Atomic structure - chemistry

a detailed description of the structure of atom including all the discoveries and inclusion of those rules in periodic classification from Dr. Raghav Samantaray phd in applied chemistry (KIIT school of Biotechnology)

1. Dr.RaghavSamantaray:
Room203-A
Email:[email protected]
Phone:[phone]
BT 1005, CHEMISTRY, 3 credits

2. Syllabus
BT1005 CHEMISTRY-I 3 CREDITS
•Unit 1-Atomic Structure: Bohr’s model of atom, Inadequacies of Bohr’s model, Sommerfold’smodification. de-Broglie’s
matter wave, Heisenberg’s uncertainty principle, photoelectric effect, concept of quantization of energy. Schrodinger
wave equation. Significance of Ψ
2
SWE, Probability and orbital, shape, Quantum numbers (mathematical derivation not
required ), Pauli’s exclusion principle, Hund’srule, Aufbau’sprinciple, effective nuclear charge.
• Unit 2-Periodic classification of elements: Atomic and ionic radii, Periodicity, Periodic properties, Size, IP, EA, EN, their variation in periodic table
and their application in periodicity and explaining chemical behavior.
• Unit 3-Chemical bonding: Ionic bond: Formation of ionic bonds, Factors affecting the formation of ionic bonds, Lattice energy, Determination of
lattice energy by Born Haber’s cycle, covalent character of ionic bond, Fajan’srule. Covalent bond: VBT, Shapes of covalent molecules (VSEPR
theory and hybridization), Limitation of VBT, Molecular orbital theory (MOT), Energy level diagrams for homonucleardiatomic molecules,
Molecular forces -Hydrogen bond, Van derWaal’s forces
•Unit 4-Chemical Equilibrium and Ionic Equilibrium: Law of mass action, law of chemical equilibrium, effect of temperature
and pressure on equilibrium, Le-Chatlierprinciple. Weak and strong electrolytes, Ionization of electrolytes, Various concepts
of acids and bases (Arrhenius, Bronsted-Lowry and Lewis) and their ionization, Degree of dissociation, Acid-base
equilibria(including multistage ionization) and ionization constants, Ionization of water, pH, Buffer solutions, Common-
Ion effect, Solubility of sparingly soluble salts and solubility products, Hydrolysis of salts.
•Unit 5-Solutions: Different methods for expressing concentration of solution: molality, molarity, mole fraction,
percentage (by volume and mass both), Vapourpressure of solutions and Raoult'sLaw, Ideal and non-ideal solutions,
vapourpressure, Composition plots for ideal and non-ideal solutions, Henry’s Law.
• Unit 6-Chemical Kinetics: Rate of reaction, Factors influencing the rate of reaction, concentration, temperature, pressure, solvent, light and catalyst,
Kinetics of zero, 1st and 2nd order reactions, Theories of reaction rates: collision and absolute reaction rate theory; Catalysis: homogeneous and
heterogeneous catalysis, factors affecting catalytic processes; Catalytic promoter and catalytic poison; Autocatalyst, Negative catalyst/inhibitor; Acid-
Base catalysis, enzyme catalysis.
• Unit 7-Classification and Nomenclature of organic compounds: IUPAC recommendation for naming different organic compounds.
• Unit 8-Reaction Mechanism–I Electron displacements in organic molecules, Inductive effect, Electromericeffect, Resonance, Heperconjugation.
Types of bond cleavage, Types of reagents, Electrophile, Nucleophile, Reactive intermediates: Carbocations, Carbanions, Free radicals, Mechanism
of organic reactions, Substitution reactions, Nucleophilic, Substitution (SN1, SN2), Electrophilicsubstitution (SE).

3. Atomic Structure
All matteris composed of atoms.
To understand the properties of matter
(inorganic, bio, organic etc..),
it is imperative to understand the
structure of atoms

4. An Atom, comparison of distances

5. Bohr’s model of Atom
➢Electrons revolve around a positively charged nucleus in discrete orbits (K, L, M or
n=1, 2, 3 respectively) with specific levels of energy.
➢Electrons positions are fixed as such, however, an electron can jump to higher or
lower energy level by absorptionor emissionof energy respectively
➢The energy levels are quantized 2
.
h
nmvr= = .hE

