Gnp ch103-lecture notes

CH103Physical Chemistry: Introduction to Bonding

G. Naresh Patwari

Room No. 215; Department of Chemistry

2576 7182 [email protected]

Physical Chemistry –I.N. LevinePhysical Chemistry – P.W. AtkinsPhysical Chemistry: A Molecular Approach – McQuarrie and Simon

Websites:http://www.chem.iitb.ac.in/~naresh/courses.html

www.chem.iitb.ac.in/academics/menu.php

IITB-Moodle http://moodle.iitb.ac.in

http://ocw.mit.edu/OcwWeb/web/courses/courses/index.htm#Chemistry

http://education.jimmyr.com/Berkeley_Chemistry_Courses_23_2008.php

Recommended Texts (Physical Chemistry)

What Do You Get to LEARN?

�Why Chemistry?

� Classical Mechanics Doesn't Work all the time!

� Is there an alternative? QUANTUM MECHANICS

�Origin of Quantization & Schrodinger Equation

� Applications of Quantum Mechanics to Chemistry

� Atomic Structure; Chemical Bonding; Molecular Structure

Why should Chemistry interest you?

Chemistry plays major role in

1. Daily use materials: Plastics, LCD displays2. Medicine: Aspirin, Vitamin supplements3. Energy: Li-ion Batteries, Photovoltaics4. Atmospheric Science Green-house gasses, Ozone depletion 5. Biotechnology Insulin, Botox6. Molecular electronics Transport junctions, DNA wires

Haber Process

Haber Process

The Haber process remains largest chemical and economic venture. Sustains third of worlds population

Transport Junctions

Quantum theory is necessary for the understanding and the development of chemical processes and molecular devices

LCD Display

Classical Mechanics

Newton's Laws of Motion

1. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

2. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. The direction of the force vector is the same as the direction of the acceleration vector

3. For every action there is an equal and opposite reaction.

Black-Body Radiation; Beginnings of Quantum Theory

Rayleigh-Jeans law was based on equipartitioning of energy

Planck’s hypothesisThe permitted values of energies are integral multiples of frequenciesE = nhν = nhc/λn = 0,1,2,…

Value of ‘h’ (6.626 x 10-34 J s) was determined by fitting the experimental curve to the Planck’s radiation law

kT4

8πρ

λ=

Planck’s

radiation law

( )hckT

hc

e5

8

1λ

πρ

λ=

−

Towards

Ultraviolet Catastrophe

Hot objects glow

Planck did not believe in the quantum theory and struggled to avoid quantum theory and make its influence as small as possible

λ =bTmax

Heat Capacities of Solids

Dulong – Petit Law

The molar heat capacity of all solids have nearly same value of ~25 kJ

Element Gram heat

capacity

J deg-1 g-1

Atomic

weight

Molar heat

capacity

J deg-1mol-

1

Bi 0.120 212.8 25.64Au 0.125 198.9 24.79Pt 0.133 188.6 25.04Sn 0.215 117.6 25.30Zn 0.388 64.5 25.01Ga 0.382 64.5 24.60Cu 0.397 63.31 25.14Ni 0.433 59.0 25.56Fe 0.460 54.27 24.98Ca 0.627 39.36 24.67S 0.787 32.19 25.30

1

,

3 3

3 25

A

mV m

V

Um N kT RT

UC R kJmol

T

−

= =

∂ = = ≈

∂


Einstein formula

Einstein considered the oscillations of atoms in the crystal about its equilibrium position with a single frequency ‘ν’ and invoked the Planck’s hypothesis that these vibrations have quantized energies nhνννν

22

2

, 3 ;

1

E

E

TE

V m E

T

e hC R

T ke

θ

θ

θ νθ

= = −

3

1

Am h

kT

N hU

eν

ν=

−


Debye formula

Averaging of all the frequencies νD

3 4

, 23 ;

( 1)

D xTD D

V m Dxo

hx eC R dx

T e k

θθ νθ

= =

− ∫

Rutherford Model of Atom

Alpha particles were (He2+) bombarded on a 0.00004 cm (few hundreds of atoms) thick gold foil and most of the alpha particles were not deflected

Rutherford Model of Atom

Positive Charge

Negatively ChargedParticles

Thompson’s model of atom is incorrect. Cannot explain Rutherford’s experimental results

Planetary model of atoms with central positively charged nucleus and electrons going around

Classical electrodynamics predicts that such an arrangement emits radiation continuously and is unstable

Atomic Spectra

Balmer Series

410.1 nm434.0 nm486.1 nm656.2 nm 2 1

Rn n

R 1 9678 x 1 nm

2 21 2

1 1 1

.0 0

λ ∞

− −

∞

= −

=

“RH is the most accurately measured fundamental physical constant”

The Rydberg-Ritz Combination Principle states that the spectral lines of any element include frequencies that are either the sum or the difference of the frequencies of two other lines.

Bohr Phenomenological Model of Atom

Electrons rotate in circular orbits around a central (massive) nucleus, and obeys the laws of classical mechanics.

Allowed orbits are those for which the electron’s angular momentum equals an integral multiple of h/2π i.e.mevr = nh/2π

Energy of H-atom can only take certain discrete values: “Stationary States”

The Atom in a stationary state does not emit electromagnetic radiation

When an atom makes a transition from one stationary state of energy Ea to another of energy Eb, it emits or absorbs a photon of light: Ea – Eb = hv

Energy expression

Bohr Model of Atom

Angular momentum quantized

n=1,2,3,...2

(2 )

ππ λ

=

=

nhmvr

r n

4

2 2 2

0

1.

8ε= − e

n

m eE

h n

Spectral lines4

2 2 2 2

1 1 , 1, 2 , 3, ...

8ν

ε

∆ = − = =

ei f

i f

m eE h n n

h n n

Explains Rydberg formula

Ionization potential of H atom 13.6 eV

42 1

2 21.09678 x 10 nm

8ε− −

∞ = =em eR

h

Bohr Model of Atom

The Bohr model is a primitive model of the hydrogen atom. As a theory, it can be derived as a first-order approximation of the hydrogen atom using the broader and much more accurate quantum mechanics

Photoelectric Effect: Wave –Particle Duality

Experimental Observations

Increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy!

Red light will not cause the ejection of electrons, no matter what the intensity!

Weak violet light will eject only a few electrons! But their maximum kinetic energies are greater than those for very intense light of longer (red) wavelengths

Electromagnetic Radiation

Wave energy is related to Intensity, I ∝ E20and is independent of ω

0 ( )ω= −E E Sin kx t

Photoelectric Effect: Wave –Particle Duality

Einstein borrowed Planck’s idea that ∆E=hν and proposed that radiation itself existed as small packets of energy (Quanta)now known as PHOTONS

Energy is frequency dependent

φ = Energy required to remove electron from surface

ν=P

E h

21

2φ φ= = + = +

P ME hv KE mv

Diffraction of Electrons : Wave –Particle Duality

Davisson-Germer Experiment

A beam of electrons is directed onto the surface of a nickel crystal. Electrons are scattered, and are detected by means of a detector that can be rotated through an angle θ. When the Bragg condition mλ = 2dsinθ was satisfied (d is the distance between the nickel atom, and m an integer) constructive interference produced peaks of high intensity


G. P. Thomson Experiment

Electrons from an electron source were accelerated towards a positive electrode into which was drilled a small hole. The resulting narrow beam of electrons was directed towards a thin film of nickel. The lattice of nickel atoms acted as a diffraction grating, producing a typical diffraction pattern on a screen

de Broglie Hypothesis: Mater waves

Since Nature likes symmetry, Particles also should have wave-like nature

De Broglie wavelength

λ = =h h

p mv

Electron moving @ 106m/s-34

10

-31 6

6.6x10 J s7 10

9.1x10 Kg 1x10 m/sλ −= = = ×

×

hm

mv

He-atom scattering

Diffraction pattern of He atoms at the speed 2347 m s-1 on a silicon nitride transmission grating with 1000 lines per millimeter. Calculated de Broglie wavelength 42.5x10-12 m

de Broglie wavelength too small for macroscopic objects


The wavelength of the electrons was calculated, and found to be in close agreement with that expected from the De Broglie equation

Wave –Particle Duality

Light can be Waves or Particles. NEWTON was RIGHT!

Electron (matter) can be Particles or Waves

Electrons and Photons show both wave and particle nature “WAVICLE”

Best suited to be called a form of “Energy”

Wave –Particle Duality

Bohr – de Broglie Atom

Constructive Interference of the electron-waves can result in stationary states (Bohr orbits)

If wavelength don’t match, there can not be any energy level (state)

Bohr condition & De Broglie wavelength

2 n=1,2,3,...

n=1,2,3,...2

π λ

λ

π

=

=

=

r n

h

mv

nhmvr

Electrons in atoms behave as standing waves

Uncertainty Principle

Uncertainty principle

.4π

∆ ∆ ≥x

hx p

Schrodinger’s philosophy

PARTICLES can be WAVES and WAVES can be PARTICLES

New theory is required to explain the behavior of electrons, atoms and molecules

Should be Probabilistic, not deterministic (non-Newtonian) in nature

Wavelike equation for describing sub/atomic systems


PARTICLES can be WAVES and WAVES can be PARTICLES

A concoction of

221

2 2

Wave is Particle

2 Particle is Wave

pE T V mv V V

m

E h

h

p k

ν ω

πλ

= + = + = +

= =

= =

�

let me start with classical wave equation

Do I need to know any Math?

Algebra

Trigonometry

Differentiation

Integration

Differential equations

[ ]1 1 2 2 1 1 2 2( ) ( ) ( ) ( )+ = +A c f x c f x c Af x c Af x

( ) ( ) ikx

Sin kx Cos kx e

2 2

2 2

∂ ∂

∂ ∂

d d

dx dx x x

( )∫ ∫b

ikx

ae dx f x dx

2 2

2 2

( ) ( ) ( ) ( )( ) ( )+ +

∂ ∂ ∂ ∂+ + + =

∂ ∂∂ ∂

f f f fm

x y x yf x nf y k

x yx y

Remember!

∂ Ψ ∂ Ψ=

∂ ∂

Ψ =

Ψ = = −

= =

= =

⋅ − ⋅ = − =

�

�

2 2

2 2 2

( , ) 1 ( , ) Classical Wave Equation

( , ) Amplitude

( , ) ; Where 2 is the phase

2

2

i

x t x tx c tx t

xx t Ce t

E h

hp k

x x p E tt

α α π νλ

ν ω

πλ

α π νλ


ix t EiCe i x t i x t

t t t

( , )( , ) ( , )α α α∂Ψ ∂ ∂ −

= ⋅ = ⋅Ψ ⋅ = ⋅ Ψ ⋅ ∂ ∂ ∂ �


x tE x t

i t

( , )( , )

− ∂Ψ= ⋅ Ψ

∂

�

i x p E tx t Ce( , ) and α α

⋅ − ⋅Ψ = =

�


∂Ψ= ⋅ Ψ

∂

� ( , )( , ) x

x tp x t

i x


⋅ − ⋅Ψ = =

�

α α α∂Ψ ∂ ∂ = ⋅ = ⋅Ψ ⋅ = ⋅Ψ ⋅ ∂ ∂ ∂ �

( , )( , ) ( , )i xpx t

iCe i x t i x tx x x

ix t EiCe i x t i x t

t t t

( , )( , ) ( , )α α α∂Ψ ∂ ∂ −

= ⋅ = ⋅Ψ ⋅ = ⋅ Ψ ⋅ ∂ ∂ ∂ �


x tE x t

i t

( , )( , )

− ∂Ψ= ⋅ Ψ

∂

�

i xpx tiCe i x t i x t

x x x( , )

( , ) ( , )α α α∂Ψ ∂ ∂ = ⋅ = ⋅Ψ ⋅ = ⋅Ψ ⋅ ∂ ∂ ∂ �

x

x tp x t

i x( , )

( , )∂Ψ

= ⋅Ψ∂

�


⋅ − ⋅Ψ = =

�

� �− ∂ ∂ ∂ ∂= = = − =

∂ ∂ ∂ ∂

� �� Operatorsxi E i p

i t t i x x

Operators

x

x t x tE x t p x t

i t i x

( , ) ( , )( , ) ( , )

− ∂Ψ ∂Ψ= ⋅ Ψ = ⋅Ψ

∂ ∂

� �

Operator

A symbol that tells you to do something to whatever follows it

Operators can be real or complex,

Operators can also be represented as matrices

xx t E x t x t p x ti t i x

( , ) ( , ) ( , ) ( , )− ∂ ∂

Ψ = ⋅Ψ Ψ = ⋅Ψ∂ ∂

� �

Operators and Eigenvalues

Operator operating on a function results in re-generating the same function multiplied by a number

The function f(x) is eigenfunction of operator Â and a its eigenvalue

( )( ) α=f x Sin x

( )( ) α α= ⋅d

f x Cos xdx

( ) ( )2

2 2

2( ) ( )α α α α α= ⋅ = − ⋅ = − ⋅

d df x Cos x Sin x f x

dx dx

is an eigenfunction of

operator and is its

eigenvalue

( )αSin x2

2

d

dx

2α−

� ( ) ( ) Eigen Value EquationA f x a f x⋅ = ⋅

The mathematical description of quantum mechanics is built upon the concept of an operator

The values which come up as result of an experiment are the eigenvalues of the self-adjoint linear operator.

