoethane Ethene (Alkene)
31
.
Explain the preparation of Benzene from coal tar ?
During fractional distillation, coal tar is heated and distills away its volatile compounds
namely benzene, toluene, xylene in the temperature range
NAME OF THE FRACTION
TEMPERATURE
NAME OF THE COMPENENTS
1, Crude light oil 350 – 443 K Benzene, Toluene, Xylenes
2, Middle oil 443 – 503 K Phenol, Naphthalene
3, Heavy oil 503 – 543 K Naphthalene , Cresol
4, Green oil 543 – 633 K Anthracene
5, Pitch 633 K Residue
32
.
Convert Acetelyne to Benzene
Acetylene on passing through a red –hot tube trimerises to give benzene
3CH CH
red –hot tube
C6H6
Acetylene benzene 33
.
How will you prepare ethylene from Kolbe’s method?
CH2 – COOK CH2 – COO
-
??????�������??????���
+ 2K
+
CH2 – COOK CH2 – COO
-
At anode
CH2 – COO
-
CH2
+ CO2 + 2e
-
CH2 – COO
-
CH2
Ethylene (Ethene)
34
.
Write a note on ortho and para directing groups.
Those which increases electron density at „ortho‟ and „para‟ position are known as ortho-
para directors
Eg :- –OH, –NH2, –NHR, –NHCOCH 3.
35
.
Write a note on meta directing groups
Those which increases electron density at „meta‟ position is known as meta director.
Eg :- –NO2, –CN, –CHO, –COOH,
With meta directing groups positive charge is created on the ortho and
para positions, but the meta positions are free of such charges, and hence
these positions are more reactive than the ortho and para position.
36
.
Structure of benzene :
Molecular formula C6H6. This indicates it is highly unsaturated.
As it does not show the properties of alkanes, it should not be a straight chain compound.
X-ray and electron diffraction studies inidicated that all carbon-carbon bonds are of equal length 1.39 Å
which is in between that of a single bond (1.54 Å) and that of a double bond (1.34 Å).
As it gives mono bromo benzene, all six hydrogen atoms are identical. It is possible only if benzene
is cyclic.
Benzene reacts with hydrogen (catalyst Raney Nickel) to give cyclohexane. So there are three
double bonds.
Benzene has resonance structures as follows:
Delocalisation of electrons in benzene
14. HALOALKANES AND HALOARENES
1. Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides.
a, CH3 – CH = CH – Cl vinylhalide
b, C6H5 – CH2 – I benzylichalide
c, CH3 – CH – CH3
Br
alkylhalide
d, CH2 = CH – Cl vinylhalide
2. Why chlorination of methane is not possible in dark?
Chlorination of methane is Free Radical substitution reaction.
Chlorine is notable to convertinto free radicals in the dark.
3. How will you prepare n propyl iodide from n-propyl bromide?
CH3 – CH2 – CH2 – Br + NaBr
Acetone
CH3 – CH2 –CH2 –I + NaBr
n -propylbromide n –propyl iodide
Reaction Name : - Finkelstein reaction.
4. Which alkyl halide from the following pair is i) chiral ii) undergoes faster SN2 reaction?
i) chiral
2-Bromo butane
ii) undergoes faster SN2 reaction
4-Chloro butane
5. How does chlorobenzene react with sodium in the presence of ether? What is the name of
the reaction?
C6H5 – Cl + 2Na + C6H5 – Cl
??????����
C6H5 – C6H5 + 2NaI
Chlorobenzene Chlorobenzene Biphenyl
Reaction Name : -Fittig Reaction
6. Give reasons for polarity of C-X bond in halo alkane.
Carbon Halogen bond is a Polar bond as halogens are more electronegative than carbon.
The carbon atom exhibits a partial positive charges (
+
) and halogen atom a partial negative
charge (
)
. +
R – CH2 – X
7. Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Grignard reagent react with water to give Alkane.
R – Mg – X + H – OH R – H + X – Mg – OH
Grignard reagent water Alkane
8. What happens when acetyl chloride is treated with excess of CH3MgI?
Acetyl chloride is react with excess of methyl magnesium iodide to give tertiary butyl alcohol.
O O–MgI O
CH3 –MgI+ CH3 –C–Cl
??????����
CH3–C–CH3
���/�
+
CH3 – C– CH3 + Mg(Cl)I
Acetyl chloride Acetone
O O–MgBr OH
CH3–MgI + CH3 –C–CH3
??????����
CH3–C–CH3
??????��/??????
+
CH3 – C– CH3 +Mg(OH)I
Acetone
CH3 CH3
tertiary butyl alcohol
9. Arrange the following alkyl halide in increasing order of bond enthalpy of RX,CH3Br,
CH3F, CH3Cl, CH3I.
The order is CH3I < CH3Br < CH3Cl < CH3F
10
.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Chloroform reacts with oxygen in the presence of sunlightto give Phoshgene.
2CHCl3 + O2
��� �����
2 COCl2 + 2HCl
Phoshgene. 11
.
Write down the possible isomers of C5H11Br and give their IUPAC and common names.
Molecular Formula Common Name IUPACName
CH3–CH2–CH2–CH2– CH2–Br n-pentyl bromide 1-Bromo pentane
CH3–CH2–CH2–CH– CH3
Br
2-pentylbromide 2- Bromo pentane
CH3–CH2–CH–CH2– CH3
Br
3-pentylbromide 3- Bromo pentane
CH3 – CH– CH2 – CH2 – Br
CH3
Iso pentylbromide 1- Bromo -3-methyl butane
CH3 – CH – CH – CH3
CH3 Br
Sec iso pentyl bromide 2- Bromo -3-methyl butane
CH3 – CH2 – CH – CH2 – Br
CH3
2-Methyl butylbromide 1- Bromo -2-methyl butane
Br
CH3 – CH2 – C – CH3
CH3
Tertiary pentyl bromide 2- Bromo -2-methyl butane
CH3
CH3 – C– CH2 –Br
CH3
1-Neo pentyl bromide 1- Bromo -2,2’-dimethyl propane
12
.
