Atomic structure and their bondings_AC.ppt

overview of chemistry

1. Quantum theory (or quantum
mechanics) was developed
mostly
between 1900 and 1930
by European physicists
including
Max Planck
Albert Einstein
Neils Bohr
Louis de Broglie
Max Born
Paul Dirac
Werner Heisenberg
Wolfgang Pauli
Erwin Schrodinger and his cat
Richard Feynman
1

2. In the 17th Century Newton came up with a theory of light.
He said that light is made of particles, each particle consisting of a
particular colour of the spectrum.
This was just after coming up with his laws of motion and gravitation
and developing calculus.
During the 18th Century, most scientists accepted this,
though a few, like Robert Hooke, Christian Huygens and
Leonhard Euler believed that light
consisted of waves rather than particles.
Then in 1803, Thomas Young did his famous double-slit experiment.
The interference Patterns produced were the same as those produced by
waves on water.
After this it was generally accepted that light was made of waves rather
than particles.
However, in 1905, Einstein produced a paper on the
photo-electric effect, showing that light really was
made of particles.
Light was used to provide the energy to dislodge
electrons from a metal.
2

3. A certain amount of energy was required to dislodge an electron.
It was expected that once the brightness of the light reached a certain level,
electrons would be released.
What he found, though, was that the brightness made no difference – what mattered
was the colour.
Very dim blue light would release electrons while very bright red light would not.
It seemed that light was made up of particles of different colour, just as Newton
had suggested.
Einstein called them photons.
Bluer colours has shorter wavelengths and therefore higher frequencies and
higher energies.
The energy of a photon was proportional to the frequency
E = hf
the constant of proportionality being called Planck’s constant and having the
value 6.6 × 10
-34
.
This showed that the energy of light came in discrete packets called quanta (plural
of quantum).
Each quantum consisted of a fixed amount of energy.
So now it seemed that light was both a wave and a particle.
3

4. In 1924 Louis de Broglie suggested that, if light could have properties of particles and
waves, then so, maybe, could electrons.
He developed a model of the atom in which the electron was a wave around the nucleus.
If the circumference of the orbit was a whole number of wavelengths, then the wave would
be a standing wave that doesn’t change with time.
Such a wave would not emit energy and so would be stable.
His model correctly predicted the energy levels of the different orbits as known
from the emission spectrum.
De-Broglie proposed that a moving particle, whatever its
nature, has waves associated with it. These waves are
called “matter waves”.
Energy of a photon is E = h
For a particle, say photon of mass, m
E = mc
2
mc
2
= hv
mc
2
= hc/
= h/mc
Suppose a particle of mass, m is moving with velocity, v
then the wavelength associated with it can be given by
 = h/mv or  = h/p
Matter wave
4

5. A wave
function
 in quantum
mechanics describes the quantum state of an
isolated
system of one or more particles
The
state of a quantum mechanical system is completely specified by a wave
function

ψ

(x,t)
that
depends
on the coordinates of the particles (x) and time t
There
is one wave function containing all the information about the entire
system
Not
a separate wave function for each particle in the system
Wave FunctionWave Function
5

6. Physically Acceptable Wave FunctionPhysically Acceptable Wave Function
Acceptable solution to the wave equation that is solutions which are physically possible,
must have certain properties;
 
ψ must be continuous.
ψ must be finite, because an infinite value at a point would mean that electron
is fixed there, which is contrary to wave representation.
ψ must be single value.
The probability of finding the electron over all the space from + ∞ to  ∞ must
be equal to one. i.e. ψ must be normalized.
 