The average value of the observable corresponding to operator Â is

The state of a system is completely specified by the wavefunction Ψ(x,y,z,t) which evolves according to time-dependent Schrodinger equation

Laws of Quantum Mechanics

ˆ* υ= Ψ Ψ∫a A d

Probability Distribution and Expectation Values

Classical mechanics uses probability theory to obtain relationships for systems composed of larger number of particles

For a probability distribution function P(x) the average value is given by

2 2

1 1

: ( ) and ( )= =

= =∑ ∑n n

j j j j j j

j j

Mean x x P x x x P x

Let us consider Maxwell distribution of speeds

The mean speed is calculated by taking the product of each speed with the fraction of molecules with that particular speed and summing up all the products. However, when the distribution of speeds is continuous, summation is replaced with an integral

RTv vf v dv

M

12

0

8( )

π

∞ = =

∫

MvRT

Mf v v e

RT

232

2 2( ) 42

ππ

− =

Probability Distribution and Expectation Values

Born Interpretation

In the classical wave equation Ψ(x,t) is the Amplitude and |Ψ(x,t)|2 is the Intensity

The state of a quantum mechanical system is completely specified by a wavefunction Ψ(x,t) ,which can be complex

All possible information can be derived from Ψ(x,t)

From the analogy of classical wave equation, Intensity is replaced by Probability. The probability is proportional to the square of the of the wavefunction |Ψ(x,t)|2 , known as probability density P(x)

Born Interpretation

P x x t x t x t2

( ) ( , ) ( , ) ( , )∗= Ψ = Ψ ⋅Ψ

Probability density

Probability

a a a aP x x x dx x t dx x t x t dx2

( ) ( , ) ( , ) ( , )∗≤ ≤ + = Ψ = Ψ ⋅Ψ

Probability in 3-dimensions

*

2

P( , , )

( , , , '). ( , , , ')

( , , , ') τ

≤ ≤ + ≤ ≤ + ≤ ≤ +

= Ψ Ψ

= Ψ

a a a a a a

a a a a a a

a a a

x x x dx y y y dy z z z dz

x y z t x y z t dxdydz

x y z t d

Normalization of Wavefunction

∞∞

x�

Ψ

∞∞

x�

Ψ

Unacceptable wavefunction

Since Ψ*Ψdτ is the probability, the total probability of finding the particle somewhere in space has to be unity

If integration diverges, i.e. � ∞: Ψ can not be normalized, and therefore is NOT an acceptable wave function. However, a constant value C ≠ 1 is perfectly acceptable.

*

*

( , , ). ( , , )

1τ

Ψ Ψ

= Ψ Ψ = Ψ Ψ =

∫∫∫

∫

all space

all space

x y z x y z dxdydz

d

Ψ must vanish at ±∞, or more appropriately at the boundaries and Ψ must be finite


�

�xx

xx

x x

d dp mv p i

i dx dx

pT

m

2

Position,

Momentum,

Kinetic Energy, 2

= = = −

=

��

�

�

�

x

yx z

dT

m dx

pp pT T

m m m m x y z

V x V x

2 2

2

22 2 2 2 2 2

2 2 2

2

Kinetic Energy, + 2 2 2 2

Potential Energy, ( ) ( )

−=

− ∂ ∂ ∂= + = + +

∂ ∂ ∂

�

�

Classical Variable QM Operator

The mathematical description of QM mechanics is built upon the concept of an operator


The values which come up as result of an experiment are the eigenvalues of the self-adjoint linear operator

In any measurement of observable associated with operator Â, the only values that will be ever observed are the eigenvalues an, which satisfy the eigenvalue equation:

Ψn are the eigenfunctions of the system and an are corresponding eigenvalues

If the system is in state Ψk , a measurement on the system will yield an eigenvalue ak

� ⋅Ψ = ⋅ Ψn n nA a


22 2

2

22 2

2

If ( ) ( )

( ) ( )

( ) ( ) ( )

If ( )

( )

( ) ( )

x

x

x

x Sin cx

dx c Cos cx

dx

dx c Sin cx c x

dx

x e

dx e

dx

dx e x

dx

α

α

α

α

α α

Ψ =

Ψ = ⋅

Ψ = − ⋅ = − ⋅Ψ

Ψ =

Ψ = ⋅

Ψ = ⋅ = ⋅Ψ

Only real eigenvalues will be observed, which will specify a number corresponding to the classical variable

There may be, and typically are, many eigenfunctions for the same QM operator!


All the eigenfunctions of Quantum Mechanical operators are “Orthogonal”

* ( ) ( ) 0 for ψ ψ ψ ψ+∞

−∞

= = ≠∫ m n m nx x dx m n


The average value of the observable corresponding to operator Â is ˆ* υ= Ψ Ψ∫a A d

From classical correspondence we can define average values for a distribution function P(x)

<a> corresponds to the average value of a classical physical quantity or observable , and Â represents the corresponding Quantum mechanical operator

2 2( ) and ( ) ∞ ∞

−∞ −∞= ⋅ = ⋅∫ ∫x xP x dx x x P x dx

� � �2 *

ˆ. ( ) = .

+∞ +∞

−∞ −∞

= Ψ ≈ Ψ Ψ = Ψ Ψ∫ ∫ ∫all space

a A P x dx A dx A dx A

Time-dependent Schrodinger equation

where

Time evolution of the wavefunction is related to the total energy of the system/particle


��

22

( , ) ( , ) ( , )2

∂Ψ = − ∇ + Ψ

∂

�� xi x t V x t x t

t m

� � �2

2

( , )2

= − ∇ +�

xH V x tm

�2

2

2

∂∇ =

∂x

x

The wavefunction Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger equation

Operators

� �xi E i p

i t t i x x

− ∂ ∂ ∂ ∂= = = −

∂ ∂ ∂ ∂

� ��

xx t E x t x t p x ti t i x

( , ) ( , ) ( , ) ( , )− ∂ ∂

Ψ = ⋅Ψ Ψ = ⋅Ψ∂ ∂

� �

Total energy operator is also known as Hamiltonian

� �E V x Hm x

2

2( )

2

− ∂= + =

∂

�

�

xpE T V V xm

ix

E V x V xm m x

2

2

2

2

( )2

( ) ( )2 2

= + = +

∂ − − ∂∂ = + = +

∂

��

Schrodinger Equation


�i x t H x t V x x tt m

22( , ) ( , ) ( ) ( , )

2

∂ −Ψ = ⋅Ψ = ∇ + Ψ

∂

��

In 3-dimensions

�i x y z t V x y z H x y z tt m

22( , , , ) ( , , ) ( , , , )

2

∂ −Ψ = ∇ + Ψ = ⋅Ψ

∂

��

x y z

2 2 22

2 2 2where

∂ ∂ ∂∇ = + +

∂ ∂ ∂

�i x t H x tt

( , ) ( , )∂

Ψ = ⋅ Ψ∂�

E and Ĥ can be interchangeably used

i x t E x tt

( , ) ( , )∂

Ψ = ⋅Ψ∂�



�i x t H x t V x x tt m

22( , ) ( , ) ( ) ( , )

2

∂ −Ψ = ⋅Ψ = ∇ + Ψ

∂

��

� �H x y z t i x y z t H V x y zt m

22( , , , ) ( , , , ) ; ( , , )

2∂ −

⋅ Ψ = Ψ = ∇ +∂

��

x y z t x y z t( , , , ) ( , , ) ( ) ψ φ ψ φΨ = ⋅ ⇒ Ψ = ⋅

Schrodinger equation in 3-dimensions

�H it

∂⋅ Ψ = Ψ

∂�

�H it

( ) ( )ψ φ ψ φ∂

⋅ = ⋅∂�


�

�

H it

Ht

( ) ( )

operates only on ψ and operates only on

ψ φ ψ φ

φ

∂⋅ = ⋅

∂

∂

∂

�

�H it

φ ψ ψ φ∂

⋅ = ∂ �

�Hit

Divide by

1

ψ φ

ψφ

ψ φ

⋅

∂ =

∂ �

LHS is a function of co-ordinates and RHS is function of time. If these two have to be equal then both functions must be equal to constant, say W


�Hi Wt

1ψφ

ψ φ

∂ = =

∂ �

��H

W H W

i W i Wt t

1

ψψ ψ

ψ

φ φ φφ

⋅= =

∂ ∂ = =

∂ ∂ � �

The solution of the differential equationiWt

i W t et

is ( )φ φ φ−∂

= =∂

��

Separation of variables


iWtt e( )φ

−

= �

iWt iWte e e

2 0 1φ φ φ−

∗= ⋅ = ⋅ = =� �

The probability distribution function

is independent of time

2 2 2 2 2ψ φ ψ φ ψΨ = ⋅ = ⋅ =

is the time independent Schrodinger Equation represents Stationary States of the system

�H Wψ ψ=


In classical mechanics Ĥ represents total energyWe can therefore write

� �H W H E as ψ ψ ψ ψ= =

�H Eψ ψ=

V x x E xm x

2 2

2( ) ( ) ( )

2ψ ψ

∂− + = ⋅

∂

�

Schrodinger equation is an eigen-value equation

There can be many solutions ψn(x) each corresponding to different energy En


In 3-dimensions the Schrodinger equation is

V x y z x y z E x y zm x y z

2 2 2 2

2 2 2( , , ) ( , , ) ( , , )

2ψ ψ

∂ ∂ ∂− + + + = ⋅

∂ ∂ ∂

�

For ‘n’ particle system the Schrodinger equation in 3-dimensions is

ψ ψ

ψ ψ

=

− − −

∂ ∂ ∂− + + + = ⋅

∂ ∂ ∂

⇐

∑�2 2 2 2

2 2 21

1 2 3 1 1 2 3 1 1 2 3 1

( , , )2

( , , ,... , , , , ,... , , , , ,... , )

n

i i i i

n n n n n n

V x y z Em x y z

x x x x x y y y y y z z z z z


�

ψ ψ

∂ ∂ ∂= − + +

∂ ∂ ∂

∂ ∂ ∂− + +

∂ ∂ ∂

∂ ∂ ∂− + + ∂ ∂ ∂

∂ ∂ ∂− + + ∂ ∂ ∂

+ + + + + +

⇐

�

�

�

�

2 2 2 2

2 2 21 1 1 1

2 2 2 2

2 2 22 2 2 2

2 2 2 2

2 2 23 3 3 3

2 2 2 2

2 2 24 4 4 4

12 13 14 23 24 34

1 2 3

2

2

2

2

( , , ,

Hm x y z

m x y z

m x y z

m x y z

V V V V V V

x x x x4 1 2 3 4 1 2 3 4, , , , , , , , )y y y y z z z z

( )1 1 1 1, ,m x y z

( )3 3 3 3, ,m x y z( )2 2 2 2, ,m x y z

( )4 4 4 4, ,m x y z

Restrictions on wavefunction

ψ must be a solution of the Schrodinger equation

ψ must be normalizable: ψ must be finite and �0 at boundaries/ ±∞

Ψ must be a continuous function of x,y,z

dΨ/dq must be must be continuous in x,y,z

Ψ must be single-valued

Ψ must be quadratically-intergrable(square of the wavefunction should be integrable)

Acceptable wavefunctions


Unacceptable because ψ is not continuous

Unacceptable because ψ is not single-valued

Unacceptable because dψ/dq is not continuous

Unacceptable because ψ goes to infinity


Because of these restrictions, solutions of the Schrodinger equations do not in general exist for arbitrary values of energy

In other words, a particle may possess only certain energies otherwise its wavefunction would be Unacceptable

The energy of a particle is quantized

Quantization?