Mention any three methods of preparation of haloalkanes from alcohols.
i, Reaction with Hydrogen halide :-
CH3–CH2–OH + HCl
Anhydrous ZnCl2 / ∆
CH3–CH2–Cl + H2O
Ethyl alcohol Ethylchloride
ii, Reaction with Phosphorous Halide :-
CH3–CH2–OH + PCl5 CH3–CH2–Cl + POCl3 + HCl
Ethyl alcohol Ethylchloride
iii, Reaction with Thionyl chloride :-
CH3–CH2–OH + SOCl2
Pyridine
CH3–CH2–Cl + SO2 + HCl
Ethyl alcohol Ethylchloride
13
.
Compare SN1 and SN2 reaction mechanisms.
SN1 SN2
1. It is a Unimolecular reaction 1. It is a bimolecular reaction
2. It follows a First order kinetic mechanism 2. It follows the second order kinetic mechanism
3. It involves two steps 3. It is a single step process
4. The rate of the reaction depends on the
concentration of the substrate.
4. The rate of the reaction is depends on the
concentration of both the substrate and
nucleophile.
5. The substrate form carbon cation
intermediate.
5. Dose not form carbon cation intermediate.
6. Optically active substrate becomes
optically inactive.
6. Inversion in configuration.
CH3 – CH2 – CH2 – Br + NaBr
Acetone
CH3 – CH2 –CH2 –I + NaBr
n -propylbromide n –propyl iodide
Reaction Name : - Finkelstein reaction.
4. Which alkyl halide from the following pair is i) chiral ii) undergoes faster SN2 reaction?
i) chiral
2-Bromo butane
ii) undergoes faster SN2 reaction
4-Chloro butane
5. How does chlorobenzene react with sodium in the presence of ether? What is the name of
the reaction?
C6H5 – Cl + 2Na + C6H5 – Cl
??????����
C6H5 – C6H5 + 2NaI
Chlorobenzene Chlorobenzene Biphenyl
Reaction Name : -Fittig Reaction
6. Give reasons for polarity of C-X bond in halo alkane.
Carbon Halogen bond is a Polar bond as halogens are more electronegative than carbon.
The carbon atom exhibits a partial positive charges (
+
) and halogen atom a partial negative
charge (
)
. +
R – CH2 – X
7. Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Grignard reagent react with water to give Alkane.
R – Mg – X + H – OH R – H + X – Mg – OH
Grignard reagent water Alkane
8. What happens when acetyl chloride is treated with excess of CH3MgI?
Acetyl chloride is react with excess of methyl magnesium iodide to give tertiary butyl alcohol.
O O–MgI O
CH3 –MgI+ CH3 –C–Cl
??????����
CH3–C–CH3
���/�
+
CH3 – C– CH3 + Mg(Cl)I
Acetyl chloride Acetone
O O–MgBr OH
CH3–MgI + CH3 –C–CH3
??????����
CH3–C–CH3
??????��/??????
+
CH3 – C– CH3 +Mg(OH)I
Acetone
CH3 CH3
tertiary butyl alcohol
9. Arrange the following alkyl halide in increasing order of bond enthalpy of RX,CH3Br,
CH3F, CH3Cl, CH3I.
The order is CH3I < CH3Br < CH3Cl < CH3F
10
.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Chloroform reacts with oxygen in the presence of sunlightto give Phoshgene.
2CHCl3 + O2
��� �����
2 COCl2 + 2HCl
Phoshgene. 11
.
Write down the possible isomers of C5H11Br and give their IUPAC and common names.
Molecular Formula Common Name IUPACName
CH3–CH2–CH2–CH2– CH2–Br n-pentyl bromide 1-Bromo pentane
CH3–CH2–CH2–CH– CH3
Br
2-pentylbromide 2- Bromo pentane
CH3–CH2–CH–CH2– CH3
Br
3-pentylbromide 3- Bromo pentane
CH3 – CH– CH2 – CH2 – Br
CH3
Iso pentylbromide 1- Bromo -3-methyl butane
CH3 – CH – CH – CH3
CH3 Br
Sec iso pentyl bromide 2- Bromo -3-methyl butane
CH3 – CH2 – CH – CH2 – Br
CH3
2-Methyl butylbromide 1- Bromo -2-methyl butane
Br
CH3 – CH2 – C – CH3
CH3
Tertiary pentyl bromide 2- Bromo -2-methyl butane
CH3
CH3 – C– CH2 –Br
CH3
1-Neo pentyl bromide 1- Bromo -2,2’-dimethyl propane
12
.
Mention any three methods of preparation of haloalkanes from alcohols.
i, Reaction with Hydrogen halide :-
CH3–CH2–OH + HCl
Anhydrous ZnCl2 / ∆
CH3–CH2–Cl + H2O
Ethyl alcohol Ethylchloride
ii, Reaction with Phosphorous Halide :-
CH3–CH2–OH + PCl5 CH3–CH2–Cl + POCl3 + HCl
Ethyl alcohol Ethylchloride
iii, Reaction with Thionyl chloride :-
CH3–CH2–OH + SOCl2
Pyridine
CH3–CH2–Cl + SO2 + HCl
Ethyl alcohol Ethylchloride
13
.
Compare SN1 and SN2 reaction mechanisms.
SN1 SN2
1. It is a Unimolecular reaction 1. It is a bimolecular reaction
2. It follows a First order kinetic mechanism 2. It follows the second order kinetic mechanism
3. It involves two steps 3. It is a single step process
4. The rate of the reaction depends on the
concentration of the substrate.
4. The rate of the reaction is depends on the
concentration of both the substrate and
nucleophile.
5. The substrate form carbon cation
intermediate.
5. Dose not form carbon cation intermediate.
6. Optically active substrate becomes
optically inactive.
6. Inversion in configuration.
14
.
Reagents and the conditions used in the reactions are given below. Complete the table by
writing down the product and the name of the reaction.