So that
So we have the Normalization condition for the wave function:
1)(
2



dxx
1
2


dxdydz
6

7. Physical Interpretation of Wave FunctionPhysical Interpretation of Wave Function
The state of a quantum mechanical system can be completely understood
with the help of the wave function ψ.
But wave function ψ can be real or imaginary
Therefore no meaning can be assigned to wave function ψ
which is called as probability density

*
2

We know that electron is definitely found somewhere in the space. The wave function ψ,
which satisfies the above condition, is called normalized wave function
Thus if ψ is the wave function of a particle within a small region of volume dv, then
Gives the probability of finding the particle within the region dv at the given instant
of time.
Normalized wave function
7

8. Acceptable or Not ??Acceptable or Not ??
8

9. Hamiltonian operator
In quantum mechanics, a
 
Hamiltonian 
is an 
operator corresponding to the sum of
the kinetic energies plus the potential energies for all the particles in the system. It is
usually denoted by H, also or Ĥ
Ȟ
Kinetic energy Potential energy
9

10. Schrödinger Wave Equation Derivation
Where E is the total energy and V is the potential energy
10

11. Particle in a box: wave functions
A particle moving along x-axis in a box of length l
•The particle moves in one dimension only: Along x-axis
•The walls are vertical and infinite height (can not be penetrated at x = 0 and x = L)
•No forces are acting on the particle inside the box. So, the potential energy of the
particle is zero, So V(x,t) = 0 for 0 < x <L.
•When the particle hits the wall, infinitely large forces push it back, so the potential
energy, So V (x,t) is infinite x ≤ 0 and x ≥ L.
• We can use the time-independent one-Dimensional Schrodinger equation within the
box
  000
8
2
2
2
2
 VE
h
m
dx
d


V=0
L
x =0 x =L
V=
V=
11

12. Therefore correct solution of the equation can be written as
x
L
n
A
n

 sin
The above equation represents Eigen functions. Where n=1,2,3,..
(n=0 is not acceptable because, for n=o the wave function ψ becomes zero for all values
of x. Then particle cannot be found anywhere)
2
22
8mL
hn
E
n

Using normalization condition:
1)(
0
2

L
dxx 1sin
0
22







L
dx
L
xn
A

L
xn
L
x

 sin
2
)(
The complete solution to the wave equation becomes
The energy E of the particle is given by
12

13. The features of the solution
1.The probability of finding the particle (ψ
2
) depends on x and E
2.The values of ‘E’ is independent of ‘x’ but depends on ‘n’
3.E can have only discrete values
4.A quantum number “n” that can take up only integral values: 1, 2, 3,…
n = 0 not acceptable: 
n
= 0 at all x
Lowest kinetic Energy = E
0
= h
2
/8mL
2
2
22
8mL
hn
E
n
13

14. Assuming an electron to be confined in a 1D box of 2.0 nm in length, find the lowest three
energy levels for the electron. Also find the wavelength of the radiation emitted during de-
excitation of electron from the 1
st
excited to ground state.
14

15. Coordinate TransformationCoordinate Transformation
Transform from Cartesian Coordinates into spherical coordinates :
•The
potential (central force)
V(r)
depends on the distance
r
between the proton and
electron
Transform
to spherical polar
coordinates
because of the radial
symmetry.
r
2
= x
2
+y
2
+z
2
15

16. Applications for conjugated molecule and nanoparticles
-electrons in a conjugated molecule to be constructed according to
the ‘aufbau’ principle.
Quantum dot:
A quantum dot is a crystal semiconductor material whose diameter is
on the order of several nanometers a size results in its free charge
carriers experiencing “quantum confinement” in all three spatial
dimensions.
The electronic properties of quantum dots are intermediate between
these of bulk semiconductors and of discrete molecules and closely
related to their size and shape.
This allows properties such as;
The bandgap, emission colour and absorption
16

17. Spectrum: Highly tunable band gap energy during fabrication.
Applications of quantum dots are:

o Transistor
o Solar cells
o LeDs
o Diode laser
17

18. Forms of H-atom Wave Functions
The wave function can be split into two parts to solve the equation:
= R(r). A Where R(r) radial part and A is angular part
Radial wave function R(r)
The radial wave function R(r) used to calculate the variation of the amplitude of
the electron or radial probability distributions of the electron from the nucleus.
It depends on the quantum number ‘n’ and ‘l’
R(r) has no physical meaning, but R
2
gives the probability of finding the
electron in a small volume ‘dv’ near the point at which ‘R’ is measured.
4
2
r
2
R
2
– This is called the radial distribution function.
18