The function f(x) = x2 can take any values

If we impose arbitrary condition that f(x) can only be multiples of three, then values if x are restricted.

Quantization!

Physically meaningful boundary conditions lead to quantization ☺

Not deterministic: Can not precisely determine many parameters in the system, but Ψ can provide all the information (spatio-temporal) of a system.

Only average values and probabilities can be obtained for classical variables, now in new form of “operators”.

Total energy is conserved, but quantization of energy levels come spontaneously from restriction on wave function or boundary condition

Final outputs tally very well with experimental results, and does not violate Classical mechanics for large value of mass.

Essence of Quantum Mechanics

Quantum Mechanics

Examples of Exactly Solvable Systems

1. Free Particle2. Particle in a Square-Well Potential3. Hydrogen Atom

Time-independent Schrodinger equation

Free Particle

�H Eψ ψ=

V x x E xm x

2 2

2( ) ( ) ( )

2ψ ψ

∂− + = ⋅

∂

�

For a free particle V(x)=0There are no external forces acting

x E xm x

2 2

2( ) ( )

2ψ ψ

∂− = ⋅

∂

�

Free Particle

( ) ( )

( )

x A kx B kx

x A kx B kx k A kx B kxdx dx

x k A kx B kx k xdx

22 2

2

( ) sin cos

( ) sin cos cos sin

( ) sin cos ( )

ψ

ψ

ψ ψ

= +

∂ ∂= + = −

∂= − + = −

x E xm x

2 2

2( ) ( )

2ψ ψ

∂− = ⋅

∂

�

m

x

Second-order linear differential equation

Let us assumeTrial Solutionx A kx B kx( ) sin cosψ = +

Free Particle

x E xm x

2 2

2( ) ( )

2ψ ψ

∂− = ⋅

∂

�

k mEk x E x E k

m m

2 2 22 2( ) ( )

2 2ψ ψ= ⋅ ⇒ = ⇒ = ±

� �

�

m

x

There are no restrictions on kE can have any valueEnergies of free particles are continuous

Free Particle

x E xm x

2 2

2( ) ( )

2ψ ψ

∂− = ⋅

∂

�

k mEk x E x E k

m m

2 2 22 2( ) ( )

2 2ψ ψ= ⋅ ⇒ = ⇒ = ±

� �

�

mE mEx A x B x

2 2( ) sin cosψ = +

� �

kE

m

2 2

2

=�

No Quantization All energies are allowed

m

xde Broglie wave

x

V x x L

x L

0

( ) 0 0

∞ <

= ≤ ≤∞ >

x V x x E xm x

2 2

2( ) ( ) ( ) ( )

2ψ ψ ψ

∂− + = ⋅

∂

�

For regions in the space x < 0 and x > L ⇒ V = ∞

( )m

x V E x xx

2

2 2

2( ) ( ) ( )ψ ψ ψ

∂= − ⋅ = ∞ ⋅

∂ �

Normalization condition not satisfied ⇒

x x L( 0) 0 and ( ) 0ψ ψ< = > =

Particle in 1-D Square-Well Potential

x V x x E xm x

2 2

2( ) ( ) ( ) ( )

2ψ ψ ψ

∂− + = ⋅

∂

�

For regions in the space 0 ≤ x ≤ L ⇒ V = 0

x E xm x

2 2

2( ) ( )

2ψ ψ

∂− = ⋅

∂

�

This equation is similar to free particle SchrodingerHowever, boundary conditions are present

Let is assumeTrial Solution

Energy

x A kx B kx( ) sin cosψ = +

kE

m

2 2

2

=�


x A kx B kx( ) sin cosψ = +

Boundary Condition x x0 ( ) 0ψ= ⇒ =

Boundary Condition

x A kx( ) sin cos0 1ψ = =∵

x L L ( ) 0ψ= ⇒ =

L A kL A kL( ) 0 sin 0 0 or sin 0ψ = ⇒ = ⇒ = =

But the wavefunction ψ(x) CANNOT be ZERO everywhere

kL kL nsin 0 n=1,2,3,4...π= ⇒ =

Wavefunction is x A kx( ) sinψ =


k nE k

m L

2 2

and 2

π= =�

n

n n hE

mL mL

2 2 2 2 2

2 2 n=1,2,3,4...

2 8π

= =�

Energy is no longer continues but has discrete values; Quantization of energy

Energy separation increases with increasing values of n

The lowest allowed energy level is for n=1

has a non zero value ⇒ Zero Point EnergyEmL

2 2

1 22π

=�


( )f ff i f i

n h n h hh E E E n n

mL mL mL

2 2 2 2 22 2

2 2 2-

8 8 8ν = ∆ = − = = −

Larger the box, smaller the energy of hν

Particle in 1-D Square-Well Potential: Spectroscopy

Wavefunction

Normalization

nx A kx A x

L( ) sin sin

πψ = =

L L nx x dx A x dx

L2 2

0 0( ) ( ) sin 1

πψ ψ∗ ⋅ ⋅ = ⋅ =∫ ∫

nA x x

L L L

2 2 ( ) sin

πψ= =

Homework Evaluate the above integral


Wavefunction nx x

L L

2( ) sin

πψ =

n=1,3.. (odd) Symmetric(even function)

n=2,4.. (even) Anti-Symmetric(odd function)

Number of Nodes (zero crossings) = n-1

Particle in 1-D Square-Well Potential: Spectroscopy

Expectation values

ψ ψ

π π

π

∗= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅

=

∫

∫

∫

0

2

0

2 2sin sin

2sin

2

L

L

x x dx

n nx x x dx

L L L L

nx x dx

L L

L

Homework Verify!

Expectation values

Homework Verify!

ψ ψ

π π

π π π

∗ ∂ = ⋅ − ⋅ ⋅

∂

∂= − ⋅ ⋅ ⋅

∂

−= ⋅ ⋅

=

∫

∫

∫

�

�

�

0

2 0

2 2sin sin

2sin cos

0

x

L

L

p i dxx

n ni x x dx

L L x L L

i n n nx x dx

L L L

Hamiltonian � � �∂ ∂= − − = +

∂ ∂

� �2 2

2 22 2x yH H H

m x m y

� ψ ψ⋅ = ⋅( , ) ( , )nH x y E x y

ψ ψ ψ= ⋅

Let us assume that

( , ) ( ) ( )x y x y


� � ( )H x y H x y( , ) ( ) ( )ψ ψ ψ⋅ = ⋅ ⋅

( ) ( )x yE E x y( , )ψ= + ⋅

� � ( )x yH H x y( ) ( )ψ ψ = + ⋅

� �x yy H x x H y( ) ( ) ( ) ( )ψ ψ ψ ψ= ⋅ ⋅ + ⋅ ⋅

x yy E x x E y( ) ( ) ( ) ( )ψ ψ ψ ψ= ⋅ ⋅ + ⋅ ⋅

x yE x y E x y( ) ( ) ( ) ( )ψ ψ ψ ψ= ⋅ ⋅ + ⋅ ⋅

( ) ( )x yE E x y( ) ( )ψ ψ= + ⋅ ⋅


Hamiltonian � � �∂ ∂= − − = +

∂ ∂

� �2 2

2 22 2x yH H H

m x m y

ψ is a product of the eigenfunctions of the parts of Ĥ

E is sum of the eigenvalues of the parts of Ĥ

� ψ ψ⋅ = ⋅( , ) ( , )nH x y E x y

ψ ψ ψ= ⋅( , ) ( ) ( )x y x y

x y x yn n n n nE E E E,= = +


x y x yn n n n

yx

x y

yxx y

x y

E E E

n hn h

mL mL

nnhn n

m L L

,

2 22 2

2 2

222

2 2

8 8

, 1,2,3,4...8

= +

= +

= + =

x x y y

x yx y

x y x y

n nx y

L L L L

n nx y

L LL L

( , ) ( ) ( )

2 2sin sin

2sin sin

ψ ψ ψ

π π

π π

= ⋅

= ⋅

= ⋅

V=0

Lx

Ly


( )

x y x yn n n n

yx

x y x y

E E E

n hn h

mL mL

hn n n n

mL

,

2 22 2

2 2

22 2

2

8 8

, 1,2,3,4...8

= +

= +

= + =

x y x y

n nx y

L L L Ln nx y

L L L

( , ) ( ) ( )

2 2sin sin

2sin sin

ψ ψ ψ

π π

π π

= ⋅

= ⋅

= ⋅

V=0

Lx

Ly

Square Box ⇒ Lx = Ly = L


x yL L Lh

E E EmL

1,2 1 2

2

1,2 1 2 2

2 2sin sin

5

8

π πψ ψ ψ= ⋅ = ⋅

= + =

V=0

Lx

Ly

x yL L Lh

E E EmL

2,1 2 1

2

2,1 2 1 2

2 2sin sin

5

8

π πψ ψ ψ= ⋅ = ⋅

= + =

⇒ are degenerate wavefunctionsE E1,2 2,1= 1,2 2,1 and ψ ψ


Square Box ⇒ Lx = Ly = L

V=0

Lx

Ly

V=0

Lx

Ly

(1,1)(2,1) (1,2)

(2,2)

(3,1) (1,3)

(1,1) (2,1)(3,1)

(1,2)(2,2)

(3,2)

(1,3)

Particle in 2-D Square-Well Potential – Symmetry

Number of nodes = nx+ny-2

Particle in a 2-D Well – Wavefunctions

Particle in a 3D-Box

yx z

x x y y z z

x y z x y z

nn nx y z

L L L L L L

( , , ) ( ) ( ) ( )

2 2 2sin sin sin

ψ ψ ψ ψ

ππ π

= ⋅ ⋅

= ⋅ ⋅

x y z x y zn n n n n n

yx zx y z

x y z

E E E E

n hn h n hn n n

mL mL mL

, ,

2 22 2 2 2

2 2 2 , , 1,2,3,4...

8 8 8

= + +

= + + =

Agrees well with the experimental value of 258 nmParticle in a box is a good model

Particle in a Box – Application in Chemistry

Hexatriene is a linear molecule of length 7.3 ÅIt absorbs at 258 nmUse particle in a box model to explain the results.

�

�

�

Six π electron fill lower three levels

( )

( )

λ

λ

∆ = − = − =

= − ≈

22 2

2

22 2

8

8251nm

f i f i

f i

h hcE E E n n

mL

mL cn n

h

Increase in bridge length increase the emission wavelength.

Predicts correct trend and gets the wavelength almost right.

Particle in a box is a good model

Particle in a Box – Application in Chemistry

Electronic spectra of conjugated molecules

λλ

= ⇒ ∝2

22

8

hc hL

mL

Β-carotene is orange because of 11 conjugated double bonds

Particle in a Box – Application in nano-science

Band gap changes due to confinement, and so does the color of emitted light

Quantum Dots have a huge application in chemistry, biology, and materials sciencefor photoemission imaging purpose, as well as light harvesting/energy science

What have we learnt?