6. Bohr’s radius and Energy levels
•Columbic Force , Centrifugal force
•And Bohr’s theory
•From above, For H atom, Z= 1, n = 1, r = 0.529 A
•E = K. E + P. E =
•For H atom, Z= 1, n = 1, E = -13.6 eV2
2
r
Ze
KF= r
mv
F
2
= 2
.
h
nmvr= 22
22
4mKZe
hn
r

= r
Ze
KmvE
2
2
2
1
−= 22
2422
2

hn
mKeZ
Eor
n

−=

7. Energies of Bohr’s stationary states

8. Electromagnetic Radiation
•Oscillating charged particles produces magnetic and electric fields which
are perpendicular to each other and both are perpendicular to the wave
of propagation.
•These waves do not require medium i.e. electromagnetic wave can travel
in vacuum.
•The energy of EMR is Jshhc
c
hhE
34
10626.6 ,
−
==== 



9. Electromagnetic Radiation Spectrum
Q. An EMR of UV in nature would corresponds to -
(a) Electron at highest energy level or
(b) Electron at lowest energy level
Q. An EMR of IR in nature would corresponds to –
(a) Electron at highest energy level or
(b) Electron at lowest energy level

10. Application of Bohr’s theory, Hydrogen Spectrum
•Bohr’s theory could explain
Hydrogen spectrum correctly.
•For H atom Z = 1
12
EEE −=
2
1
2
2








−−








−=
n
R
n
R
E
HH 1018.2Constant
18
JRydbergR
H
−
==
11
2
2
2
1








−=
nnh
R
H

Q. In which case you expect a highest EMR frequency for Balmerseries (a) hydrogen
atom (b) He
+
(c) Li
+2
(d) Be
+3
? What would be ratios of their frequencies?

11. Q. The ratio of the energy of a photon of 2000 A
wavelength radiation to that of 10000A radiation is
(a)¼ (c) 5
(b)½ (d) 3
Answer is (C)

12. Photoelectric effect
•Metallic surfaces when irradiated by light (monochromatic), electrons are
emitted.This phenomenon is called Photoelectric Effect. The ejected
electrons are called as photoelectrons.
•For Photoelectric Effect to take place, energy
of incident light photons should be greater than
or equal to the work function of the metal.
•Work function
•This proves the particle nature of atoms.2
0
2
1
mVhhE +==  serghh .10626.6
28
0
−
==

13. Inadequacies of Bohr’s model
•It violates theHeisenberg's uncertainty principle. Bohr’s theory predicts
radius (position) and momentum (energy) at the same time, which is
impossible according to Heisenberg.
•Bohr’s spectral predictions for unielectronspecies (hydrogen, He
+
, Li
2+
etc)
were good butspectral predictions for larger atoms were poor.
•Bohr’s model could not explain the Zeeman effect (Bohr’s model could not
in the presence of a magnetic field).
•Bohr’s model could not explain the Stark effect (Bohr’s model could not in
the presence of a electric field)

14. Q.Whatarethewavelengthofaphotonemittedduringa
transitionfromn=5statetothen=2stateintheLi
2+
ion?
Calculateradiusoforbitsassociatedwiththetransition.
25
EEE −= 3 ZLi, For ,
2
6.13
5
6.13
2
2
2
2
=








−−








−=
ZZ
E
2
3
6.13
5
3
6.13
2
2
2
2








−−








−=E ( )( ) eV 25.7046.30896.4 =−−−=E A
J
mJ
E
hc
483
106.1704.25
sec)/103()10626.6(
19
834
=


=

=
−
−
 24
22
4mKZe
hn
r

= nm 44or 40.4
3
5
529.0
2
5
AAr =








= nm 7.05or 705.0
3
2
529.0
2
3
AAr =








=

15. Sommerfeld’smodification
•Sommerfeldsuggested electron revolves around the nucleus, in an
elliptical path with the nucleus at one of its foci.
✓Thus the velocity of the electron in an elliptical orbit would vary
and also causes the relativistic variation of the mass of the electron.
Sommerfeldcould explain multi-electron spectra to some extend.
✓The electron would have two axis, one longer and the other shorter
•Sommerfeldintroduced azimuthalquantum number and modified
electron energy levels number Quantum Azimuthal l ,
2
. ==

h
lmvr

16. Dual behavior of Atom, de-Broglie Law
•De-Broglie proposed that the Atom shows both particle and wave
behavior.
•Atoms behave as a particle as it obeys photoelectric effect and also
behave as a wave as it undergoes diffraction (a property that only
wave can have)
•Also de-Broglie suggested that every particle has a particle and
wave behavior. Coonst. Planks h ===
mv
h
p
h