19. Radial distribution functions from various orbitals in the H-atom
Possible number of
radial nodes = (n-l-1).
Several points emerge from these plots:-
i.The probability distribution is zero at the nucleus.
ii.The distribution vary according to the ‘n’ (Principal quantum No.) i.e.
the most probable distance increases markedly as the ‘n’
increases.
iii.For 2s and 2p or 3s,3p and 3d the most probable radius
decreases slightly as the ‘l’ (Subsidiary or Azimuthal quantum no.)
increases.
19

20. Angular wave function (A)
The ‘A’ which depends on the quantum numbers ‘m’ (Magnetic quantum
no.) and ‘l’. A depends only on the direction but it is independence of the
distance from the nucleus.
Thus A
2
m,l
(θ,φ) is the probability of finding an electron at a given
direction θ,φ at any distance from the nucleus to infinity.
Angular part of the wave function A(θ,φ) for 1s, 2p and 3d orbitals.
Shape of s orbital
Shape of p orbitals
20

21. Shape of d orbitals
21

22. Louis de Broglie (1892–1987)
• de Broglie proposed that particles could have wave-like character
• the wavelength of a particle was inversely proportional to its momentum
= h/(mv)
Heisenberg’s Uncertainty Principle:
The product of the uncertainties in both the position and speed of a particle was
inversely proportional to its mass…

Δx .(m Δ v) ≥ h/(4π)
x = position, Δ x = uncertainty in position
v = velocity, Δ v = uncertainty in velocity
m = mass

This means that the more accurately you know the position of a small particle,
such as an electron, the less you know about its speed and vice-versa
22

23. Chemical Bonding
•Two existing theories,
• Molecular Orbital Theory (MOT)
• Valence Bond Theory (VBT)
Molecular Orbital Theory
•MOT starts with the idea that the quantum mechanical principles applied
to atoms may be applied equally well to the molecules.
The molecular orbitals created via the
 linear
combinations of atomic orbitals
(LCAOs)
 approximation were created from the
sum and the difference of two atomic orbitals
23

24. MOT (Hund's & Mulliken)
Each atom combining with other atomic orbitals (AOs) with same or nearly same
energy and appropriate symmetry to form a completely new set of orbitals known
as molecular orbital (MO).
The number of MOs formed is equal to the number od AOs participating in
linear combination.
The MO formed by constructive overlap between two atomic orbitals
corresponds to bonding AOs.
Conversely, subtracting one atomic orbital from another corresponds to
destructive overlap between two waves, which reduces the internuclear electron
probability density is called antibonding AOs.
Unlike AOs, MO is polycentric.
The shape of MOs depend on the shape of AOs from which they are generated.
Filling of MOs follows same rules used for filling AOs: Aufbau Principle
(Building Principle) - Pauli Exclusion Principle.
24

25. Rules for linear combinationRules for linear combination
•1. Atomic orbitals must be roughly of the same energy.
•2. The orbital must overlap one another as much as possible- atoms must be
close enough for effective overlap.
•3. In order to produce bonding and antibonding MOs, either the symmetry of
two atomic orbital must remain unchanged when rotated about the
internuclear line or both atomic orbitals must change symmetry in identical
manner.
Linear combination of atomic orbitalsLinear combination of atomic orbitals
25

26. Formation of MOs from s-orbitals, p-orbitals and d-orbitals.
 
Bonding and anti-bonding orbitals.
26

27. A B

A 
B
Linear Combination of Atomic Orbitals (LCAO)Linear Combination of Atomic Orbitals (LCAO)
The wave function for the molecular orbitals can be approximated by taking linear
combinations of atomic orbitals.
+.+.
..
+
bonding
+.-. ..
node
antibonding
*
+ -
27