Reaction Product Name of the reaction
CH3–CH2–OH + SOCl2
��????????????�??????��
CH3–CH2–Cl + SO2 + HCl
CH3–CH2–Cl
Ethyl chloride
Darzen’s reaction.
CH3 – CH2 –Br + AgF
CH3 – CH2 –F + AgBr
CH3 – CH2 –F
Ethyl fluride
Swartz reaction
C6H5–Cl +2Na+ C6H5– Cl
��ℎ�??????
C6H5 –C6H5 + 2NaI
C6H5 –C6H5
Bi phenyl
Fittig reaction.
15
.
Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.
Chlorobenzene does not give nucleophilic substitution reaction.
C- Cl bond in chlorobenzene is Short and Strong.
So it is substituted by OH
-
, NH2
-
(or) CN
-
with appropriate nucleophilic reagents at high
temperature and pressure.
Ex :-
Chlorobenzene Phenol
Chlorobenzene Phenylcyanide
16
.
Account for the following
(i) tertiary-butyl chloride reacts with aqueous KOH by SN1 mechanism while
n-butyl chloride reacts with SN2 mechanism.
SN1 reaction proceeds through the formation of carbocation.
The tertiary-butylchloride readily loses Cl
-
ion to form stable tertiary carbocation.
It prefers to undergo SN1 mechanism.
n-Butyl chloride does not undergo ionization to form n-butyl carbocation,because it is unstable.
It prefers to undergo SN2 mechanism.
(ii) p-dichloro benzene has higher melting point than those of o-and m-dichloro benzene.
It is due to its symmetry which leads to more close packing of its molecules in the crystal lattice
and consequently strong intermolecular attractive force which requires more energy for melting.
17
.
In an experiment ethyl iodide in ether is allowed to stand over magnesium pieces.
Magnesium dissolves and product is formed
a) Name the product and write the equation for the reaction.
C2H5I + Mg
��ℎ�??????
C2H5MgI (Methyl magnesium iodide)
b) Why all the reagents used in the reaction should be dry? Explain
Because the reagent react with water to give alkane.
c) How is acetone prepared from the product obtained in the experiment.
O O–MgBr O
CH3–MgI + CH3 –C–Cl
??????����
CH3–C–CH3
���/�
+
CH3– C– CH3 + Mg(Cl)I
Acetyl chloride Acetone
18
.
Write a chemical reaction useful to prepare the following:
i) Freon-12 from Carbon tetrachloride
Carbon tetrachloride Freon-12
ii) Carbon tetrachloride from carbon disulphide
Carbon disulphide Carbon tetrachloride
19
.
What are Freons? Discuss their uses and environmental effects
Freons :-The chlorofluro derivatives of methane and ethane are called freons.
Uses :-
Freons are used as a refrigerants in refrigerators and air conditioners.
It is used as a propellant for aerosols and foams.
It is used as propellant for foams to spray out deodorants, shaving creams, and
insecticides.
Environmental effects :- Highly toxic in nature and affects nervous system.
20
.
Predict the products when bromoethane is treated with the following i) KNO2 ii) AgNO2
i) KNO2 :-
CH3 – CH2 – Br + KNO2 CH3 – CH2 – O – N = O + KBr
Bromoethane Ethyl nitride
ii) AgNO2 :-
CH3 – CH2 – Br + AgNO2 CH3 – CH2 –NO2 + AgBr
Bromoethane Nitro ethane
21
.
Explain the mechanism of SN1 reaction by highlighting the stereochemistry behind it
It is a Unimolecular reaction
It follows a First order kinetic mechanism.
It involves two steps.
The rate of the reaction depends on the concentration of the substrate.
The substrate form carbon cation intermediate.
Optically active substrate becomes optically inactive.
Eg :- Reaction between tertiary butyl bromide with aqueous KOH.
Step -1 :-
Carbon cation
Step -2 :-
22
.
Write short notes on the the following
i) Raschig process ii) Dows Process iii) Darzens process
i) Raschig process
C6H5–H + HCl + ½O2
����
� /����
C6H5 – Cl + H2O
Benzene Chlorobenzene
ii) Dows Process C6H5 –Cl + NaOH
��� ��� /���
�
�
C6H5 – OH + NaCl
Chlorobenzene Phenol
iii) Darzens
process
C6H5 – OH + SOCl2
�??????������
C6H5 –Cl + SO2 + HCl
Phenol Chlorobenzene
.
Reagents and the conditions used in the reactions are given below. Complete the table by
writing down the product and the name of the reaction.
Reaction Product Name of the reaction
CH3–CH2–OH + SOCl2
��????????????�??????��
CH3–CH2–Cl + SO2 + HCl
CH3–CH2–Cl
Ethyl chloride
Darzen’s reaction.
CH3 – CH2 –Br + AgF
CH3 – CH2 –F + AgBr
CH3 – CH2 –F
Ethyl fluride
Swartz reaction
C6H5–Cl +2Na+ C6H5– Cl
��ℎ�??????
C6H5 –C6H5 + 2NaI
C6H5 –C6H5
Bi phenyl
Fittig reaction.
15
.
Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.
Chlorobenzene does not give nucleophilic substitution reaction.
C- Cl bond in chlorobenzene is Short and Strong.
So it is substituted by OH
-
, NH2
-
(or) CN
-
with appropriate nucleophilic reagents at high
temperature and pressure.
Ex :-
Chlorobenzene Phenol
Chlorobenzene Phenylcyanide
16
.
Account for the following
(i) tertiary-butyl chloride reacts with aqueous KOH by SN1 mechanism while
n-butyl chloride reacts with SN2 mechanism.
SN1 reaction proceeds through the formation of carbocation.
The tertiary-butylchloride readily loses Cl
-
ion to form stable tertiary carbocation.
It prefers to undergo SN1 mechanism.
n-Butyl chloride does not undergo ionization to form n-butyl carbocation,because it is unstable.
It prefers to undergo SN2 mechanism.
(ii) p-dichloro benzene has higher melting point than those of o-and m-dichloro benzene.