Formulate a correct Hamiltonian(energy) Operator H

Solve TISE HΨΨΨΨ=E ΨΨΨΨby separation of variables and intelligent trial wavefunction

Impose boundary conditions for eigenfunctions and obtain

Quantum numbers

Eigenstatesor Wavefunctions: Should be “well behaved” -

Normalization of Wavefunction

Probabilities and Expectation Values

Hydrogen Atom

�

πε= − ∇ − ∇ −

� �22 2

2 2

0

1

2 2 4N e

N eN e eN

Z Z eH

m m r

� � � �N e N eH T T V -= + +

(xe,ye,ze)

= − + − + −2 2 2( ) ( ) ( )eN e N e N e Nr x x y y z z

(xN,yN,zN)

N eN N N e e ex y z x y z

2 2 2 2 2 22 2

2 2 2 2 2 2

∂ ∂ ∂ ∂ ∂ ∂∇ = + + ∇ = + +

∂ ∂ ∂ ∂ ∂ ∂

Two particle central-force problem

Completely solvable – a rare example!

Hydrogen Atom

− ∇ − ∇ − Ψ = ⋅Ψ

� �2 2 2

2 2

2 2N e Total Total TotalN e eN

QZeE

m m r


Total N N N e e ex y z x y z( , , , , , )Ψ = Ψ

�

πε= − ∇ − ∇ −

� �22 2

2 2

0

1

2 2 4N e

N eN e eN

Z Z eH

m m r

�

πε

= − ∇ − ∇ −

= = =

� �2 2 2

2 2

0

2 2

1 with 1 and

4

N eN e eN

N e

QZeH

m m r

Z Z Z Q

Hydrogen Atom: Relative Frame of Reference

Separation of Ĥ into Center of Mass and Internal co-ordinates

x

z

y

-re

RrN

me(xe,ye,ze)

CMreNMN

(xN,yN,zN) ( )

( )

( )

= −

= −

= −

= + +

= + +

= = −

= + +

2 2 2

2 2 2

2 2 2

e N

e N

e N

e e e e

N N N N

eN e N

x x x

y y y

z z z

r x y z

r x y z

r r r r

x y z

− ∇ − ∇ − Ψ = ⋅Ψ

� �2 2 2

2 2


QZeE

m m r


Separation of Ĥ into Center of Mass and Internal co-ordinates

x

z

y

-re

RrN

me(xe,ye,ze)

CMreNMN

(xN,yN,zN)

+=

+

+=

+

+=

+

+=

+

e e N n

e N

e e N n

e N

e e N n

e N

e e N N

e N

m x m xX

m m

m y m yY

m m

m z m zZ

m m

m r m rR

m m

− ∇ − ∇ − Ψ = ⋅Ψ

� �2 2 2

2 2


QZeE

m m r

( )

( )

( )

= −

= −

= −

= + +

= + +

= = −

= + +

2 2 2

2 2 2

2 2 2

e N

e N

e N

e e e e

N N N N

eN e N

x x x

y y y

z z z

r x y z

r x y z

r r r r

x y z


µ

µ

− ∇ − ∇ − Ψ = ⋅Ψ

= + =+

� �2 2 2

2 2

2 2

where and

R r Total Total Total

e Ne N

e N

QZeE

M r

m mM m m

m m

⇓

− ∇ − ∇ − Ψ = ⋅Ψ

� �2 2 2

2 2


QZeE

m m r

Checkout Appendix-1

Hydrogen Atom: Separation to Relative Frame

Hydrogen atom has two particles the nucleus and electron with co-ordinates xN,yN,zN and xe,ye,ze

The potential energy between the two is function of relative co-ordinates x=xe-xN, y=ye-yN, z=ze-zN

= + +

= − = − = −

= + +

+ + += = =

+ + +

, ,

, ,

e N e N e N

e e N n e e N n e e N n

e N e N e N

r ix jy kz

x x x y y y z z z

R iX jY kZ

m x m x m y m y m z m zX Y Z

m m m m m m

x

z

y

-re

RrN

me(xe,ye,ze)

CMreNMN

(xN,yN,zN)

Appendix-1

+=

+

= = −

= −+

= −+

e e N N

e N

eN e N

Ne

e N

eN

e N

m r m rR

m m

r r r r

mr R r

m m

mr R r

m m


x

z

y

-re

RrN

me(xe,ye,ze)

CMreNMN

(xN,yN,zN)

Appendix-1

( )

µ µ

= +

= − ⋅ −

+ +

+ − ⋅ −

+ +

= + +

+

= + = + =+

� �

� ��

� ��

� �

� �

2 2

2 2

2 2

1 1

2 2

1

2

1

2

1 1

2 2

1 1 where and

2 2

e e N N

N Ne

e N e N

e ee

e N e N

e Ne N

e N

e Ne N

e N

T m r m r

m mT m R r R r

m m m m

m mm R r R r

m m m m

m mT m m R r

m m

m mT M R r M m m

m m


=

=

=

=

�

�

�

�

ee

NN

drr

dtdr

rdtdr

rdtdR

Rdt

Appendix-1

µ

µ

= +

= +

� �2 2

2 2

1 1

2 2

2 2R r

T M R r

p pT

M


In the above equation the first term represent the kinetic energy of the center of mass (CM) motion and second term represents the kinetic energy of the relative motion of electron and

�

µ

µ

⋅= + −

⋅= − ∇ − ∇ −� �

2 2

2 22 2

2 2

2 2

N eR r

N eR r

Z Zp pH

M r

Z ZH

M r

Appendix-1

Free particle!Kinetic energy of the atom

Hydrogen Atom: Separation of CM motion

� χ χ χ

= − ∇ =

�2

2

2N N R N N NH E

M

=�2 2

2N

kE

M

χ ψΨ = ⋅Total N e� � �= +N eH H H = +Total N eE E E

µ

− ∇ − ∇ − Ψ = ⋅Ψ

� �2 2 2

2 2

2 2R r Total Total Total

QZeE

M r

� �

µ− ∇ = − ∇ − =� �2 2 2

2 2 2 2

N eR r

QZeH H

M r

Hydrogen Atom: Electronic Hamiltonian

( )ψ ψ

µ

ψ

∂ ∂ ∂− + + −

∂ ∂ ∂ + +

= ⋅

�2 2 2 2 2

2 2 2 2 2 2( , , ) ( , , )

2

( , , )

e e

e e

QZex y z x y z

x y z x y z

E x y z

Not possible to separate out into three different co-ordinates. Need a new co-ordinate system

r

� ψ ψ ψµ

ψ ψ

⋅ = − ∇ − = ⋅

⇒

�2 2

2

2

( , , )

e e r e e e

e e

QZeH E

r

x y z

Spherical Polar Co-ordinates

θ φ

θ φ

θ

=

=

=

sin cos

sin sin

cos

x r

y r

z r

( )

θ

φ

−

= + +

=

= −

2 2 2

1cos

tan 1

r x y z

zr

y

x

τ θ θ φ= ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅2 sind dx dy dz r dr d d

‘r’ ranges from 0 to ∞

‘θ’ ranges from 0 to π

‘φ’ ranges from 0 to 2π

Spherical Polar Co-ordinates

ψ ψ θ φ ψ⇒ ⇐( , , ) ( , , )e e er x y z

θθ θ θ θ φ

∂ ∂ ∂+ +

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ = + +

∂ ∂ ∂ ∂ ∂

2 2 2

2 2 2

22

2 2 2 2 2

1 1 1sin

sin sin

fx y z

f f fr

r r r r r

ψ ψ ψθ

µ θ θ θ θ φ

ψ ψ

∂ ∂ ∂∂ ∂ − + + ∂ ∂ ∂ ∂ ∂

− =

�22

22 2 2 2 2

2

1 1 1sin

2 sin sine e e

e e e

rr r r r r

QZeE

r


ψ ψ ψθ

µ θ θ θ θ φ

ψ ψ

∂ ∂ ∂∂ ∂ − + + ∂ ∂ ∂ ∂ ∂

− =

�22

22 2 2 2 2

2

1 1 1sin

2 sin sine e e

e e e

rr r r r r

QZeE

r

µ−

�

2

2

2Multiply with

r

ψ ψ ψθ

θ θ θ θ φ

µ µψ ψ

∂ ∂ ∂∂ ∂ + + ∂ ∂ ∂ ∂ ∂

+ + =� �

22

2 2

2 2

2 2

1 1sin

sin sin

2 20

e e e

e e e

rr r

rQZe rE


( )ψ θ φ θ φ

ψ

⇒ ⋅Θ ⋅Φ

⇒ ⋅Θ ⋅Φ

( , , ) ( ) ( )e

e

r R r

R

θθ θ θ θ φ

µ µ

∂ ∂ ⋅Θ⋅Φ ∂ ∂ ⋅Θ⋅Φ ∂ ⋅Θ⋅Φ + +

∂ ∂ ∂ ∂ ∂

+ ⋅Θ⋅Φ + ⋅Θ⋅Φ =� �

22

2 2

2 2

2 2

( ) 1 ( ) 1 ( )sin

sin sin

2 2( ) ( ) 0e

R R Rr

r r

rQZe rR E R

ψ ψ ψθ

θ θ θ θ φ

µ µψ ψ

∂ ∂ ∂∂ ∂ + + ∂ ∂ ∂ ∂ ∂

+ + =� �

22

2 2

2 2

2 2

1 1sin

sin sin

2 20

e e e

e e e

rr r

rQZe rE


θθ θ θ θ φ

µ µ

∂ ∂ ⋅Θ⋅Φ ∂ ∂ ⋅Θ⋅Φ ∂ ⋅Θ⋅Φ + +

∂ ∂ ∂ ∂ ∂

+ ⋅Θ⋅Φ + ⋅Θ⋅Φ =� �

22

2 2

2 2

2 2

( ) 1 ( ) 1 ( )sin

sin sin

2 2( ) ( ) 0e

R R Rr

r r

rQZe rR E R

θθ θ θ θ φ

µ µ

∂ ∂ ∂ ∂Θ ∂ Φ Θ⋅Φ + ⋅Φ + ⋅Θ

∂ ∂ ∂ ∂ ∂

+ ⋅Θ⋅Φ + ⋅Θ⋅Φ =� �

22

2 2

2 2

2 2

1 1( ) ( ) sin ( )

sin sin

2 2( ) ( ) 0e

Rr R R

r r

rQZe rR E R

Rearrange


∂ ∂ ∂ ∂Θ ∂ Φ Θ⋅Φ + ⋅Φ + ⋅Θ

∂ ∂ ∂ ∂ ∂

+ ⋅Θ⋅Φ + ⋅Θ⋅Φ =� �

22

2 2

2 2

2 2

1 1( ) ( ) sin ( )