28. Sigma
pi
S orbital
p orbital
28

29. Sigma bonding
Sigma antibonding
pi antibonding
pi bonding
29

30. Diatomic molecules (HDiatomic molecules (H
22))
1s
2
H
E
n
e
r
g
y
HH
2
1s 1s

*
Bond order =
½ (bonding electrons – antibonding electrons)
Bond order: 1
30
AO MOAO

31. 1s
2
, 
*
1s
2
He
E
n
e
r
g
y
HeHe2
1s 1s

*
Molecular Orbital theory is powerful because it allows us to predict
whether molecules should exist or not and it gives us a clear picture of the
of the electronic structure of any hypothetical molecule that we can
imagine.
Diatomic molecules: The bonding in He
2
Bond
order: 0
31
AO MOAO

32. Diatomic molecules (LiDiatomic molecules (Li
22))
1s
2
, 
*
1s
2
, 2s
2
Bond order: 1
Li
E
n
e
r
g
y
LiLi
2
1s 1s
1
1*
2s 2s
2
2*
32
AO MOAO

33. 1s
2
, 
*
1s
2
, 2s
2
, 
*
2s
2

Bond
order: 0
Be
E
n
e
r
g
y
BeBe2
1s 1s
1
1*
2s 2s
2
2*
Diatomic molecules: Homonuclear Molecules (Be
2
)
33
AO MOAO

34. 34

35. 35
Z axis is the molecular axis

36. 36

37. 37
B
2
N
2
E
n
e
r
g
y
AO MO AOAO MO AO
Z axis is the molecular axis

38. Co-ordination compounds
Double salts: Double salts 
are salts containing more than one cation or anion,
and are obtained by combination of two different salts which lose their identity in
solution and dissociates into simple ions. [Examples: K
2
SO
4
.Al
2
(SO
4
)
3
.24H
2
O]
Complexes: Coordination complex consists of a central atom or ion, which is
usually metallic and is called the coordination centre, and a surrounding array
of bound molecules/ions, that are known as ligands/complexing agents.
Complexes retain their identity in solution. (Examples: [Co(NH
3)
6]Cl
3
Werner’s work:
Transition metals form coordination compounds very readily because they have
vacant ‘d’ orbitals which can accommodate these electrons pairs.
Bonding in Transition metals complexes and structures and properties:
Valence Bond Theory (VBT):
The Ligand must have a lone pair of electrons and the metal must have an
empty orbital of suitable energy available for bonding.
Limitations:
i.Most transition metals complexes are coloured, but the theory provides no
explanation for their electronic spectra.
ii.Why magnetic properties vary with temperature.
38

39. Crystal Field Theory
•The The
relationship relationship
between colors between colors
and complex and complex
metal ionsmetal ions
400 500 600 800
39

40. Crystal Field Model
•The attraction between the central metal and ligand in the complex is
considered to be purely electrostatic. (ion – ion interaction, ion – dipole
interaction)
•Ligands is considered as point charges.
•No interaction between metal orbitals and Ligand orbitals.
•‘d’ orbital is degenerate in free atom, when complex is formed orbitals now
have different energies.
40
CFT is remarkably successful in explaining the electronic spectra
and magnetism of transition metal complexes.

41. d-orbitals: look attentively along the axis
Linear combination of
d
z
2
-d
x
2
and d
z
2
-d
y
2

d
2z
2
-
x
2
-
y
2
41

42. Octahedral Field
42

43. •We assume an octahedral array of negative charges placed around the
metal ion (which is positive).
•The ligand and orbitals lie on the same axes as negative charges.
–Therefore, there is a large, unfavorable interaction between ligand (-)
and these orbitals.
–These orbitals form the degenerate high energy pair of energy levels.
•The d
xy
, d
yz
, and d
xz
orbitals bisect the negative charges.
–Therefore, there is a smaller repulsion between ligand and metal for
these orbitals.
–
These orbitals form the degenerate low energy set of energy levels.
43