It is due to its symmetry which leads to more close packing of its molecules in the crystal lattice
and consequently strong intermolecular attractive force which requires more energy for melting.
17
.
In an experiment ethyl iodide in ether is allowed to stand over magnesium pieces.
Magnesium dissolves and product is formed
a) Name the product and write the equation for the reaction.
C2H5I + Mg
��ℎ�??????
C2H5MgI (Methyl magnesium iodide)
b) Why all the reagents used in the reaction should be dry? Explain
Because the reagent react with water to give alkane.
c) How is acetone prepared from the product obtained in the experiment.
O O–MgBr O
CH3–MgI + CH3 –C–Cl
??????����
CH3–C–CH3
���/�
+
CH3– C– CH3 + Mg(Cl)I
Acetyl chloride Acetone
18
.
Write a chemical reaction useful to prepare the following:
i) Freon-12 from Carbon tetrachloride
Carbon tetrachloride Freon-12
ii) Carbon tetrachloride from carbon disulphide
Carbon disulphide Carbon tetrachloride
19
.
What are Freons? Discuss their uses and environmental effects
Freons :-The chlorofluro derivatives of methane and ethane are called freons.
Uses :-
Freons are used as a refrigerants in refrigerators and air conditioners.
It is used as a propellant for aerosols and foams.
It is used as propellant for foams to spray out deodorants, shaving creams, and
insecticides.
Environmental effects :- Highly toxic in nature and affects nervous system.
20
.
Predict the products when bromoethane is treated with the following i) KNO2 ii) AgNO2
i) KNO2 :-
CH3 – CH2 – Br + KNO2 CH3 – CH2 – O – N = O + KBr
Bromoethane Ethyl nitride
ii) AgNO2 :-
CH3 – CH2 – Br + AgNO2 CH3 – CH2 –NO2 + AgBr
Bromoethane Nitro ethane
21
.
Explain the mechanism of SN1 reaction by highlighting the stereochemistry behind it
It is a Unimolecular reaction
It follows a First order kinetic mechanism.
It involves two steps.
The rate of the reaction depends on the concentration of the substrate.
The substrate form carbon cation intermediate.
Optically active substrate becomes optically inactive.
Eg :- Reaction between tertiary butyl bromide with aqueous KOH.
Step -1 :-
Carbon cation
Step -2 :-
22
.
Write short notes on the the following
i) Raschig process ii) Dows Process iii) Darzens process
i) Raschig process
C6H5–H + HCl + ½O2
����
� /����
C6H5 – Cl + H2O
Benzene Chlorobenzene
ii) Dows Process C6H5 –Cl + NaOH
��� ��� /���
�
�
C6H5 – OH + NaCl
Chlorobenzene Phenol
iii) Darzens
process
C6H5 – OH + SOCl2
�??????������
C6H5 –Cl + SO2 + HCl
Phenol Chlorobenzene
23
.
Starting from CH3MgI, How will you prepare the following?
i) Acetic acid, ii) Acetone, iii) Ethyl acetate, iv) Iso propyl alcohol, v) Methyl cyanide
i) Acetic acid:-
O O – MgI OH
CH3 –MgI + C = O CH3 – C = O
??????��/??????
+
CH3 – C = O + Mg(OH)I
Carbon dioxide Acetic acid
ii) Acetone:-
O O– MgI O
CH3 –MgI + CH3–C –Cl CH3 – C – Cl
??????��/??????
+
CH3 – C – CH3 + Mg(Cl)I
Acetyl chloride CH3 Acetone
iii) Ethyl acetate:-
O O– MgI O
CH3 –MgI + C2H5O–C –Cl C2H5O– C – Cl
??????��/??????
+
C2H5O – C – CH3
Ethyl chloroformate CH3 Ethyl acetate + Mg(Cl)I
iv) Iso propyl alcohol:-
O O –MgI OH
CH3 –MgI + CH3–C –H CH3 – C – H
??????��/??????
+
CH3 – C – H + Mg(OH)I
Acetaldehyde CH3 CH3 Iso propyl alcohol
v) Methyl cyanide:-
CH3 –MgI + CNCl CH3 – CN + Mg(Cl)I
Cyanogen chloride Methyl cyanide
24
.
Complete the following reactions
i) CH3 – CH = CH2 + HBr
Peroxide
?
CH3 – CH = CH2 + HBr
Peroxide
CH3 – CH2 – CH2 –Br (n-propyl bromide)
ii) CH3 – CH2 –Br + NaSH
Alcohol / H
2O
?
CH3 – CH2 –Br + NaSH
Alcohol / H
2O
CH3 – CH2 – SH + NaBr
(Ethyl thioalcohol)
iii) C6H5Cl + Mg
�????????????
?
C6H5Cl + Mg
�????????????
C6H5MgCl (Phenyl magnesium chloride)
iv) CHCl3 + HNO3
∆
?
CHCl3 + HNO3
∆
CCl3NO2 + H2O (Chloropicrin)
iv) CCl4 + H2O
∆
?
CCl4 + H2O
∆
COCl2 + 2HCl (Phoshgene)
25
.
Explain the preparation of the following compounds
i) DDT, ii) Chloroform, iii) Biphenyl, iv) Chloropicrin&v) Freon-12
i) DDT :-
(p,p‟-DichloroDiphenylTrichloroethane)
ii) Chloroform:-
This reaction proceeds in three steps as shown below.
Step -1 :- (Oxidation)
CH3 – CH2 – OH + Cl2 CH3 – CHO + 2HCl
Ethyl alcohol Acetaldehyde
Step -1 :- (Chlorination)
CH3 – CHO + 3Cl2 CCl3 – CHO + 3HCl
Acetaldehyde Trichloro acetaldehyde
Step -1 :- ( Hyrolysis)
CCl3 – CHO + Ca(OH)2 2CHCl3 + (HCOO)2Ca
Trichloro acetaldehyde Chloroform
iii) Biphenyl:-
C6H5–Cl +2Na+ C6H5– Cl
Ether
C6H5 –C6H5 + 2NaI (Fittig Reaction)
Chloro benzene Biphenyl
v) Freon-12:-
Carbon tetrachloride Freon - 12
26
.