sin sin

2 2( ) ( ) 0e

Rr R R

r r

rQZe rR E R

θθ θ θ θ φ

µ µ

⋅Θ ⋅Φ

1Multiply with

R

∂ ∂ ∂ ∂Θ ∂ Φ + +

∂ ∂ Θ ∂ ∂ Φ ∂

+ + =� �

22

2 2

2 2

2 2

1 1 1 1 1sin

sin sin

2 20e

Rr

R r r

rQZe rE

θθ θ θ θ φ

µ µ


∂ ∂ ∂ ∂Θ ∂ Φ + +

∂ ∂ Θ ∂ ∂ Φ ∂

+ + =� �

22

2 2

2 2

2 2

1 1 1 1 1sin

sin sin

2 20e

Rr

R r r

rQZe rE

θθ θ θ θ φ

µ µ

Rearrange

∂ ∂ + + +

∂ ∂

∂ ∂Θ ∂ Φ = − +

Θ ∂ ∂ Φ ∂

� �

2 22

2 2

2

2 2

1 2 2

1 1 1 1sin

sin sin

e

R rQZe rr E

R r r

µ µ

θθ θ θ θ φ

LHS = f(r)=f(θ ,φ) =RHS⇒ f(r)=f(θ ,φ) =constant=β


∂ ∂ ∂ ∂Θ ∂ Φ + +

∂ ∂ Θ ∂ ∂ Φ ∂

+ + =� �

22

2 2

2 2

2 2

1 1 1 1 1sin

sin sin

2 20e

Rr

R r r

rQZe rE

θθ θ θ θ φ

µ µ

Rearrange

∂ ∂ + + +

∂ ∂

∂ ∂Θ ∂ Φ = − + =

Θ ∂ ∂ Φ ∂

� �

2 22

2 2

2

2 2

1 2 2

1 1 1 1sin

sin sin

e

R rQZe rr E

R r r

µ µ

θ βθ θ θ θ φ

LHS = f(r)=f(θ ,φ) =RHS⇒ f(r)=f(θ ,φ) =constant=β


∂ ∂ + + + =

∂ ∂

∂ ∂Θ ∂ Φ + = −

Θ ∂ ∂ Φ ∂

� �

2 22

2 2

2

2 2

1 2 2

1 1 1 1sin

sin sin

e

R rQZe rr E

R r rµ µ

β

θ βθ θ θ θ φ

θ βθ θ θ θ φ

∂ ∂Θ ∂ Φ + = −

Θ ∂ ∂ Φ ∂

2

2 2

1 1 1 1sin

sin sin

Let us consider

θ2Multiply with sin and rearrange

θθ β θ

θ θ φ

∂ ∂Θ ∂ Φ + = −

Θ ∂ ∂ Φ ∂

22

2

sin 1sin sin

θθ β

θ θ

∂ ∂Θ + =

Θ ∂ ∂

2sinsin m

φ

∂ Φ= −

Φ ∂

22

2

1m


θθ β θ

θ θ φ

∂ ∂Θ ∂ Φ + = −

Θ ∂ ∂ Φ ∂

22

2

sin 1sin sin

LHS = f(θ)=f(φ) =RHS⇒ f(θ)=f(φ) =constant=m2

∂ ∂ + + + =

∂ ∂ � �

2 22

2 2

1 2 2e

R rQZe rr E

R r r

µ µβ

θθ β θ

θ θ

∂ ∂Θ + =

Θ ∂ ∂

2 2sinsin sin m

φ

∂ Φ= −

Φ ∂

22

2

1m


We have separated out all the three variables r, θ and φ

Solution to ΦΦΦΦ part

φ

φ φ

φφ

φ

∂ Φ+ =

Φ ∂

∂ Φ= − Φ

∂

22

2

22

2

1 ( )0

( )

( )( )

m

m

Let is assume

as trial solution

φφ ±Φ =( ) imAeφ

φ

∂Φ= ± Φ

∂

∂ Φ= − Φ

∂

22

2

0im

m

Wavefunction has to be continuous

φ π φ⇒ Φ + = Φ( 2 ) ( )

‘φ’ ranges from 0 to 2π

Solution to ΦΦΦΦ part

φ π φ φ π φ

π π

+ − + −

− −

−

= =

= =

( 2 ) ( ) ( 2 ) ( )

(2 ) (2 )

and

1 and 1

im im im imm m m m

im im

A e A e A e A e

e e

True only if m=0, ±1, ±2, ±3, ±4,….m is the “magnetic quantum” number

m is restricted by another quantum number (orbital Angular momentum), l, such that |m|<l

φ π φ⇒ Φ + = Φ( 2 ) ( )

The ΘΘΘΘ and the R part

∂ ∂ + + + =

∂ ∂ � �

2 22

2 2

1 2 2e

R rQZe rr E

R r r

µ µβ

θθ β θ

θ θ

∂ ∂Θ + =

Θ ∂ ∂

2 2sinsin sin m

∂ ∂ + + − =

∂ ∂ �

2 22

2

( ) 2( ) ( ) 0e

R r r QZer E R r R r

r r rµ

β

θθ θ β θ

θ θ θ θ

∂ ∂Θ − Θ + Θ =

∂ ∂

2

2

1 ( )sin ( ) ( ) 0

sin sin

m

Rearrange

Solve to get Θ(θ)

Need serious mathematical skill to solve these two equations. We only look at solutions

The ΘΘΘΘ and the R part

∂ ∂ + + − =

∂ ∂ �

2 22

2

( ) 2( ) ( ) 0e


r r rµ

β

θθ θ β θ

θ θ θ θ

∂ ∂Θ − Θ + Θ =

∂ ∂

2

2

1 ( )sin ( ) ( ) 0

sin sin

m

Solve to get R(r)

Restriction on m are due this this equation

The ΘΘΘΘ part

are known as Associated Legendre Polynomials

The new quantum number is ‘l’ called orbital / Azimuthalquantum number

Restriction on m≤lis due to this equation

θθ θ β θ

θ θ θ θ

∂ ∂Θ − Θ + Θ =

∂ ∂

2

2

1 ( )sin ( ) ( ) 0

sin sin

m

θ θ θ

θ θ β

+

+

−

−= − −

−= − = +

+

2 22( 1)

(cos ) (1 cos ) (cos 1)2 !

( )!(cos ) ( 1) (cos ) with ( 1)

( )!

m l mmm ll l l m

m m ml l

dP

l dx

l mP P l l

l m

Solution to Θ(θ) are

θ(cos )mlP

l=0,1,2,3…

The angular (ΘΘΘΘ·ΦΦΦΦ) part

The angular part of the solution

are called spherical harmonicsθ φ θ φ⇒Θ ⋅Φ( , ) ( ) ( )mlY

φθ φ θπ

+ −=

+

(2 1) ( )!( , ) (cos )

4 ( )!m m iml l

l l mY P e

l m

l=0,1,2,3…m=0, ±1, ±2, ±3… and |m|≤l

The R part

∂ ∂ + + − =

∂ ∂ �

2 22

2

( ) 2( ) ( ) 0e


r r rµ

β

( )

( )

+−

+

+

− − = − +

0

12 3

22 1

3

1 ! 2 2( )

2 !

lZrnal l

nl n l

n l Z ZrR r r e L

na nan n l

Solution to R(r) are

Where are called associated Laguerre functions

The new quantum number is ‘n’ called principal quantum number

+

+

2 1

0

2ln l

ZrL

na

= =��22

02 2

4aQ e e

πε

µ µ

Restriction on l<n

Energy of the Hydrogen Atom

( )= − = − = − ≈

−=

�

2 2 4 2 4 2 4

2 2 2 2 2 20 0 0

2

2

8 8

13.6

n e

n

Q Z e Z e Z eE m

n h n a n

eVE

n

µ µµ

ε πε

Energy is dependent only on ‘n’

Energy obtained by full quantum mechanical treatment is equal to Bohr energy

Potential energy term is only dependent on the Radialpart and has no contribution from the Angular parts

Quantum Numbers of Hydrogen Atom

n Principal Quantum number

Specifies the energy of the electron

l Orbital Angular Momentum Quantum number

Specifies the magnitude of the electron's orbital angular momentum

m Z-component of Angular Momentum Quantum number

Specifies the orientation of the electron's orbital angular momentum

s Orbital Angular Momentum Quantum number

Specifies the orientation of the electron's spin angular momentum

Orbital Angular Momentum Quantum Number

l=0 ⇒⇒⇒⇒ s-Orbital

l=1 ⇒⇒⇒⇒ p-Orbital

l=2 ⇒⇒⇒⇒ d-Orbital

l=3 ⇒⇒⇒⇒ f-Orbital

Normalization

( )

( ) ( )

( )

∞

∗

∞ ∞

= ⋅

⇒

= ⋅ = ⋅

=

= =

=

∫ ∫ ∫

∫ ∫ ∫ ∫

∫ ∫

, , ,

2 22 2

, , , ,

2 22, ,0 0 0

2 22

0 0 0 0

22, ,0 0

( , , ) ( , )

( , , ) ( , ) ( , )

sin ( , , ) 1

sin ( , ) sin ( , ) ( , ) 1

mn l m n l l

m mn l m n l l n l l

n l m

m m ml l l

n l n l

r R r Y

r R r Y R r Y

r dr d d r

d d Y d d Y Y

r dr R r dr R

π π

π π π π

ψ θ φ θ φ

ψ θ φ θ φ θ φ

θ θ φ ψ θ φ

θ θ φ θ φ θ θ φ θ φ θ φ

( ) ( )∗

= , 1n lr R r

Normalize the Radial and Angular parts separately

Spherical Harmonics Ylm

( )

φ

φ

φ

π

θπ

θπ

θπ

θ θπ

θπ

±

±

±

= =

= =

= = ±

= = −

= = ±

= = ±

∓

∓

1 2

1 2

1 2

1 22

1 2

1 2

2 2

10; 0

4

31; 0 cos

4

31; 1 sin

8

32; 0 3cos 1

8

152; 1 cos sin

8

152; 2 sin

32

i

i

i

l m

l m

l m e

l m

l m e

l m e

( )

( ) φ

φ

φ

θ θπ

θ θπ

θ θπ

θπ

±

±

±

= = −

= = ± −

= = ±

= = ±

∓

∓

1 23

1 2

2

1 2

2 2

1 2

3 3

73; 0 5cos 3cos

16

213; 1 5cos 1 sin

64

1053; 2 sin cos

32

353; 3 sin

64

i

i

i

l m

l m e

l m e

l m e

φθ φ θπ

+ −=

+

(2 1) ( )!( , ) (cos )

4 ( )!m m iml l

l l mY P e

l m

Radial Functions

( )

( )

( )

ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ ρ

ρ

ρ

−

−

−

−

−

−

= =

= = −

= =

= = − −

= = −

= =

3 2

2

3 2

212

3 2

212

3 22 2

12

3 2

212

3 22 2

12

1; 0 2

12; 0 2

8

12; 1

24

13; 0 6 6

243

13; 1 4

486

13; 2

2430

Zn l e

a

Zn l e

a

Zn l e

a

Zn l e

a

Zn l e

a

Zn l e

a

ρ

πε

µ

µ

=

=

= =

�2

02

0

2

4

(for )e

Zr

na

ae

a a m

( )

( )

+−

+

+

− − = − +

0

12 3

22 1

3

1 ! 2 2( )

2 !

l Zrnal l

nl n l

n l Z ZrR r r e L

na nan n l

Radial Functions of Hydrogen Atom

−

−

−

−

= =

= = −

= =

= = − −

= =

3 2

3 2

212 0

3 2

212 0

3 2 2

3

0 0

12

11; 0 2

1 12; 0 2

8

1 12; 1

24

1 2 23; 0 2 1

3 3 27

1 13; 1

486

o

o

o

o

ra

o

ra

o

ra

o

ra

o

o

n l ea

rn l e

a a

rn l e

a a

r rn l e

a a a

n la

−

−

−

= =

3 2

3

0

3 2 2

312 0

24

3

1 1 23; 2

32430

o

o

ra

ra

o

re

a

rn l e

a a

( )

( )

+−

+

+

− − = − +

0

1 322

2 13

0 0

1 ! 2 2( )

2 !

lrnal l

nl n l

n l rR r r e L

na nan n l

ρ

πε

µ

µ

=

=

= =

�2

02

0

2

4

(for )e

Zr

na

ae

a a m

Wavefunctions of Hydrogen Atom

φ

ψ ψπ

ψ ψπ

ψ ψ θπ

ψ ψ θπ

ψ ψπ

+

−

−

−

−

−

+

−

= =

= = −

= =

= =

= =

1

1

3 2

1,0,0 1

3 2

22,0,0 2

0

3 2

22,1,0 2

0

3 2

22,1, 1 2

0

3 2

2,1, 1 20

1 1

1 12

4 2

1 1cos

4 2

1 1sin

8

1 1

8

o

o

o

z

o

ra

so

ra

so

ra

po

ra i

po

po

ea

re

a a

re

a a

re e

a a

r

a aφθ

−−

2 sin o

ra ie e

( )ψ θ φ θ φ= ⋅, , ,( , , ) ( , )mn l m n l lr R r Y

f(r)

f(r)

f(r,θ)

f(r,θ,φ)

f(r,θ,φ)

1s and 2s Orbitals

ψ ψπ

ψ ψπ

−

−

= =

= = −

3 2

1,0,0 1

3 2

22,0,0 2

0

1 1

1 12

4 2

o

o

ra

so

ra

so

ea

re

a a

Functions of only ‘r’

φ

φ

ψ ψ θπ

ψ ψ θπ

ψ ψ θπ

+

−

−

−

+

−−

−

= =

= =

= =

1

1

3 2

22,1,0 2

0

3 2

22,1, 1 2

0

3 2

22,1, 1 2

0

1 1cos

4 2

1 1sin

8

1 1sin

8

o

z

o

o

ra

po

ra i

po

ra i

po

re

a a

re e

a a

re e

a a

2p Orbitals

Functions of ‘r’, ‘θ’ and ‘φ’

φ

φ

ψ ψ θπ

ψ ψ θπ

ψ ψ θπ

+

−

−

−

+

−−

−

= =

= =

= =

1

1

3 2

22,1,0 2

0

3 2

22,1, 1 2

0

3 2

22,1, 1 2

0

1 1cos

4 2

1 1sin

8

1 1sin

8

o

z

o

o

ra

po

ra i

po

ra i

po

re

a a

re e

a a

re e

a a

2p Orbitals

( )

( )

ψ θ φ ψ ψπ

ψ θ φ ψ ψπ

−

+ −

−

+ −

= +

= −

3 2

22 2,1, 1 2,1, 1

0

3 2

22 2,1, 1 2,1, 1

0

1 1 1sin cos =

32 2

1 1 1sin sin =

32 2

o

x

o

y

ra

po

ra

po

re

a a

re

a a i

Linear combination

Radial functions

ρψ −′=1001s N e ( )

ρ

ψ ρ−

′′= −200 22 2s N e

ρ =0

r

a

ρ

ψ ρ θ−

′′′=210 22 coszpN e

For s-Orbitals the maximum probability denisty of finding the electron is on the nucleus

For s-Orbitals the probability of finding the electron on the nucleus zero

Surface plots

Surface plot of the ΨΨΨΨ2s ; 2s wavefunction (orbital) of the hydrogen atom. The

height of any point on the surface above the xy plane (the nuclear plane)

represents the magnitude of the ΨΨΨΨ2s function at the at point (x,y) in the

nuclear plane. Note that there is a negative region (depression) about the

nucleus; the negative region begins at r=2a0 an goes asymptotically to zero at

r=∞∞∞∞.