44. 44

45. 45

46. 
o
3/5 o
2/5 o

o
is the crystal field splitting
t
2g
e
g
E(t
2g
) = -0.4
o
x 3 = -1.2o
E(e
g
) = +0.6
o
x 2 = +1.2o
Splitting of d orbitals in an octahedral field
Barycentre
46

47. Crystal Field Theory: Splitting of the 5 d orbitals
Consider the response of the energy of
the d orbitals to the approach of 6
negatively charged ligands (a “crystal
field”) along the x, y and z axes of the
metal
The two d orbitals (d
x2-y2 and d
z2) that
are directed along the x, y and z axes
are affected more than the other three
d orbitals (d
xy
, d
xz
and d
yz
)
The result is that the d
x2-y2 and d
z2
orbital increase in energy relative
to the d
xy
, d
xz
and d
yz
orbitals (D
0

is called the “crystal field energy
splitting”
t
2g
orbitals
e
g
orbitals
crystal
field
energy
splitting
47

48. The magnitude of the splitting
(ligand effect)
Strong
field
Weak
field
•Spectrochemical Series: An order of ligand field
strength based on experiment: I
-
 Br
-
 S
2-
 SCN
-

Cl
-
 NO
3
-
 F
-
 C
2
O
4
2-
 H
2
O NCS
-
 CH
3
CN NH
3
 en
 bipy phen NO
2
-
 PPh
3
 CN
-
 CO
N N
2,2'-bipyridine (bipy)
NH
2
NH
2
Ethylenediamine (en)
N
N
1.10 - penanthroline (phen)
48

49. 49
Factors that affect the magnitude of 
o
Oxidation state of the transition metal: Increasing oxidation state of the metal also
increases Δo. As the oxidation state of the transition metal (effectively the charge on
the metal) is increased, the surrounding ligands are attracted more closely to the metal
centre. The orbitals on the ligands interact more strongly with the d orbitals and Δo
increases
Row of the transition metal: On going from the first to the second row of a transition
metal, there is approximately a 50% increase in Δo, and another 50% increase on
going from the second row to the third row
The nature of the ligands: The size of Δo depends on the ligand(s) present according to
the spectrochemical series, which is a list of ligands in order of decreasing crystal field
strength. CN
-
= CO = C
2
H
4
> PR
3
> NO
2
-
= phen > bipy > SO
3
2-
> en = py = NH3 > edta
4-
> NCS
-
> H
2O > C
2O
4
2-
> ONO
2
-
> OSO
3
2-
> OH
-
= ONO
-
> F
-
> Cl
-
= SCN
-
> Br
-
> I
-
Number of ligands and geometry 
t
< 
o

t
= 4/9
o

50. The magnitude of the splitting (metal ion effect)
50
For STRONG Filed ligand:
 
The magnitude of Δ is high, the Ligand will form low spin complexes.
 
The number of unpaired electrons for d
6
system will be zero (t
2g
6
)
 
Examples: [Co(NH
3
)
6
]
3+
and [Co(H
2
O)
6
]
3+
 
For WEAK Filed ligand:
 
The magnitude of Δ is low, the Ligand will form High spin complexes.
 
The number of unpaired electrons for d
6
system will be four (t
2g
4
e
g
2
)
 
Examples: [CoF
6
]
3-
.
High Spin (HS) – No additional pairing in metal d orbital
Low Spin (LS) – Spin pairing in the metal d orbitals

51. d
1
d
2
d
3 d
4
Placing electrons in d orbitals
Strong field Weak field
Strong field Weak field
51

52. d
4
Strong field =
Low spin
(2 unpaired)
Weak field =
High spin
(4 unpaired)
 < 
o  > 
o
When the 4
th
electron is assigned it will either go into the higher energy e
g orbital at
an energy cost of 
0
or be paired at an energy cost of , the pairing energy.
52

53. 53
[Fe (CN)
6
]
3-
Magnetic Moment
BM
= 1.73 BM n = 1

54. 54
[FeF
6]
3-
Magnetic Moment = 5.91 BM n = 5

55. 55

56. Splitting of d orbitals in a tetrahedral
field
t
2
e
t

t = 4/9
o
Always weak field (high spin)
56

57. In coordination chemistry, a
 
ligand 
is an ion or molecule (functional
group) that binds to a central metal atom to form a coordination
complex. The bonding with the metal generally involves formal
donation of one or more of the
 
ligand's 
electron pairs
Monodentate ligands 
are Lewis bases that donate a single pair
("mono") of electrons to a metal atom.
 