An organic compound (A) with molecular formula C2H5Cl reacts with KOH gives
compounds (B) and with alcoholic KOH gives compound (C). Identify (A),(B), and (C).
CH3 – CH2– Cl
dil KOH
CH3 – CH2– OH + KCl
(A) (B)
CH3 – CH2– OH
Alcoholic KOH
CH2 = CH2 + KCl + H2O
(B) (C)
Compound Molecular Formula Name
(A) CH3 – CH2 – Cl Ethyl chloride
(B) CH3 – CH2 – OH Ethyl alcohol
(C) CH2 = CH2 Ethylene
27
.
Simplest alkene (A) reacts with HCl to form compound (B).Compound (B) reacts with
ammonia to form compound (C) of molecular formula C2H7N.Compound (C) undergoes
carbylamine test. Identify (A), (B), and (C).
CH2 = CH2 + HCl CH3 – CH2– Cl
(A) (B)
CH3 – CH2– Cl + NH3 CH3 – CH2 – NH2 + HCl
(B) (C)
.
Starting from CH3MgI, How will you prepare the following?
i) Acetic acid, ii) Acetone, iii) Ethyl acetate, iv) Iso propyl alcohol, v) Methyl cyanide
i) Acetic acid:-
O O – MgI OH
CH3 –MgI + C = O CH3 – C = O
??????��/??????
+
CH3 – C = O + Mg(OH)I
Carbon dioxide Acetic acid
ii) Acetone:-
O O– MgI O
CH3 –MgI + CH3–C –Cl CH3 – C – Cl
??????��/??????
+
CH3 – C – CH3 + Mg(Cl)I
Acetyl chloride CH3 Acetone
iii) Ethyl acetate:-
O O– MgI O
CH3 –MgI + C2H5O–C –Cl C2H5O– C – Cl
??????��/??????
+
C2H5O – C – CH3
Ethyl chloroformate CH3 Ethyl acetate + Mg(Cl)I
iv) Iso propyl alcohol:-
O O –MgI OH
CH3 –MgI + CH3–C –H CH3 – C – H
??????��/??????
+
CH3 – C – H + Mg(OH)I
Acetaldehyde CH3 CH3 Iso propyl alcohol
v) Methyl cyanide:-
CH3 –MgI + CNCl CH3 – CN + Mg(Cl)I
Cyanogen chloride Methyl cyanide
24
.
Complete the following reactions
i) CH3 – CH = CH2 + HBr
Peroxide
?
CH3 – CH = CH2 + HBr
Peroxide
CH3 – CH2 – CH2 –Br (n-propyl bromide)
ii) CH3 – CH2 –Br + NaSH
Alcohol / H
2O
?
CH3 – CH2 –Br + NaSH
Alcohol / H
2O
CH3 – CH2 – SH + NaBr
(Ethyl thioalcohol)
iii) C6H5Cl + Mg
�????????????
?
C6H5Cl + Mg
�????????????
C6H5MgCl (Phenyl magnesium chloride)
iv) CHCl3 + HNO3
∆
?
CHCl3 + HNO3
∆
CCl3NO2 + H2O (Chloropicrin)
iv) CCl4 + H2O
∆
?
CCl4 + H2O
∆
COCl2 + 2HCl (Phoshgene)
25
.
Explain the preparation of the following compounds
i) DDT, ii) Chloroform, iii) Biphenyl, iv) Chloropicrin&v) Freon-12
i) DDT :-
(p,p‟-DichloroDiphenylTrichloroethane)
ii) Chloroform:-
This reaction proceeds in three steps as shown below.
Step -1 :- (Oxidation)
CH3 – CH2 – OH + Cl2 CH3 – CHO + 2HCl
Ethyl alcohol Acetaldehyde
Step -1 :- (Chlorination)
CH3 – CHO + 3Cl2 CCl3 – CHO + 3HCl
Acetaldehyde Trichloro acetaldehyde
Step -1 :- ( Hyrolysis)
CCl3 – CHO + Ca(OH)2 2CHCl3 + (HCOO)2Ca
Trichloro acetaldehyde Chloroform
iii) Biphenyl:-
C6H5–Cl +2Na+ C6H5– Cl
Ether
C6H5 –C6H5 + 2NaI (Fittig Reaction)
Chloro benzene Biphenyl
v) Freon-12:-
Carbon tetrachloride Freon - 12
26
.
An organic compound (A) with molecular formula C2H5Cl reacts with KOH gives
compounds (B) and with alcoholic KOH gives compound (C). Identify (A),(B), and (C).
CH3 – CH2– Cl
dil KOH
CH3 – CH2– OH + KCl
(A) (B)
CH3 – CH2– OH
Alcoholic KOH
CH2 = CH2 + KCl + H2O
(B) (C)
Compound Molecular Formula Name
(A) CH3 – CH2 – Cl Ethyl chloride
(B) CH3 – CH2 – OH Ethyl alcohol
(C) CH2 = CH2 Ethylene
27
.
Simplest alkene (A) reacts with HCl to form compound (B).Compound (B) reacts with
ammonia to form compound (C) of molecular formula C2H7N.Compound (C) undergoes
carbylamine test. Identify (A), (B), and (C).
CH2 = CH2 + HCl CH3 – CH2– Cl
(A) (B)
CH3 – CH2– Cl + NH3 CH3 – CH2 – NH2 + HCl
(B) (C)
Compound Molecular Formula Name
(A) CH2 = CH2 Ethylene
(B) CH3 – CH2– Cl Ethyl chloride
(C) CH3 – CH2– NH2 Ethyl amine
28
.
A hydrocarbon C3H6 (A) reacts with HBr to form compound (B). Compound (B) reacts
with aqueous potassium hydroxide to give (C) of molecular formula C3H6O.what are (A)
(B) and (C). Explain the reactions.