Surface plot of the |ΨΨΨΨ2s|2; the probability density associated with

the 1s wavefunction of the hydrogen atom. Note that the negative

region of the 2s plot on the left now appears as positive region.

Surface plot of the 1s wavefunction (orbital) of the hydrogen atom. The height

of any point on the surface above the xy plane (the nuclear plane) represents

the magnitude of the ΨΨΨΨ1s function at the at point (x,y) in the nuclear plane.

The nucleus is located in the xy place immediately below the ‘peak’

Surface plot of the |ΨΨΨΨ1s|2; the probability density associated with

the 1s wavefunction of the hydrogen atom.

1s

2s

(1s)2

(2s)2

Surface plots

R(2pz)

(2pz)2

Surface plot of radial portion of a 2p wavefunction of the hydrogen

atom. The gird lines have been left transparent so that the inner

‘hollow’ portion is visible.

Profile of the radial portion of a 2p wavefunction of the hydrogen atom.

Profile of the 2pz orbital along the z-azis. Surface plot of the 2pz wavefunction (orbital)

in the xz (or yz) plane for the hydrogen atom.

The ‘pit’ represents the negative lobe and the

‘hill’ the positive lobe of a 2p orbital.

Surface plot of the (2pz)2; the probability density

associated with the 2pz wavefunction of the

hydrogen atom. Each of the hills represents and

area in the xz (or yz) plane where the probability

density is the highest, The probability density

along the x (or y) axis passing through the nucleus

(0,0) is everywhere zero.

2pz2pz

R(2pz)

Surface plots

Surface plot of the 3dz2 wavefunction (orbital) in the xz (or yz) plane for the

hydrogen atom. The large hills correspond to the positive lobes and the

small pits correspond to the negative lobes.

Surface plot of the (3dz2 )2 the probability density associated with the 3dz2

orbital of the hydrogen atom. This figure is rotated with respect to the

figure on the left so that the small hill will be clearly visible. Another

smaller hill is hidden behind the large hill.

Surface plot of the 3dxy wavefunction (orbital) in the xz plane for the

hydrogen atom. The hills and the pits have same amplitude. Surface plot of the (3dxy )2 the probability density associated with the 3dxy

orbital of the hydrogen atom. Pits in the figure to the left appear has hills.

Radial and Radial Distribution Functions

π

π π

→

→ →

2 2

2

2 2 2

Probability of finding the electron

anywhere in a shell of thickness

at radius is 4 ( ) (for )

increasing function

4 ( ) 0 as 4 0

nl

nl

dr r r R r dr s

r

r R r dr r dr

Radial Distribution Functions

π 2 24 ( )nlr R r

3s: n=3, l=0Nodes=2

3p: n=3, l=1Nodes=1

3d: n=3, l=2Nodes=2

= Ψ Ψns nsr r

Number of radial nodes = n-l-1

Shapes and Symmetries of the Orbitals

s-Orbitals

ψ ψπ π

− − = = −

3 2 3 2

21 2

0

1 1 1 1 2

4 2o o

r ra a

s so o

re e

a a a

Function of only r; No angular dependence⇒⇒⇒⇒Spherical symmetric

n-l-1=0l=0

n-l=0

radial nodesangular nodesTotal nodes

n-l-1=1l=0n-l=1

Shapes and Symmetries of the Orbitals

p-Orbitals

Function of only r , θθθθ (and φφφφ)⇒⇒⇒⇒Not Spherical symmetric

2pz Orbital: No φφφφ dependence⇒⇒⇒⇒Symmetric around z-axis

radial nodesangular nodesTotal nodes

n-l-1=0l=1

n-l=1

ψ ψ θπ

− = =

3 2

2210 2

0

1 1cos

4 2o

z

ra

po

re

a a

xy nodal planeZero amplitude at nucleus

Angular Distribution Functions

p-Orbitals

ψ ψ θπ

− = = =

3 2

2210 2

0

1 1cos 0 case

4 2o

z

ra

po

re m

a a

+

–

θθθθ cosθθθθ

0 1.000

30 0.866

60 0.500

90 0.000

120 -0.500

150 -0.866

180 -1.000

210 -0.866

240 -0.500

270 0.000

300 0.500

330 0.866

360 1.000

ρ

ψ ψ ρ θ−

= = 2210 2 cos

zpN e

Angular part: Polar plot of 2pz --- cosθ

x

z

p-Orbitals

ρ

ψ ψ ρ θ−

= = 2210 2 cos

zpN e

ρ

ρ

ρ

ψ ρ θ

ψ ρ θ φ

ψ ρ θ φ

−

−

−

=

=

=

22

22

22

cos

sin cos

sin sin

z

x

x

p

p

p

N e

N e

N e

Color/shading are related to sign of the wavefunction

d-Orbitals

ρ

ρ

ρ

ρ

ρ

ψ ρ θ

ψ ρ θ θ φ

ψ ρ θ θ φ

ψ ρ θ φ

ψ ρ θ φ

−

−

−

−

−

−

= −

=

=

=

=

2

2 2

2 2 313

2 33 2

2 33 3

2 2 33 4

2 2 33 5

(3cos 1)

(sin cos cos )

(sin cos sin )

(sin cos2 )

(sin sin2 )

z

xz

yz

x y

xy

d

d

d

d

d

N e

N e

N e

N e

N e

Angular part

Blue: -veYellow: +ve

Angular + Radial

n=3; l=2; m=0,±1, ±2

f-Orbitals

n=4; l=3; m=0,±1, ±2, ±3

Green: -veRed: +ve

Cross-sections of Orbitals

Hydrogen Wavefunctions

Orbitals: External Structure

Orbitals: Internal Stucture

Hydrogen atom & Orbitals

Hydrogen atom has only one electron, so why bother about all these orbitals?

1. Excited states2. Spectra3. Many electron atoms

Many Electron Atoms

Helium is the simplest many electron atom

+

-

-

r1r2

r12= r1- r2

�

πε

= − ∇ − ∇ − ∇ − + −

� � �2 22 2 2 2

2 2 21 2

0 1 2 12

12 2 2 4

N NN

N e e

Z e Z e eH

m m m r r r

KE of Nucleus

KE of Electron1

KE of Electron2

Attraction between nucleus and Electron1

Attraction between nucleus and Electron1

Repulsion between Electron1 and Electron2

Helium Atom

�

�

� �

� �

= − ∇ − ∇ − ∇ − + −

= − ∇ − ∇ − − ∇ − + =

= − ∇ = − ∇ − − ∇ − +

=

� � �

� � �

� � �

2 22 2 2 22 2 2

1 20 1 2 12

2 22 2 2 22 2 2

1 21 2 12 0

2 22 2 2 22 2 2

1 21 2 12

1

2 2 2 4

1;

2 2 2 4

2 2 2

N NN

N e e

N NN

N e e

N NN eN

N e e

N eN n N

Z e Z e eH

m m m r r r

QZ e QZ e QeH Q

m m r m r r

QZ e QZ e QeH H

m m r m r r

H E H

πε

πε

χ χ ψ =e e eE ψ

Helium Atom

�

� � �

� �

= − ∇ − − ∇ − +

= + +

= − ∇ − = − ∇ −

� �

� �

2 22 2 22 21 2

1 2 12

2

1 2

12

2 22 22 2

1 21 21 2

2 2

and 2 2

N Ne

e e

e

N N

e e

QZ e QZ e QeH

m r m r r

QeH H H

r

QZ e QZ eH H

m r m r

The Hamiltonians Ĥ1 and Ĥ1 are one electron Hamiltonians similar to that of hydrogen atom

� � �= +

+

1 21 1 1 2 2 2 1 1 1 2 2 2 1 1 1 2 2 2

2

1 1 1 2 2 212

( , , , , , ) ( , , , , , ) ( , , , , , )

( , , , , , )

e e e e

e

H r r H r r H r r

Qer r

r

ψ θ φ θ φ ψ θ φ θ φ ψ θ φ θ φ

ψ θ φ θ φ

Orbital Approximation

ψ θ φ θ φ ψ θ φ ψ θ φ=1 1 1 2 2 2 1 1 1 1 2 2 2 2( , , , , , ) ( , , ) ( , , )e e er r r r

ψ φ φ φ φ≈ ⋅ ⋅ ⋅⋅⋅ ⋅ ⋅(1,2,3,... ) (1) (2) (3) ( )e n n

Orbital is a one electron wavefunction

The total electronic wavefunction of n number of electrons can be written as a product of n one electron wavefunctions

� � �= +

+

1 21 1 1 2 2 2 1 1 1 2 2 2 1 1 1 2 2 2

2

1 1 1 2 2 212

( , , , , , ) ( , , , , , ) ( , , , , , )

( , , , , , )

e e e e

e

H r r H r r H r r

Qer r

r

ψ θ φ θ φ ψ θ φ θ φ ψ θ φ θ φ

ψ θ φ θ φ

ψ θ φ θ φ ψ θ φ ψ θ φ=1 1 1 2 2 2 1 1 1 1 2 2 2 2( , , , , , ) ( , , ) ( , , )e e er r r r

� � �= +

+

1 21 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2

2

1 1 1 1 2 2 2 212

( , , ) ( , , ) ( , , ) ( , , )

( , , ) ( , , )

e e e e e e

e e

H H r r H r r

Qer r

r

ψ ψ θ φ ψ θ φ ψ θ φ ψ θ φ

ψ θ φ ψ θ φ

Helium Atom: Orbital Approximation

� � �ψ ψ θ φ ψ θ φ ψ θ φ ψ θ φ

ψ θ φ ψ θ φ

= +

+

1 21 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2

2

1 1 1 1 2 2 2 212

( , , ) ( , , ) ( , , ) ( , , )

( , , ) ( , , )

e e e e e e

e e

H H r r H r r

Qer r

r


� ψ ε ψ θ φ ψ θ φ ε ψ θ φ ψ θ φ

ψ θ φ ψ θ φ

= +

+

1 1 1 1 1 2 2 2 2 2 1 1 1 1 2 2 2 2

2

1 1 1 1 2 2 2 212

( , , ) ( , , ) ( , , ) ( , , )