57

58. Crystal Field Splitting Energy (CFSE)
•In Octahedral field, configuration is: t
2g
x
e
g
y

•Net energy of the configuration relative to the average
energy of the orbitals is:
= (-0.4x + 0.6y)
O

O = 10 Dq
BEYOND d
3
•In weak field: 
O  P, => t
2g
3
e
g
1
•In strong field 
O  P, => t
2g
4
•P - paring energy
58

59. 59

60. 
B
is the Bohr magneton
•Since each unpaired electron has a spin ½,
•S = (½)n, where n = no. of unpaired
electrons
 = {n(n+2)}
1/2

B

•In d
4
, d
5
, d
6
, and d
7
octahedral complexes,
magnetic measurements can very easily
predict weak versus strong field.
•Tetrahedral complexes - only high spin
complexes result, for 
t  
O.
60

61. Magnetic properties of metal complexes
Diamagnetic complexes
very small repulsive interaction with
external magnetic field
no unpaired electrons
Paramagnetic complexes
attractive interaction with external
magnetic field
some unpaired electrons
)2(nns
For example: [Ti(H
2
O)
6
]
3+
, where Ti
3+
is 3d
1
system and has one unpaired electron
So that the 
s
= 1.732 BM
[CoF
6
]
3-
where Co
3+
is 3d
6
system and F
-
is a weak field Ligand.
Therefore the complex ion has four unpaired electrons.
So that the 
s
=
= 4.90 BM24
61
BM

62. Aromaticity
An unsaturated cyclic or polycyclic molecule or ion may be classified as
aromatic if all the annular atoms participate in a conjugated system.
Test of Aromaticity:
I.The compound must be cyclic and planar. The planarity is essential
because overlapping of  and p-orbitals are possible only when they are
parallel.
II.Complete delocalization of the  electrons in the ring.
III.The compound must have obey Huckle Rule i.e. presence of (4n+2) 
electrons in the ring, where n is an integer (n = 0,1,2,3,….)
Benzoid Aromatics:
62

63. 63
Pyrrole

64. Heterocyclies Aromatics
Non-benzoid Aromatics:
Cyclopentadienylium ion
64

65. 65
Cyclooctatetraene
Non Planar (Non Aromatic)
Tub Shape

66. Conductors, Semiconductor and Insulator
66
Conductors:
Conductor is a substance in which the electrons move easily from atom to atom
with the application of voltage.
Example: Copper, steel, gold, Al, Silver, brass, mercury
Semiconductor:
Semiconductor is a substance which acts as a good conductor under some
conditions and poor conductor under some conditions.
Conductivity increases with increase in temperature.
Example: Silicon and Germanium

67. Doping of Semiconductor:
Pure Si and Ge can be made semiconducting in a controlled way by adding
impurities which act as charge carriers.
n-type semiconductor:
Example: As (pentavalent impurity) is
added deliberately added to the Ge-
crystal. The 5
th
electron is localized on
the As atom. Since the current is carried
by excess electrons, it is n-type
semiconductor.
p-type semiconductor:
Example: In (trivalent impurity) is added
deliberately added to the Si-crystal. When
B is doped, one electron is missing in Si-
crystal. Missing electron is called positive
hole. Here current is carried by the positive
hole, it is p-type semiconductor.
67

