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. Raghav Samantaray:
Room 203-A
Email: rsamantray@gmail.com
Phone: 9778826785
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’s modification. de-Broglie’s
matter wave, Heisenberg’s uncertainty principle, photoelectric effect, concept of quantization of energy. Schrodinger
wave equation. Significance of Ψ2 SWE, Probability and orbital, shape, Quantum numbers (mathematical derivation not
required ), Pauli’s exclusion principle, Hund’s rule, Aufbau’s principle, 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’s rule. Covalent bond: VBT, Shapes of covalent molecules (VSEPR
theory and hybridization), Limitation of VBT, Molecular orbital theory (MOT), Energy level diagrams for homonuclear diatomic molecules,
Molecular forces - Hydrogen bond, Van der Waal’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-Chatlier principle. 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), Vapour pressure of solutions and Raoult's Law, Ideal and non-ideal solutions,
vapour pressure, 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, Electromeric effect, 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), Electrophilic substitution (SE).

3. Atomic Structure
All matter is 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 absorption or emission of 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 eV
2
2
r
Ze
KF =
r
mv
F
2
=
2
.
h
nmvr =
22
22
4 mKZe
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
JRydbergRH
−
==
11
2
2
2
1








−=
nnh
RH

Q. In which case you expect a highest EMR frequency for Balmer series (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 the Heisenberg'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 unielectron species (hydrogen, He+, Li 2+ etc)
were good but spectral 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. What are the wavelength of a photon emitted during a
transition from n = 5 state to the n = 2 state in the Li2+ ion?
Calculate radius of orbits associated with the transition.
25 EEE −=
3ZLi,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
( ) ( ) eV25.7046.30896.4 =−−−=E
A
J
mJ
E
hc
483
106.1704.25
sec)/103()10626.6(
19
834
=


=

= −
−

24
22
4 mKZe
hn
r

= nm44or40.4
3
5
529.0
2
5 AAr =





=
nm7.05or705.0
3
2
529.0
2
3 AAr =





=

15. Sommerfeld’s modification
• Sommerfeld suggested 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.
Sommerfeld could explain multi-electron spectra to some extend.
✓ The electron would have two axis, one longer and the other shorter
• Sommerfeld introduced azimuthal quantum number and modified
electron energy levels
numberQuantumAzimuthall,
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.Planksh ===
mv
h
p
h


17. Q. What would be the approximate ratios of de Broglie wavelength
of an electron of hydrogen to oxygen if the electron revolve with the
same velocity
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
giveseqnBrogliedein thevofvaluethePutting
mv
h
=
2
,

h
nmvrBut =
2
,
mr
h
nvhence

=
2  nr =

19. Q. A bullet of weight 10 g is fired from a gun moved with a
velocity 1000 [m/s]. Calculate the wavelength of the Bullet.
λ = h/mv = 6.6x10-34 [J s] / (0.01 [kg])(1000 [m/s])
= 6.6x10-35 m

20. Microscope and Size of object
In visible 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 microscope uses the
wave behavior of electrons for 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 both the 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 = 0.01 p = 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. A bowler bowls a ball of 100 g at 144 km/hour. If the momentum can be
measured to an accuracy of 1 percent, calculate the uncertainty in the position of the
ball. What would be the uncertainty in the position if the cricket ball is replaced by
an electron at the same speed.

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 Bohrs model is that it ignored the dual
behavior of the atoms. Bohrs model was classical in nature.
- Schrodinger’s equation determines position and momentum
accurately, hence the probability of finding an electron.
- Solutions to Schrodingers equation 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, ml
- 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 Schrodingers equation results in many wave functions, Ψ.
- Ψ 2 vs distance 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 equn 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
−−=
dimentiononeineqnrSchrodingeisThis0,).(.
8
2
2
2
2
=−+ 

EPE
h
m
dx
d
0,).(.
8
becomes,equnthedimention,For three
2
2
2
2
2
2
2
2
=−+++ 

EPE
h
m
dz
d
dy
d
dx
d
Schrodinger’s equn and model Contd…

31. 0,).(.
8
becomes,equnthedimention,For three
2
2
2
2
2
2
2
2
=−+++ 

EPE
h
m
dz
d
dy
d
dx
d
OperatorLaplacianasknownis
0,).(.
8
2
2
2
2

=− 

 EPE
h
m
ValueEigenasknownisE
functionEigenasknownis
Schrodinger’s equn 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?
ni = 6, nf = 3
( )
=






= −
−
J10x1.816-
s
m
10x2.998J.s10x6.626
λ 19
834
= -1.816 x 10-19 J
I 1.094 × 10-6 m I
λ
Δ
hc
E =
E
=
hc
λ
( ) 





−−= −
22
E
6
1
3
1
J10x2.179Δ 18








−−= 2
i
2
f
H
11
Δ
nn
RE
Rydeberg Costant RH = 2.179 × 10-18 J
1.094 × 10-6 m

34. Quantum Mechanical model and significance of Ψ and Ψ2
0).(.
8
2
2
2
=− 

 EPE
h
m
)4(
Vatom,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.
- Ψ2 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 Ψ2 is high then electron density is high, i.e., the probability of finding an electron
is high.
- If Ψ2 is not uniform around the nucleolus.
4D,bygivenisfuntiononDistributi 22
 r=