17. Q.WhatwouldbetheapproximateratiosofdeBrogliewavelength
ofanelectronofhydrogentooxygeniftheelectronrevolvewiththe
samevelocity
16
1
)( =
H
O
a

 16 )( =
H
O
b


8
1
)( =
H
O
c

 8 )( =
H
O
d



18. Dual behavior of Atom, de-Broglie Law
•Electrons orbits the nucleus in a wave like motion giveseqn Broglie de in the vof value thePutting
mv
h
=
2
,

h
nmvrBut =
2
,
mr
h
nvhence

= 2 nr=

19. Q.Abulletofweight10gisfiredfromagunmovedwitha
velocity1000[m/s].CalculatethewavelengthoftheBullet.
λ= h/mv= 6.6x10
-34
[J s] / (0.01 [kg])(1000 [m/s])
= 6.6x10
-35
m

20. Microscope and Size of object
Invisible light microscope, we would be
able to see objects whose size is larger than ~
0.5*10
-6
m= 0.5 μm= 500 nm.
Reson: visible light, with a wavelength of ~ 500 nm cannot
resolve the image properly as it is smaller than its wavelength.
Bacteria, as viewed
using visible light
Bacteria, as viewed
using electrons!

21. Scanning Electron Microscope
A Scanning Electron Microscope (SEM) image.
SEM can resolve pictures as small as 5 nm.
Approx. 100 times better than normal visible
light microscopes
The electron microscopeuses the
wave behavior of electronsfor images

22. -For very small particles, to measure any property, we need interaction of
light with it
-The process is -shining light on electron and detecting
reflected light using a microscope
-For very small particles there is a
uncertainty in the result
-Heisenberg suggested that one can never
measure all the properties exactly
-Minimum uncertainty in position
is given by the wavelength of the light
Heisenberg Uncertainty Principle

23. -As per Planck’s law, a photon with a shorter wavelength would have a large
energy
-To locate the electron accurately, it is necessary to use light with a short
wavelength
-But, it would give a large ‘kick’(momentum transfer) to the electron
-But to determine its momentum accurately, electron must only be shined
with a long wavelength
-Thus, it is impossible to know boththe position and momentum exactly, i.e.,
Δx=0 and Δp=0
-It rules out existence of definite paths or trajectories of electrons and other
similar particles.
Heisenberg Uncertainty Principle

24. 100 g= 0.1-kg 144 km/hr = 40 m/s
Momentum = p = 0.1 x 40 = 4 kg m/s
Δp = 0.01p= 4 x 10
-2
kg m/s
An electron has mass 9.11 x 10
-31
kg and if the speed 144 km/ hr
momentum = 3.6 x 10
-29
kg m/s
uncertainty in momentum = 3.6 x 10
-31
kg m/s
The uncertainty in position is then
Q.Abowlerbowlsaballof100gat144km/hour.Ifthemomentumcanbe
measuredtoanaccuracyof1percent,calculatetheuncertaintyinthepositionofthe
ball.Whatwouldbetheuncertaintyinthepositionifthecricketballisreplacedby
anelectronatthesamespeed.

25. Black body radiation
-Blackbody radiation was one of the first signature of Quantum phenomenon
-Intensity shifts to the left (less energy) with rise in temperature

26. Quantum mechanics answered many questions in chemistry and Physics
–Explained periodic table
–Explained behavior of atoms, Molecules, chemical bonding
–Provides the theoretical basis for lasers, Spectra, and other countless
atomic and molecular phenomenon
-Quantum mechanics deals with the study of the motions of the
microscopic objects that have both observable wave like and particle
like properties. Note: It follows the laws of motion.
-Quantum mechanics was developed by Werner Physicist Heisenberg
and mathematician Erwin Schrodinger.
Quantum Mechanical model of Atom

27. Quantum Mechanical model of Atom
-The reason for failure of Bohrsmodel is that it ignored the dual
behavior of the atoms. Bohrsmodel was classical in nature.
-Schrodinger’s equation determines position and momentum
accurately, hence the probability of finding an electron.
-Solutions to Schrodingersequation results in wave functions, Ψ.
-Ψ
2
represents an orbital, a probability distribution map in an atom
where the electron resides.
-Quantized the energy states of the electron further with the following
quantum number
❖Angular momentum quantum number, l
❖Magnetic quantum number, m
l
-Thus, the size, shape, and orientation in space of an orbital are
determined by these quantum numbers, based on the wave function.