68. Assignments:
1. What is aromaticity? Write down the characteristic features for aromaticity
with examples.
2. Which one is aromatic and why: Cyclobutadiene and benzene.
3. Why benzene and naphthalene are aromatic and why?
4. Predict the aromaticity of Furan.
5. Predict the aromaticity of pyridine.
6. Why metals are good conductor of electricity?
7. For a semiconductor, the conductivity _____ with increase in
temperature
A.Increases, B. Decreases, C. Remains same and D. None
8. n-type semiconductor is formed due to doping of Si with
A. Phosphorus
B. Sodium
C. Boron
D. Oxygen
9. Why cyclobutadiene is not-aromatic?
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69. 10. Assuming an electron to be confined in a one-dimensional box 2.0 nm
in length. If the 1st energy level E1 = 1.506  10-20 J. Find out the E2 =
A)3.012  10-20 J, B) 1.506  10-18 J, C) 0.753  10-20 J and D) 6.024 
10-20 J
11. Assuming an electron to be confined in a one-dimensional box 3.0 nm
in length. If the 1st energy level E1 = 2.820  10-20 J. Find out the E
2?
12. Write the Schrödinger’s time-independent wave equation.
13. Mention the conditions for acceptable wave function.
14. Write down the significance of ψ
2
.
15. Assuming an electron to be confined in a one-dimensional box 1.0 nm in
length. Find the wavelength of the radiation emitted during de-excitation of
electron from the first excited state to ground state.
16. Which one is incorrect statement in the solution of a particle in a box
using Schrödinger equation
A) The probability of finding the particle depends on X (position) and E
(energy),
B) E can have only discrete values.
C) The value of E is dependent on X.
D) The value of E depends on n (Principal quantum number).
69

70. 17. The relation α = h/mv (where λ = wavelength, h = plank’s
constant, m = mass and v = velocity of the particle) is derived
from
A) Heisenberg’s uncertainty principle
B) De Broglie’s concept
C) Photoelectric effect
D) Schrödinger equation
18. Particle in motion has dual character, i.e., both particle and wave
character: this postulate was described by
A) Neils Bohr B) Dalton C) Max plank D) Louis De Broglie
19. How many radial nodes are present in 3p sub-shell?
20. How many radial nodes are present in 3d sub-shell?
21. Why Li
2 molecule contains a single covalent bond? Explain using
MOT.
22. Why C
2 molecule does not show paramagnetic in nature? Explain
using MOT.
23. Illustrate schematically of π-bonding and π-anti-bonding of p – p –
orbitals combination.
70

71. 24. Which one is incorrect statement of MOT
A.The number of MOS formed is equal to the number of AOS
participating in linear combination.
B.The shapes of MOS does not depend on the shapes of AOS from which
they are generated.
C.Unlike AOS, MOS is polycentric.
D.Filling of MOS follows the same rules used for filling of AOS.
25. This molecule does not exist according to MOT
A.Li
2
B.Be
2
C.B
2

D.C
2
26. The number of unpaired electrons in O2 molecule is
A.0, B. 1, C. 2 and D. 3
27. The number of unpaired electrons in N2 molecule is
A.0, B. 1, C. 2 and D. 3
2. The number of unpaired electrons in O
2
+
molecular ion.
A. 0, B. 1, C. 2 and D. 3
71

72. 29. What are the factors affecting magnitude of energy gap (Δ) in transition metal
complexes?
30. Which of the following transition metal complexes is not paramagnetic?
A. [Ti(H2O)6]3+; B. CoF63-; C. NiCl4]2-; D. [Co(NH3)6]3+
31. Calculate the spin magnetic moment for [Co(H2O)6]3+.
32.. Find the number of unpaired electrons in the [Co(NH3)6]3+complex ion.
33. Find the value of magnetic moment in the [Co(NH3)6]3+complex ion.
34. Whai is a monodentate ligand? Give examples.
35. Find the number of unpaired electrons in the [Fe(Cl)6]3-complex ion.
36. Find the value of magnetic moment in the [Fe(Cl)6]3-complex ion.
37. What is complex salt? What are the differences between complex salt and
double salt?
3. Which of the following ions has the maximum magnetic moment? Ti2+ and Fe2+
72