( , , ) ( , , )

e e e e e e

e e

H r r r r

Qer r

r

� [ ]ψ ε ε ψ θ φ ψ θ φ

= + +

2

1 2 1 1 1 1 2 2 2 212

( , , ) ( , , )e e e e

QeH r r

r

µε ε

ε

−= = − =

2 4 2

1 2 2 2 2 20

13.68Z e Z

eVh n n


� [ ]ψ ε ε ψ θ φ ψ θ φ

= + +

2

1 2 1 1 1 1 2 2 2 212

( , , ) ( , , )e e e e

QeH r r

r

If we ignore the term 2

12

Qe

r

� ( )[ ]ψ ε ε ψ θ φ ψ θ φ= +1 2 1 1 1 1 2 2 2 2( , , ) ( , , )e e e eH r r

ε ε+ = −He 1 2E = 108.8eV

ψ ψ ψπ π

− − = ⋅ = ⋅

3 2 3 2

1 1

1 1(1) (2)o o

Zr Zra a

e s so o

Z Ze e

a a


ε ε+ = − + = −

= − + = −

He 1 2

He

E = (54.4 54.4) 108.8

E (24.59 54.4) 78.99 (Experimental)

eV eV

eV eV

Ignoring is not justified! Need better approximation2

12

Qe

r

Many Electron Atoms

�

α = = = >

= − ∇ − ∇ − +∑ ∑ ∑∑� �2 2

2 2 2 2

1 1 1

1 1

2 2

n n n n

N i Ni i i j ie i ij

H QZ e Qem m r r

Nuclei are static

�

�

= = = >

= = >

= − ∇ − +

= +

∑ ∑ ∑∑

∑ ∑∑

�2

2 2 2

1 1 1

2

1 1

1 1

2

1

n n n n

e i Ni i i j ie i ij

n n n

e ii i j i ij

H QZ e Qem r r

H H Qer

Inter-electron repulsion term leads to deviation from the hydrogen atom. Unfortunately CANNOT be ignored

≠ +

1 1 1

ij i j

f g hr r r

term in the Hamiltonian is not separable1

ijr

Many Electron Atoms

�

�

= = = >

= = >

= − ∇ − +

= +

∑ ∑ ∑∑

∑ ∑∑

�2

2 2 2

1 1 1

2

1 1

1 1

2

1

n n n n


n n n

e ii i j i ij

H QZ e Qem r r

H H Qer

Hamiltonian is no longer spherically symmetric and the Time-Independent Schrodinger Equation (TISE) cannot be solved using analytical techniques

Numerical methods must be used solve the TISE

Many Electron Atoms: Orbital Approximation

He atom result indicate that neglecting the inter-electron interaction is not a good idea

Improvement

The term in the Hamiltonian represents the interaction

between the electrons. Which mean the electron move in the potential provided by the nucleus and rest of the electron.

Since the electron and nucleus have opposite charges, it can be thought that the rest of the electrons reduce the charge felt by a particular electron ⇒ Shielding

1

ijr

ψ φ φ φ φ≈ ⋅ ⋅ ⋅⋅⋅ ⋅⋅(1,2,3,... ) (1) (2) (3) ( )e n n

Effective Nuclear Charge

�

�

= = = >

= =

= − ∇ − +

= − ∇ −

∑ ∑ ∑∑

∑ ∑

�

�

22 2 2

1 1 1

22 2

1 1

1 12

12

n n n n


n neff

e i Ni ie i

H QZ e Qem r r

H QZ em r

Effective Nuclear Charge Zeff

For Helium atom

ψ ψ ψπ π

− − ′ ′= ⋅ = ⋅

3 2 3 2

1 1

1 1(1) (2)

eff eff

o o

Z r Z reff effa a

e s so o

Z Ze e

a a

� = − ∇ − ∇ − −� �

2 22 22 21 2

1 22 2

eff effN N

e

e N

QZ e QZ eH

m m r r


Due to Shielding, the electrons do not see the full nuclear charge Z, but Zeff = Z–σσσσ (σ = Shielding Constant)

( )

σ

σ

=

= −

−= ⋅

= ⋅ +

∑2

1

2 2

eff

Ni

Hatomi i

He Hatom eff eff

Z Z

ZE E

N

E E Z Z

For Helium atom

σ= − =

=

1.69

1

effZ Z

n

=

= ⋅

− = −

−

∑2

2

1

13.6 5.712 77.68

Compare with 78.99

effHe Hatom

i i

ZE E

n

X eVThere are methods such as Perturbation Theory and Variational Method to estimate Zeff


Due to Shielding, the electrons do not see the full nuclear charge Z, but Zeff = Z–σσσσ (σ = Shielding Constant)

σ

σ

=

= −

−= ⋅

∑

2

1

eff

Ni

Hatomi i

Z Z

ZE E

N

Effective nuclear charge is same for electrons in the same orbital, but varies greatly for electrons of different orbitals(s,p,d,f) and n.

Zeff determines chemical properties of many electron atoms

Building-up (Aufbau) Principle

Effective nuclear charge varies for electrons of different orbitals. Different orbitals corresponding to same n. are no longer degenerate

How do we get 2p energy higher than 2s?

How does Radial distributions change?

How does Zeff affect atomic properties?

Orbital Angular Momentum

e-x

y

z

L= L r X p

Orbital Angular Momentum ‘L’

= +

=

≤ −

=

= ± ± ± ±

�

�

( 1)

orbital angular momentum quantum number

1

0, 1, 2, 3,....,z

L l l

l

l n

L m

m l

Spin Angular Momentum

Stern-Gerlach Experiment A beam of silver atoms (4d105s1) thorough an inhomogeneous magnetic field and observer that the beam split into two of quantized components

Classical, "spinning" particles, would have truly random distribution of their spin angular momentum vectors. This would produce an even distribution on screen.

But electrons are deflected either up or down by a specific amount.

Uhlenbeck-GoudsmitSuggested intrinsic spin angular momentum for electrons

Spin Angular Momentum

Spin Angular Momentum ‘S’’ = +

=

=

=

=

�

�

( 1)

orbital angular momentum quantum number

1 2

1 2z s

s

S s s

s

s

L m

m

Electrons are spin-1⁄2 particles. Only two possible spin angular momentum values. “spin-up” (or α) and “spin-down” (or β)The exact value in the z direction is ms= +ħ/2 or −ħ/2

Not a result of the rotating particles, otherwise would be spinning impossibly fast (GREATER THAN SPEED OF LIGHT)

Spin S(ω) where ω is an unknown coordinate

Hydrogen Atom Wavefunctions: Redefined

Incorporate “spin” component to each of the 1-electron wavefunctions. Each level is now doubly degenerate

1-Electron wavefunctions are now called SPIN ORBITALS

Total wavefunctions is a product of spatial and spin parts

H-atom wavefunctions now can be written as

Which are orthogonal and normalized. Quantum numbers are n,l,m,ms

θ φ ω ψ θ φ ω α ω ψ θ φ ω β ω

ψ α ψ βπ π

− −

−

Ψ = ⋅ ⋅

= =

3 2 3 2

1 11,0,0, 1,0,0,2 2

( , , , ) ( , , , ) ( ) or ( , , , ) ( )

1 1 1 1 o o

r ra a

o o

r r r

e ea a

Spin Orbitals and Exclusion Principle

Spin should always be included for systems with more than one electron

Two electron wavefunctions should include four spin functions

The last two wavefunctions are strictly not allowed because the two electron can be distinguished.

α α β β α β β α(1) (2) (1) (2) (1) (2) (1) (2)

Indistinguishability

Exchange Operator

Ψ = ±Ψ(1,2) (2,1)[ ]

[ ]

α α β β α β β α

α β β α

+

−

1(1) (2) (1) (2) (1) (2) (1) (2)

2

1(1) (2) (1) (2)

2

Symmetric

Anti-symmetric

He atom wavefunction

Spin Orbitals and Exclusion Principle

Ψ = −Ψ(1,2) (2,1)

[ ]ψ ψ ψ α β β α= ⋅ −1 1

1(1) (2) (1) (2) (1) (2)

2He s s

Pauli’s Exclusion Principle (by Dirac!)The complete wavefunction (both spin and spatial coordinates) of a system of identical fermions (i.e. electrons) must be anti-symmetric with respect to interchange of all their coordinates (spatial and spin) of any two particles

If the two electrons in 1s orbital had same spin then the wavefunction would be symmetric and hence it is not allowed

Helium Atom: Excited States

[ ] [ ]

α α

α β β α

β β

= =

⋅ − ⋅ + = = = = −

(1) (2) ( 1; 1)

1 11 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 1; 0)

2 2(1) (2) ( 1; 1)

s

s

s

s m

s s s s s m

s m

[ ][ ]α β β α⋅ + ⋅ − = =1 1

1 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 0; 0)2 2

ss s s s s m

If the second electron is in the 2s orbital then it could have the same spin or the opposite spin.

He excited state 1s1.2s1 (triplet)

He excited state 1s1.2s1 (singlet)

1s (1)1s (2) The spatial part is symmetric

1s (1)2s (2) or 1s (2)2s (1) symmetric nor anti-symmetric

1s (1)2s (2) + 1s (2)2s (1) Symmetric1s (1)2s (2) - 1s (2)2s (1) Anti-symmetric

[ ] [ ]

α α

α β β α

β β

= =

⋅ − ⋅ + = = = = −

(1) (2) ( 1; 1)

1 11 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 1; 0)

2 2(1) (2) ( 1; 1)

s

s

s

s m

s s s s s m

s m

[ ][ ]α β β α⋅ + ⋅ − = =1 1

1 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 0; 0)2 2

ss s s s s m

Helium Atom: Excited States

Helium Atom

1s (1)1s (2) The spatial part is symmetric

1s (1)2s (2) or 1s (2)2s (1) symmetric nor anti-symmetric

1s (1)2s (2) + 1s (2)2s (1) Symmetric1s (1)2s (2) - 1s (2)2s (1) Anti-symmetric

[ ] [ ]

α α

α β β α

β β

= =

⋅ − ⋅ + = = = = −

(1) (2) ( 1; 1)

1 11 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 1; 0)

2 2(1) (2) ( 1; 1)

s

s

s

s m

s s s s s m

s m

[ ][ ]α β β α⋅ + ⋅ − = =1 1

1 (1) 2 (2) 1 (2) 2 (1) (1) (2) (1) (2) ( 0; 0)2 2

ss s s s s m

Homework – Write the correct wavefunctions

Bonding

Bonding: H2

+and H2molecules

+

-

+RHA HB

rA rBr

e-

+

-

+RHA HB

r1A r1Br1

e-

-e-r2Br2A

r2

� ( )+ = − ∇ − ∇ − ∇

− − +

� � �2 2 2

2 2 22

2 2 2

2 2 2

A B eA B e

A B

H Hm m m

e e eQ Q Qr r R

� ( ) = − ∇ − ∇

− ∇ − ∇

− − −

+ +

� �

� �

2 22 2

2

2 22 21 2

2 2 2 2

1 1 2 2

2 2

12

2 2

2 2

A BA B

e ee e

A B A B

H Hm m

m m

e e e eQ Q Q Qr r r r

e eQ Qr R

Born – Oppenheimer Approximation

� ( )+ = − ∇ − ∇ − ∇ − − +� � �2 2 2 2 2 2

2 2 22 2 2 2A B e

A B e A B

e e eH H Q Q Q

m m m r r R

Nuclei are STATIONARY with respect to electrons

� ( )+ = − ∇ − − +�2 2 2 2

22 2 e

e A B

e e eH H Q Q Q

m r r R

� ( )+ = − ∇ − ∇ − ∇ − − +� � �2 2 2 2 2 2

2 2 22 2 2 2A B e

A B e A B

e e eH H Q Q Q

m m m r r R

ignore

Born – Oppenheimer Approximation

� ( ) = − ∇ − ∇ − ∇ − ∇

− − − + +

� � � �2 2 2 2

2 2 2 22 1 2

2 2 2 2 2 2

1 1 2 2 12

2 2 2 2

A B e eA B e e

A B A B

H Hm m m m

e e e e e eQ Q Q Q Q Qr r r r r R

� ( ) = − ∇ − ∇ − − − + +� �2 2 2 2 2 2 2 2

2 22 1 2

1 1 2 2 122 2e ee e A B A B

e e e e e eH H Q Q Q Q Q Q

m m r r r r r R

ignore

Bonding: H2

+Molecule

� ( )+ = − ∇ − − +�2 2 2 2

22 2 e

e A B

e e eH H Q Q Q

m r r R

� ( ) ψ ψ+ ⋅ = ⋅2 ( , ) ( ) ( , )H H r R E R r R

Difficult; but can be solved using elliptical polar co-ordinates

Bonding: H2molecule

� ( ) ψ ψ⋅ = ⋅2 ( , ) ( ) ( , )H H r R E R r R

CANNOT be Solved

� ( ) = − ∇ − ∇ − − − + +� �2 2 2 2 2 2 2 2

2 22 1 2

1 1 2 2 122 2e ee e A B A B


m m r r r r r R

For all the molecules except the simplest molecule H2+ the Schrodinger equation cannot be solved.