CH3 –CH= CH2 + HBr CH3 – CHBr– CH3
(A) (B)
CH3 – CHBr– CH3
dil KOH
CH3 – CHOH– CH3 + KBr
(B) (C)
Compound Molecular Formula Name
(A) CH3 –CH= CH2 Propylene
(B) CH3 – CHBr– CH3 Isopropyl bromide
(C) CH3 – CHOH– CH3 Isopropyl alcohol
29
.
Two isomers (A) and (B) have the same molecular formula C2H4Cl2. Compound (A) reacts
with aqueous KOH gives compound (C) of molecular formula C2H4O. Compound (B)
reacts with aqueous KOH gives compound (D) of molecular formula C2H6O2. Identify
(A),(B),(C) and (D).
CH3-CHCl2 + 2KOH CH3 – CH(OH)2 + 2KCl
(A)-H2O
CH3 – CHO (C)
CH2–Cl CH2–OH
CH2–Cl + 2KOH CH2–OH + 2KCl
(B) (D)
Compound Molecular Formula Name
(A) CH3-CHCl2 Ethylidene chloride
(B) CH2Cl– CH2Cl Ethylene dichloride
(C) CH3 – CHO Acetaldehyde
(D) CH2OH– CH2OH Ethylene glycol
15. ENVIRONMENTAL CHEMISTRY
1. Dissolved oxygen in water is responsible for aquatic life. What processes are responsible
for the reduction in dissolved oxygen in water?
The microorganisms which oxidize organic matter also used oxygen dissolved in water. more over
during night, photosynthesis stops but the aquatic plants continue to respire, resulting in dissolved
oxygen in water
2. What would happen, if the green house gases were totally missing in the earth’s
atmosphere?
In the absence of green house gases, the average earth temperature will decrease. At very
low temperature, life is impossible..
3. Define smog.
Smog is a combination of smoke and fog which forms droplets that remain suspended in the
air.
4. Explain Particulate pollutants Particulate pollutants are small solid particles and liquid droplets
suspended in air. Many of particulate pollutants are hazardous. Examples: dust, pollen, smoke,
soot and liquid droplets (aerosols) etc,.
They are blown into the atmosphere by volcanic eruption, blowing of dust,
Incomplete combustion of fossil fuels induces soot. Combustion of high ash fossil fuels creates fly ash and finishing of metals throws metallic
particles into the atmosphere.
5. Which is considered to be earth’s protective umbrella? Why?
Ozone layer is earth‟s protective umbrella because it prevents theultraviolet radiation from
the sun.
6. What are degradable and non-degradable pollutants?
Degradable Pollutants :-
The pollutants which can bedecomposed by natural biological processesEx :- Plants wastes,
animal wastes.
Non -Degradable Pollutants :-
The pollutants which cannot be decomposed by natural biological processesEx :- Metal wastes,
Plastics, nuclear wastes.
7. From where does ozone come in the photo chemical smog?
The NO2 present in the pollutants emitted by automobiles, industries etc..,gives NO and oxygen
atoms.
These oxygen atoms combine with atmospheric dioxygen to give ozone.
NO2
����????????????ℎ�
NO + (O)
(O) + O2 O3
8. A person was using water supplied by corporation. Due to shortage of water he started
using underground water. He felt laxative effect. What could be the cause?
Excessive concentration ( > 500ppm ) of sulphates in drinking water causes laxative effect.
9. What is green chemistry?
Green chemistry is a chemical philosophy encouraging the design of products and processes that
reduce or eliminate the use and generation of hazardous substances.
10. What is Eutrophication ?
Eutrophication is a process by which water bodies receive excess nutrients that stimulates
excessive plant growth ( algae, other plant weeds).
11. Explain how does greenhouse effect cause global warming
The heating up of the earth‟s surface due to trapping of infrared radiations reflected by earth‟s
surface by CO2 layer in the atmosphere is known as green house effect.
Hence we can understand that green house effect causes global warming.
12. Mention the standards prescribed by BIS for quality of drinking water
S.No Characteristics Desirable limit
I Physico-chemical Characteristics
i) P
H
6.5 to 8.5
ii) Total Dissolved Solids(TDS) 500 ppm
iii) Total Hardness (asCaCO3) 300 ppm
iv) Nitrate 45ppm
v) Chloride 250 ppm
vi) Sulphate 200 ppm
vii) Fluoride 1 ppm
II Biological Characteristics
i) Escherichia Coli Not at all
ii) Coliforms Not to exceed 10 (In 100 mlwater sample)
13. How does classical smog differ from photochemical smog?
Classical smog Photochemical smog
occurs in cool humid climate occurs in warm and dry climate
combination of SO2, SO3 and humidity. combination of smoke, fog, dustsand air
pollutants
reducing smog oxidising smog
causes acid rain and bronchial irritation causes irritation of eyes, skin andlungs
(A) CH2 = CH2 Ethylene
(B) CH3 – CH2– Cl Ethyl chloride
(C) CH3 – CH2– NH2 Ethyl amine
28
.
A hydrocarbon C3H6 (A) reacts with HBr to form compound (B). Compound (B) reacts
with aqueous potassium hydroxide to give (C) of molecular formula C3H6O.what are (A)
(B) and (C). Explain the reactions.
CH3 –CH= CH2 + HBr CH3 – CHBr– CH3
(A) (B)
CH3 – CHBr– CH3
dil KOH
CH3 – CHOH– CH3 + KBr
(B) (C)
Compound Molecular Formula Name
(A) CH3 –CH= CH2 Propylene
(B) CH3 – CHBr– CH3 Isopropyl bromide
(C) CH3 – CHOH– CH3 Isopropyl alcohol
29
.
Two isomers (A) and (B) have the same molecular formula C2H4Cl2. Compound (A) reacts
with aqueous KOH gives compound (C) of molecular formula C2H4O. Compound (B)
reacts with aqueous KOH gives compound (D) of molecular formula C2H6O2. Identify
(A),(B),(C) and (D).