35. Probability and Orbitals
- Ψ2 gives the probability density.
- Orbitals are the space around the nucleous where electron stays for
maximum time, while in constant high speed motion
- For s orbital maximum at the nucleus?
– Decreases as you move away from
the nucleus
- Nodes are seen in the orbitals where
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
Bohrs radius of H atom.
1s orbital
52.9 pm

37. r
-2
1
.
24
18
bygivenisatom,HForQ.
a
-r
e
0
3/2
0
2 















aa
s


Where a0 is Bohr’s radius. Let the radial node in 2s be at r0. Calculate r in terms of a0
02arAns. =

38. Q. An atomic gas absorbs a photon of 355 nm and emits at two
wavelengths. If one of the emissions is at 680 nm then calculate
the wavelength of the other emissions
Ans. 743 nm

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. 24.8 M V

40. Q. A photochemical reaction is given by
Cl2(g) → 2Cl(g), △H=245 KJ/mol
What should be maximum wavelength of light that can initiate
the reaction?
Ans. 480 nm

41. m
2eV
v =
m
sec
v = 0.77108
vmax =31018
maxeV = h
Q. High voltage, 6kV, is applied between two electrodes installed in the X-ray
facility. Electrons are drawn out from cathode, and collide with anode at high
speed. Calculate the velocity of electrons. From your answer guess the maximum
limit of frequency that can be effectively used for x-ray experiments.
2
2
1
mveV =
eVmv 22
=

42. Q. If Electromagnetic radiations of wavelength 242nm can ionize
sodium atom, then find out the ionization energy of sodium in
KJ/mol.
Ans. 495 KJ/mol

43. Q. In a gas chamber of hydrogen gas hydrogen atoms collide head on and end
up with zero kinetic energy. Each atom then emits a photon of wavelength
121.6nm. Predict the energy state corresponding to this wavelength? Also
calculate the speed of the hydrogen atoms traveling before collision?
Given: RH=1.097×107m−1 and mH = 1.67 × 10−27 Kg








−= 2
2
2
1
111
nn
RH


hc
mv =2
2
1
Ans. n2 =2
Ans. v = 4.4 x 10 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 × 10–27erg sec]
Ans. 13.2 pm

45. 1. Principal Quantum Numbers (QN)
Azimuthal momentum QN:
Magnetic QN:
Spin QN:
Quantum numbers
.....4,3,2,1,0=l
.....4,3,2,1=n
10ofValue −= ntol
ltolm +−=ofValuesPossible
2,1,0,1,2,2For −−== lml
2
1
2
1
ofValuesPossible −+== ors

46. The energy of an orbital of an uni-electron atom (hydrogen and similar ions) only
depends on the value of n
- All orbitals in one shell has same principal QN, n
- All orbitals in Sub-shell has same value of n and l
- An orbital can be fully identified by three quantum numbers, n, l, and ml
Each shell of QN = n contains n
sub-shells
n = 1, one subshell
n= 2, two subshells, etc
Each subshell of 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
- Electrons are distributed in the subshells/orbitals of an atom in different
ways and that produces different electron configurations.
- The most stable configuration has the lowest energy.
Note: Lowest energy corresponds to the condition when electrons are
close to the nucleus as possible and electrons stay as far away from each
other as possible.
- We will discuss the following rules to describe E.C.
- Pauli Exclusion Principle
- Hund’s Rule
- and Aufbau Principle

49. Electron Configurations Contd….
Pauli exclusion principle
- All four quantum numbers of two electrons can not
be alike (similar)
Hund’s rule
- In case orbitals of have identical energy (known as
degenerate orbitals) are available, electrons (initially)
occupy these orbitals singly
Aufbau rule
- Electrons occupy orbitals in 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, for example O atom

51. The Aufbau’s rule to fill electrons in subshells

52. Q. Write (a) two possible sets of the four quantum numbers of
outermost electron of Na atom (b) the four quantum numbers
of outermost electron of Cl- ion.
Soln. (a) For Na, n = 3, l=0, m =0, s = + ½
and n = 3, l=0, m =0, s = - ½
(b) For Cl, n = 3, l=1, m = 0 (could be -1 and 1 as well), s = + ½
n = 3, l=1, m = -1 (could be 0 and 1 as well), s = + ½

53. Q. Which rule the following energy diagram violate?
Soln. First fig: Pauli’s Exclusion Principle
Second fig: Aufbau rule

54. The (n+l) Rule for EC – for Cr and Cu
3d54s1
3d104s1

55. EC and the Periodic Table

56. Q. Suppose someone is capable of putting the contents of library on
a smooth postcard. Assume the library has 100 books with average
number of pages 500. Will it be readable with the electron
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 orbitals decreases
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/np valence 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/nf valence 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: (1s2)(2s22p4)
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. What is the effective nuclear charge
experienced by a valence p- electron in boron?
B: 1s2 2s2 2p1 .
The valence p- electron in boron
resides in the 2p subshell.
S for [2p] = 0.85(2) + 0.35(2) =
2.40
Z = 5, Zeff = 2.60
S and p
Orbitals
d and f
Orbitals
Shell
Number

61. Q. What is the shielding constant experienced
by a 2p electron in the nitrogen atom?
N S[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: (1s2)(2s22p6)(3s23p6)(3d8)(4s2)
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