28. Schrodinger’s equation and model of Atom
-Schrödinger’s equation is
-Solutions to Schrodingersequation results in many wave functions, Ψ.
-Ψ
2
vsdistance plot produces Orbital
ˆ
EH=
2



x
SinA=
4
2
2
2
2



−=
dx
d
22



 x
CosA
dx
d
=
24
2
2
2
2



 x
SinA
dx
d
−=

29. Schrodinger’s equ
n
and model Contd…
ˆ
EH= -(1)-----
4
2
2
2
2



−=
dx
d mv
2
1
.
2
=EK
mv
h
=
22
2
2
vm
h
=
2
2
22

h
vm=
m
.
2
2

h
EK=
/
4
),1(
22
2
2
dxd
EqnApplying


−= .
4
.
2
1
.
2
2
2
2
dx
dh
m
EK


−= .
8
.
2
2
2
2
dx
d
m
h
EK


−=

30. ..)( EPEKEEnergyTotal += .
8
.
2
2
2
2
dx
d
m
h
EK


−= .. EPEEK −= .
8
.
2
2
2
2
dx
d
m
h
EPE


−=− ).( .
8
2
2
2
2


EPE
h
m
dx
d
−−= dimention onein eqn r Schrodinge is This 0, ).( .
8
2
2
2
2
=−+ 

EPE
h
m
dx
d 0, ).( .
8
becomes,equn thedimention, For three
2
2
2
2
2
2
2
2
=−+++ 

EPE
h
m
dz
d
dy
d
dx
d Schrodinger’s equ
n
and model Contd…

31. 0, ).( .
8
becomes,equn thedimention, For three
2
2
2
2
2
2
2
2
=−+++ 

EPE
h
m
dz
d
dy
d
dx
d OperatorLaplacian asknown is
0, ).( .
8
2
2
2
2

=− 

 EPE
h
m ValueEigen asknown is E functionEigen asknown is  Schrodinger’s equ
n
and model Contd…

32. Quantum Mechanical Explanation of Atomic Spectra
-Electron(ic) transition occurs between orbitals. Each wavelength of the
spectrum corresponds to the energy difference of orbital of an atom
-Electron transitions occur from an orbital in a lower energy level to an
orbital in a higher energy level –Absorption Spectra
-When an electron relaxes, electron jumps from an orbital in a higher
energy level to an orbital in a lower energy level -Emission Spectra
-In emission Spectra, a photon of light is released whose energy equals the
energy difference between the orbitals.

33. Q. Calculate the wavelength of the electromagnetic radiation emitted
when the electron in a hydrogen atom undergoes a transition from n =
6 to n= 3?
n
i= 6, n
f= 3( )
=






=
−
−
J10x1.816-
s
m
10x2.998J.s10x6.626
λ
19
834
= -1.816 x 10
-19
J
I 1.094 ×10
-6
m Iλ
Δ
hc
E= E
=
hc
λ ( ) 





−−=
−
22
E
6
1
3
1
J10x2.179Δ
18 







−−=
2
i
2
f
H
11
Δ
nn
RE
RydebergCostantR
H = 2.179 ×10
-18
J
1.094 ×10
-6
m

34. Quantum Mechanical model and significance of Ψand Ψ
2 0 ).( .
8
2
2
2
=− 

 EPE
h
m
)4(
V atom, HFor
0
2

me
−=
-The values of energy obtained from this model agrees well with experimental
results, and also with those calculated by Bohr model of atom.
-The position of electron are defined by polar coordinates (r, θ, φ), Since the atom
has spherical symmetry.
-Ψ
2
at any point around the nucleus gives measure of the electronic charge density
at that point in an extremely small volume and is a direct measure of probability of
finding an electron
-If Ψ
2
is high then electron density is high, i.e., the probability of finding an electron
is high.
-If Ψ
2
is not uniform around the nucleolus. 4D ,by given isfuntion on Distributi
22
r=