CH3-CHCl2 + 2KOH CH3 – CH(OH)2 + 2KCl
(A)-H2O
CH3 – CHO (C)
CH2–Cl CH2–OH
CH2–Cl + 2KOH CH2–OH + 2KCl
(B) (D)
Compound Molecular Formula Name
(A) CH3-CHCl2 Ethylidene chloride
(B) CH2Cl– CH2Cl Ethylene dichloride
(C) CH3 – CHO Acetaldehyde
(D) CH2OH– CH2OH Ethylene glycol
15. ENVIRONMENTAL CHEMISTRY
1. Dissolved oxygen in water is responsible for aquatic life. What processes are responsible
for the reduction in dissolved oxygen in water?
The microorganisms which oxidize organic matter also used oxygen dissolved in water. more over
during night, photosynthesis stops but the aquatic plants continue to respire, resulting in dissolved
oxygen in water
2. What would happen, if the green house gases were totally missing in the earth’s
atmosphere?
In the absence of green house gases, the average earth temperature will decrease. At very
low temperature, life is impossible..
3. Define smog.
Smog is a combination of smoke and fog which forms droplets that remain suspended in the
air.
4. Explain Particulate pollutants Particulate pollutants are small solid particles and liquid droplets
suspended in air. Many of particulate pollutants are hazardous. Examples: dust, pollen, smoke,
soot and liquid droplets (aerosols) etc,.
They are blown into the atmosphere by volcanic eruption, blowing of dust,
Incomplete combustion of fossil fuels induces soot. Combustion of high ash fossil fuels creates fly ash and finishing of metals throws metallic
particles into the atmosphere.
5. Which is considered to be earth’s protective umbrella? Why?
Ozone layer is earth‟s protective umbrella because it prevents theultraviolet radiation from
the sun.
6. What are degradable and non-degradable pollutants?
Degradable Pollutants :-
The pollutants which can bedecomposed by natural biological processesEx :- Plants wastes,
animal wastes.
Non -Degradable Pollutants :-
The pollutants which cannot be decomposed by natural biological processesEx :- Metal wastes,
Plastics, nuclear wastes.
7. From where does ozone come in the photo chemical smog?
The NO2 present in the pollutants emitted by automobiles, industries etc..,gives NO and oxygen
atoms.
These oxygen atoms combine with atmospheric dioxygen to give ozone.
NO2
����????????????ℎ�
NO + (O)
(O) + O2 O3
8. A person was using water supplied by corporation. Due to shortage of water he started
using underground water. He felt laxative effect. What could be the cause?
Excessive concentration ( > 500ppm ) of sulphates in drinking water causes laxative effect.
9. What is green chemistry?
Green chemistry is a chemical philosophy encouraging the design of products and processes that
reduce or eliminate the use and generation of hazardous substances.
10. What is Eutrophication ?
Eutrophication is a process by which water bodies receive excess nutrients that stimulates
excessive plant growth ( algae, other plant weeds).
11. Explain how does greenhouse effect cause global warming
The heating up of the earth‟s surface due to trapping of infrared radiations reflected by earth‟s
surface by CO2 layer in the atmosphere is known as green house effect.
Hence we can understand that green house effect causes global warming.
12. Mention the standards prescribed by BIS for quality of drinking water
S.No Characteristics Desirable limit
I Physico-chemical Characteristics
i) P
H
6.5 to 8.5
ii) Total Dissolved Solids(TDS) 500 ppm
iii) Total Hardness (asCaCO3) 300 ppm
iv) Nitrate 45ppm
v) Chloride 250 ppm
vi) Sulphate 200 ppm
vii) Fluoride 1 ppm
II Biological Characteristics
i) Escherichia Coli Not at all
ii) Coliforms Not to exceed 10 (In 100 mlwater sample)
13. How does classical smog differ from photochemical smog?
Classical smog Photochemical smog
occurs in cool humid climate occurs in warm and dry climate
combination of SO2, SO3 and humidity. combination of smoke, fog, dustsand air
pollutants
reducing smog oxidising smog
causes acid rain and bronchial irritation causes irritation of eyes, skin andlungs
14. Even though the use of pesticides increases the crop production, they adversely affect the
living organisms. Explain the function and the adverse effects of the pesticides.
Pesticides are the chemicals that are used to kill or stop the growth of unwantedorganisms. But
these pesticides can affect the health of human beings.These are further classified as
Pesticides Example Function Effects
Insecticides: DDT, BHC, aldrin
etc.,
It can stay in soil
for long period of time and
are absorbed by soil.
They contaminate root crops
like carrot,raddish etc.,
Fungicide: Organo mercury
compounds
It used as most common
fungicide.
Theydissociate in soil to
produce mercury which
is highly toxic.
Herbicides: Sodium chlorate +
Sodium arsenide
It is used to control
unwanted plants.
Toxic to mammals.
15. Ethane burns completely in air to give CO2, while in a limited supply of air gives CO.
The same gases are found in automobile exhaust. Both CO and CO2 are atmospheric
pollutants
i) What is the danger associated with these gases
(a) Carbon monoxide: Poisonous gas produced as a result of incomplete combustion of coal are
firewood.
(b) Carbon dioxide:Responsible for global warming.
ii) How do the pollutants affect the human body?
(a) Carbon monoxide:
It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen
transport by blood and hence the oxygen carrying capacity of blood is reduced.
This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring
of eye sight and cardiac arrest.
(b) Carbon dioxide:It causes headache and nausea.
16. On the basis of chemical reactions involved, explain how do CFC’s cause depletion of
ozone layer in stratosphere?
In the presence of uv radiation, CFC‟s break up into chlorine free radical
CF2Cl2
h
CF2Cl + Cl
.
CFCl3
h
CFCl2 + Cl
.
Cl
.
+ O
3 → ClO
.
+ O2
ClO
.
+ O → Cl
.
+ O2
Chlorine radical is regenerated in the course of reaction. Due to this continuous attack of Cl
.
thinning of ozone layer takes place which leads to formation of ozone hole.
17. How is acid rain formed? Explain its effect
The oxides of sulphur and nitrogen is absorbed by the water in the cloudsand converted into
sulphuric acid and nitric acid.
2SO2 + O2 + 2H2O 2H2SO4
4NO2 + O2 + 2H2O 4HNO3
The pH of rain waterbecomes5.6. This is called as Acid rain.
Effects of acid rain:
Damage to building (stone leprosy)
Affects plants and animal life in aquatic ecosystem
Harmful for agriculture as it removes nutrients
Corrodes water pipes
Respiratory ailment in humans and animals. 18. Explain how oxygen deficiency is caused by carbon monoxide in our blood? Give its
effect
Poisonous gas produced as a result of incomplete combustion of coal are firewood.
It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen
transport by blood and hence the oxygen carrying capacity of blood is reduced.
This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring
of eye sight and cardiac arrest.
19. What are the various methods you suggest to protect our environment from pollution?
Waste management and proper disposal of wastes
Recycling waste materials and reusing them.
Growing more trees
Controlling vehicle emission
Using fuels with low sulphur content
20. Differentiate BOD and COD
Biochemical Oxygen Demand (BOD) Chemical oxygen demand (COD)
The total amount of oxygen in milligrams
consumed by microorganisms in
decomposing the waste in one litreof water
at 20̊ C for a period of 5 days and its value
is expressed in ppm.
It is defined as the amount of oxygen required by the
organic matter in a sample of water for its oxidation
by a strong oxidizing agent like K2Cr2O7 in acid
medium for a period of 2 hrs.
21. Differentiate Viable and non-viable particulate pollutants
Viable particulate pollutants non-viable particulate pollutants
The viable particulates are the small size
living organisms such as bacteria, fungi,
moulds, algae, etc. Which are dispersed in
air.
The non-viable particulates are small solid particles
and liquid droplets suspended in air.
Example:Smoke, dust, mists, fumes etc
V.SURESHKANNA , PG.ASST, G.H.S.S ,THIRUMANJOLAI . SIVAGANGAI – DT.
living organisms. Explain the function and the adverse effects of the pesticides.
Pesticides are the chemicals that are used to kill or stop the growth of unwantedorganisms. But
these pesticides can affect the health of human beings.These are further classified as
Pesticides Example Function Effects
Insecticides: DDT, BHC, aldrin
etc.,
It can stay in soil
for long period of time and
are absorbed by soil.
They contaminate root crops
like carrot,raddish etc.,
Fungicide: Organo mercury
compounds
It used as most common
fungicide.
Theydissociate in soil to
produce mercury which
is highly toxic.
Herbicides: Sodium chlorate +
Sodium arsenide
It is used to control
unwanted plants.
Toxic to mammals.
15. Ethane burns completely in air to give CO2, while in a limited supply of air gives CO.
The same gases are found in automobile exhaust. Both CO and CO2 are atmospheric
pollutants
i) What is the danger associated with these gases
(a) Carbon monoxide: Poisonous gas produced as a result of incomplete combustion of coal are
firewood.
(b) Carbon dioxide:Responsible for global warming.
ii) How do the pollutants affect the human body?
(a) Carbon monoxide:
It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen
transport by blood and hence the oxygen carrying capacity of blood is reduced.
This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring
of eye sight and cardiac arrest.
(b) Carbon dioxide:It causes headache and nausea.
16. On the basis of chemical reactions involved, explain how do CFC’s cause depletion of
ozone layer in stratosphere?
In the presence of uv radiation, CFC‟s break up into chlorine free radical
CF2Cl2
h
CF2Cl + Cl
.
CFCl3
h
CFCl2 + Cl
.
Cl
.
+ O
3 → ClO
.
+ O2
ClO
.
+ O → Cl
.
+ O2
Chlorine radical is regenerated in the course of reaction. Due to this continuous attack of Cl
.
thinning of ozone layer takes place which leads to formation of ozone hole.
17. How is acid rain formed? Explain its effect
The oxides of sulphur and nitrogen is absorbed by the water in the cloudsand converted into
sulphuric acid and nitric acid.
2SO2 + O2 + 2H2O 2H2SO4
4NO2 + O2 + 2H2O 4HNO3
The pH of rain waterbecomes5.6. This is called as Acid rain.
Effects of acid rain:
Damage to building (stone leprosy)
Affects plants and animal life in aquatic ecosystem
Harmful for agriculture as it removes nutrients
Corrodes water pipes
Respiratory ailment in humans and animals. 18. Explain how oxygen deficiency is caused by carbon monoxide in our blood? Give its
effect
Poisonous gas produced as a result of incomplete combustion of coal are firewood.
It binds with haemoglobin and form carboxy haemoglobin which impairs normal oxygen
transport by blood and hence the oxygen carrying capacity of blood is reduced.
This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring
of eye sight and cardiac arrest.
19. What are the various methods you suggest to protect our environment from pollution?
Waste management and proper disposal of wastes
Recycling waste materials and reusing them.
Growing more trees
Controlling vehicle emission
Using fuels with low sulphur content
20. Differentiate BOD and COD
Biochemical Oxygen Demand (BOD) Chemical oxygen demand (COD)
The total amount of oxygen in milligrams
consumed by microorganisms in
decomposing the waste in one litreof water
at 20̊ C for a period of 5 days and its value
is expressed in ppm.
It is defined as the amount of oxygen required by the
organic matter in a sample of water for its oxidation
by a strong oxidizing agent like K2Cr2O7 in acid
medium for a period of 2 hrs.
21. Differentiate Viable and non-viable particulate pollutants
Viable particulate pollutants non-viable particulate pollutants
The viable particulates are the small size
living organisms such as bacteria, fungi,
moulds, algae, etc. Which are dispersed in
air.
The non-viable particulates are small solid particles
and liquid droplets suspended in air.
Example:Smoke, dust, mists, fumes etc
V.SURESHKANNA , PG.ASST, G.H.S.S ,THIRUMANJOLAI . SIVAGANGAI – DT.
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