35. Probability and Orbitals
-Ψ
2
gives the probability density.
-Orbitalsare the space around the nucleouswhere electron stays for
maximum time, while in constant high speed motion
-For sorbital maximum at the nucleus?
–Decreases as you move away from
the nucleus
-Nodes are seen in the orbitalswhere
the probability drops to 0.
Nodes

36. Probability density and S Orbitals
-The most probable distance of the electron in a
1s orbital of H atom is 52.9 pm, agrees well with
Bohrsradius of H atom.
1s orbital
52.9 pm

37. r
-2
1
.
24
18
by given is atom, HFor Q.
a
-r
e
0
3/2
0
2 















aa
s

 Where a
0is Bohr’s radius. Let the radial node in 2s be at r
0. Calculate r in terms of a
0 0
2ar Ans.=

38. Q.Anatomicgasabsorbsaphotonof355nmandemitsattwo
wavelengths.Ifoneoftheemissionsisat680nmthencalculate
thewavelengthoftheotheremissions
Ans.743nm

39. Q. What is the electric potential that you should apply to a
proton beam to get an effective wavelength of 0.0005 A.
Ans.24.8MV

40. Q. A photochemical reaction is given by
Cl
2(g) → 2Cl(g), △H=245KJ/mol
Whatshould be maximum wavelength of light that can initiate
the reaction?
Ans.480nm

41. m
2eV
v=
m
sec
v =0.7710
8
v
max=310
18
max
eV =h
Q.Highvoltage,6kV,isappliedbetweentwoelectrodesinstalledintheX-ray
facility.Electronsaredrawnoutfromcathode,andcollidewithanodeathigh
speed.Calculatethevelocityofelectrons.Fromyouranswerguessthemaximum
limitoffrequencythatcanbeeffectivelyusedforx-rayexperiments.2
2
1
mveV= eVmv2
2
=

42. Q.IfElectromagneticradiationsofwavelength242nmcanionize
sodiumatom,thenfindouttheionizationenergyofsodiumin
KJ/mol.
Ans.495KJ/mol

43. Q.Inagaschamberofhydrogengashydrogenatomscollideheadonandend
upwithzerokineticenergy.Eachatomthenemitsaphotonofwavelength
121.6nm.Predicttheenergystatecorrespondingtothiswavelength?Also
calculatethespeedofthehydrogenatomstravelingbeforecollision?
Given:R
H=1.097×10
7
m
−1
andm
H=1.67×10
−27
Kg







−=
2
2
2
1
111
nn
R
H
 
hc
mv=
2
2
1
Ans.n
2=2
Ans.v=4.4x10
4
m/sec

44. Q. The speed of a proton is one hundredth of the speed of light in
vacuum. What is its de-Broglie wavelength ? Assume that one mole
of protons has a mass equal to one gram [h = 6.626 ×10
–27
erg sec]
Ans.13.2pm

45. 1. Principal Quantum Numbers (QN)
Azimuthalmomentum QN:
Magnetic QN:
Spin QN:
Quantum numbers.....4,3,2,1,0=l .....4,3,2,1=n 10 of Value −= ntol ltolm +−= of Values Possible 2,1,0,1,2,2For −−==
l
ml 2
1
2
1
of Values Possible −+== ors

46. Theenergyofanorbitalofanuni-electronatom(hydrogenandsimilarions)only
dependsonthevalueofn
-All orbitalsin one shell has same principal QN, n
-All orbitalsin Sub-shell has same value of n and l
-An orbital can be fully identified by three quantum numbers, n, l, and m
l
Each shell of QN = n contains n
sub-shells
n = 1, one subshell
n= 2, two subshells, etc
Each subshellof QN = l, contains
2l+ 1 orbitals
l= 0, 2(0) + 1 = 1
l= 1, 2(1) + 1 = 3
Quantum numbers

47. Orbital Shape
S orbital
p orbital
d orbital

48. Electron Configurations
-Electronsaredistributedinthesubshells/orbitalsofanatomindifferent
waysandthatproducesdifferentelectronconfigurations.
-Themoststableconfigurationhasthelowestenergy.
Note:Lowestenergycorrespondstotheconditionwhenelectronsare
closetothenucleusaspossibleandelectronsstayasfarawayfromeach
otheraspossible.
-We will discuss the following rules to describe E.C.
-Pauli Exclusion Principle
-Hund’sRule
-and AufbauPrinciple

49. Electron Configurations Contd….
Pauli exclusion principle
-All four quantum numbers of two electrons can not
be alike (similar)
Hund’srule
-In case orbitals of have identical energy (known as
degenerate orbitals) are available, electrons (initially)
occupy these orbitalssingly
Aufbaurule
-Electrons occupy orbitalsin such a way that the
energy of the atom is minimised

50. Q. Can an atom with an even number of electrons have
unpaired electrons?
Ans.Yes,forexampleOatom

51. The Aufbau’srule to fill electrons in subshells

52. Q.Write(a)twopossiblesetsofthefourquantumnumbersof
outermostelectronofNaatom(b)thefourquantumnumbers
ofoutermostelectronofCl
-
ion.
Soln.(a)ForNa,n=3,l=0,m=0,s=+½
andn=3,l=0,m=0,s=-½
(b)ForCl,n=3,l=1,m=0(couldbe-1and1aswell),s=+½
n=3,l=1,m=-1(couldbe0and1aswell),s=+½

53. Q. Which rule the following energy diagram violate?
Soln.Firstfig:Pauli’sExclusionPrinciple
Secondfig:Aufbaurule

54. The (n+l) Rule for EC –for Cr and Cu
3d
5
4s
1
3d
10
4s
1

55. EC and the Periodic Table

56. Q.Supposesomeoneiscapableofputtingthecontentsoflibraryon
asmoothpostcard.Assumethelibraryhas100bookswithaverage
numberofpages500.Willitbereadablewiththeelectron
microscope?Explain.
Solution: Let there are 100 books in the library,
and each book has 500 pages,
and each page is as large as two postcards.
the magnification should be = 2 ×500 ×100 ≈ 10000,
An electron microscope have resolution of 5 nm.
So the magnification is of the order of 800,000,
Thus, the postcard readable.

57. Shielding
-Predicts exact order of electron filling
-reduces the attraction towards the nucleus,
–Each e-acts as a shield and put e farther from
nucleus
–Hence with increasing E
-Energy ordering
–n is most important
–does change order in a multi-electron systems
-With increase in Z, the attraction for e-
increases
-and the Energy of the orbitalsdecreases
Shielding

58. Slater’s Rule: Z* = effective nuclear attraction = Z –S
–Grouping: 2s,2p/3s,3p/3d/4s,4p/4d/4f/5s,5p
–e in higher groups don’t shield lower groups
Slater’s Rule, Shielding Contd…
S and p
Orbitals
d and f
Orbitals
Shell
Number
–For ns/npvalence e-
-e in same group contributes 0.35 (1s = 0.3)
-e in n-1 group contributes 0.85
-e in n-2 group contributes 1.00
-For nd/nfvalence electrons
e in same group contributes 0.35
e-in groups to the left contribute 1.00
S = sum of all contributions

59. Slater’s Rule, Shielding Contd…
S and p
Orbitals
d and f
Orbitals
Shell
Number
Oxygen: (1s
2
)(2s
2
2p
4
)
Z* = Z –S
= 8 –2(0.85) –5(0.35)
= 4.55
Last electron detains only = 57% (4.55/8.00) of force for 8+ proton
nucleus

60. Q.Whatistheeffectivenuclearcharge
experiencedbyavalencep-electroninboron?
B: 1s
2
2s
2
2p
1
.
The valence p-electron in boron
resides in the 2psubshell.
S for [2p]= 0.85(2) + 0.35(2) =
2.40
Z= 5, Z
eff= 2.60
S and p
Orbitals
d and f
Orbitals
Shell
Number

61. Q. What is the shielding constantexperienced
by a 2pelectron in the nitrogen atom?
NS[2p]= 1.00(0) + 0.85(2) + 0.35(4) = 3.10

62. Slater’s Rule, Shielding Contd…
S and p
Orbitals
d and f
Orbitals
Shell
Number
Nickel: (1s
2
)(2s
2
2p
6
)(3s
2
3p
6
)(3d
8
)(4s
2
)
Z* = 28 –10(1.00) –16(0.85) –
1(0.35)
= 4.05 for 4s electron
-3s,3p 100% shield 3d because their probability is higher than 3d near nucleus
= shielding
-2s,2p only shield 3s,3p 85% because 3s/3p have significant regions of high
probability near the nucleus
For 3d electron,
σ = (0.35 ×9) + (1 ×18) = 21.15
Z* = Z –σ = 28–21.15 = 6.85