We have approximate solutions

Bonding

For all the molecules except the simplest molecule H2+ the Schrodinger equation cannot be solved.

We have only approximate solutions

Valance-Bond Theory & Molecular Orbital Theory

are two different models that solve the Schrodinger equation in different methods

Valance Bond Theory

ψ ψΨ = ⋅(1) (2)A Bψ ψ ψ ψΨ = ⋅ + ⋅(1) (2) (2) (1)A B A B

ψ (1)A

ψ (2)B

R=∞ R= Re

( ) ( )ψ ψ ψ ψ λ ψ ψ λ ψ ψ

λ λ+ − − +

Ψ = ⋅ + ⋅ + ⋅ + ⋅

Ψ = Ψ + Ψ + Ψ

(1) (2) (2) (1) (1) (2) (1) (2)

cov

A B A B A A B B

H H H H

Resonance

H−−−−−−−−H ←→←→←→←→ H+−−−−−−−−H

−−−−←→←→←→←→ H

−−−−−−−−−−−−H

+

Inclusion of Ionic terms

Valance Bond Theory

R= Re

λ λ+ − − +Ψ = Ψ + Ψ + Ψcov H H H H

+

-

+RHA HB

rA rBr

e-

A molecular orbital is analogous concept to atomic orbital but spreads throughout the molecule

It’s a polycentric one-electron wavefunction (Orbital!)

It can be produced by Linear Combination of Atomic OrbitalsLCAO-MO

ψ ψ

− ∇ − − + = ⋅

�2 2 2 2

2

2 ee A B

e e eQ Q Q E

m r r R

Molecular Orbital Theory of H2+

+

-

+RHA HB

rA rBr

e-

ψ ψ

− ∇ − − + = ⋅

�2 2 2 2

2

2 ee A B

e e eQ Q Q E

m r r R

LCAO-MO

ψ φ φ= +1 1 2 1A BMO s sC C

( )ψ φ φ φ φ= + +2 2 2 2 2

1 1 2 1 1 2 1 12A B A BMO s s s sC C C C

= ⇒ = ±2 21 2 1 2

Symmetry requirement

C C C C


+

-

+RHA HB

rA rBr

e-= ⇒ = ±2 2

1 2 1 2

Symmetry requirement

C C C C

( ) ( )ψ φ φ

= =

= + = +

1 2

1 1 1 1 1A B

a

a s s a A B

C C C

C C s s ( ) ( )ψ φ φ

= − =

= − = −

1 2

2 1 1 1 1A B

b

b s s b A B

C C C

C C s s

+ +

( )ψ = +1 1Bonding a A BC s s

+ -

( )ψ − = −1 1Anti bonding b A BC s s


( )ψ = +1 1Bonding a A BC s s ( )ψ − = −1 1Anti bonding b A BC s s


Bracket Notation

� �

φ φ τ φ φ δ

φ φ τ φ φ

∗

∗

= =

= =

∫

∫

i

i

j i j ijallspace

j i j ijallspace

d

A d A A

= =

= ≠

1 (for )

0 (for )

i j

i j

Normalization

( ) ( )

[ ]

[ ]

[ ]

ψ ψ φ φ φ φ

φ φ φ φ φ φ φ φ

= = + +

= + + +

= +

=+

=−

21 1 1 1 1 1

21 1 1 1 1 1 1 1

2

1

1

1 2 2

1

2 2

Similarly

1

2 2

A B A B

A A B B A B B A

a s s s s

a s s s s s s s s

a

a

b

C

C

C S

CS

CS

1

S is called Overlap-Integral

φ φ φ φ

φ φ φ φ

= =

= =

1 1 1 1

1 1 1 1

1A A B B

A B B A

s s s s

s s s sS

Overlap Integral

Overlap-Integral S can be positive or negative or zero

[ ]( )

[ ]( )

ψ φ φ

ψ φ φ

= ++

= −−

1 1 1

2 1 1

1

2 2

1

2 2

A B

A B

s s

s s

S

S

�

�

ψ ψ

ψ ψ

=

=

1 1 1

1 2 2

E H

E H


�

[ ]( ) �

[ ]( )

[ ]( ) � ( )

[ ]� � � �

ψ ψ

φ φ φ φ

φ φ φ φ


=

= + ++ +

= + ++

= + + ++

1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1

2 2 2 2

1

2 2

1

2 2

A B A B

A B A B

A A B B A B B A

s s s s

s s s s

s s s s s s s s

E H

E HS S

E HS

E H H H HS


�

[ ]( ) �

[ ]( )

[ ]( ) � ( )

[ ]� � � �

ψ ψ

φ φ φ φ

φ φ φ φ


=

= − −− +

= − −−

= + − −−

2 2 2

2 1 1 1 1

2 1 1 1 1

2 1 1 1 1 1 1 1 1

1 1

2 2 2 2

1

2 2

1

2 2

A B A B

A B A B

A A B B A B B A

s s s s

s s s s

s s s s s s s s

E H

E HS S

E HS

E H H H HS


[ ]( )

[ ]( )

ψ φ φ

ψ φ φ

= ++

= −−

1 1 1

2 1 1

1

2 2

1

2 2

A B

A B

s s

s s

S

S

� �ψ ψ ψ ψ= =1 1 1 1 2 2 E H E H


[ ]� � � �φ φ φ φ φ φ φ φ= + − −

−2 1 1 1 1 1 1 1 1

1

2 2 A A B B A B B As s s s s s s sE H H H HS

[ ]� � � �φ φ φ φ φ φ φ φ= + + +

+1 1 1 1 1 1 1 1 1

1


[ ]� � � �φ φ φ φ φ φ φ φ= + + +

+1 1 1 1 1 1 1 1 1

1



� �

� �

φ φ φ φ

φ φ φ φ

φ φ φ φ

= = =

= = =

= = =

1 1 1 1

1 1 1 1

1 1 1 1

i i j j

i j j i

i j j i

s s ii jj s s

s s ij ji s s

s s ij ji s s

H H H H

H H H H

S S

Ĥ is Hermitian

[ ]� � � �φ φ φ φ φ φ φ φ= + − −

−2 1 1 1 1 1 1 1 1

12 2 A A B B A B B As s s s s s s sE H H H H

S

+ += =

+ +

− −= =

− −

1

2

2 2

2 2 1

2 2

2 2 1

ii ij ii ij

ij ij

ii ij ii ij

ij ij

H H H HE

S S

H H H HE

S S


+ += =

+ +

− −= =

− −

1

2

2 2

2 2 1

2 2

2 2 1

ii ij ii ij

ij ij

ii ij ii ij

ij ij

H H H HE

S S

H H H HE

S S


�

�

� �

= − ∇ − − +

= − ∇ − − +

= − +

�

�

2 2 2 22

2 2 2 22

2 2

1

2

2

ee A B

ee A B

e

B

e e eH Q Q Q

m r r R

e e eH Q Q Q

m r r R

e eH H Q Q

r R

�

�

φ φ

φ φ φ φ φ φ

= =

= + −

1 1

2 211 1 1 1 1 1

(or )

1 1

i i

i i i i i i

ii AA BB s s

es s s s s sB

H H H H

H Qe QeR r


�

�

�

φ φ

φ φ φ φ φ φ

φ φ φ φ φ φ

= =

= + −

= + −

= + − ⋅

1 1

2 211 1 1 1 1 1

22

11 1 1 1 1 1

22

1

(or )

1 1

1

i i

i i i i i i

i i i i i i

ii AA BB s s

eii s s s s s sB

eii s s s s s sB

ii s

H H H H

H H Qe QeR r

QeH H Qe

R r

QeH E Qe J

R φ φ

φ φ

=

=

1 1

1 1

1

1

i i

i i

s s

s sB

Jr

Constant at Fixed Nuclear Distance

J ⇒⇒⇒⇒ Coulomb Integral


�

�

�

φ φ

φ φ φ φ φ φ

φ φ φ φ φ φ

= =

= + −

= + −

= + − ⋅

1 1

2 211 1 1 1 1 1

22

11 1 1 1 1 1

22

1

(or )

1 1

1

i j

i j i j i j

i j i j i j

ij AB BA s s

eij s s s s s sB

eij s s s s s sB

ij s

H H H H

H H Qe QeR r

QeH H Qe

R r

QeH E S S Qe K

R φ φ

φ φ

=

=

1 1

1 1

1

i j

i j

s s

s sB

S

Kr

K ⇒⇒⇒⇒ Exchange IntegralResonance Integral

Constant

K is purely a quantum mechanical concept. There is no classical counterpart


[ ]

[ ][ ] [ ] [ ]

[ ][ ]

[ ]

[ ][ ]

+ = = + − + + − + +

= + + + − +

+

+= + −

+

− = = + − − − − − −

= −−

2 21 1 1

22

1 1

22

1 1

2 22 1 1

2 1

1 1

11

11 1

1

1

1 1

11

11

1

ii ijs s

ij

s

s

ii ijs s

ij

s

H H SE E Qe J E S Qe K

S R RS

QeE E S S Qe J K

S R

Qe J KQeE E

R S

H H SE E Qe J E S Qe K

S R RS

E E SS

[ ] [ ]

[ ][ ]

+ − − −

−= + −

−

22

22

2 1

1

1s

QeS Qe J K

R

Qe J KQeE E

R S


[ ]( )

[ ][ ]

[ ]( )

[ ][ ]

ψ φ φ

ψ φ φ

= ++

+= + −

+

= −−

−= + −

−

1 1 1

22

1 1

2 1 1

22

2 1

1

2 2

1

1

2 2

1

A B

A B

s s

s

s s

s

S

Qe J KQeE E

R S

S

Qe J KQeE E

R S


[ ][ ]

[ ][ ]

+= + −

+

−= + −

−

≤ ≤ < <

22

1 1

22

2 1

1

1

0 1; 0& 0

s

s

Qe J KQeE E

R S

Qe J KQeE E

R S

S J K

Destabilization of Anti-bonding orbital is more than Stabilization of Bonding orbital

J - Coulomb integral -

interaction of electron

in 1s orbital around A

with a nucleus at B

K - Exchange integral

– exchange (resonance)

of electron between the

two nuclei.


E1

1E2

Sigma Bonding with 1s Orbitals

Bonding with 2p Orbitals

Note the signs, symmetries and nodes

Sigma Bonding with 2p Orbitals

Pi Bonding with 2p Orbitals

Note the signs, symmetries and nodes

Symmetry of Orbitals

Hydrogen molecule ion:

Bonding: Symmetric→ σg

Anti-bonding: Antisymmetric→ σu

Gerade (g) → SymmetricUngarade (u) →Antisymmetric

Types of Bonds

Overlap Integrals

Molecular Orbital Theory of H2

� ( ) = − ∇ − ∇ − ∇ − ∇

− − − + +

� � � �2 2 2 2

2 2 2 22 1 2

2 2 2 2 2 2

1 1 2 2 12

2 2 2 2

A B e eA B e e

A B A B

H Hm m m m

e e e e e eQ Q Q Q Q Qr r r r r R

� ( )

� ( )

� ( ) � ( ) � ( )

= − ∇ − ∇ − − − + +

= − ∇ + − ∇ + − + +

= + − − + +

� �

� �

2 2 2 2 2 2 2 22 2

2 1 21 1 2 2 12

2 2 2 2 2 2 2 22 2

2 1 21 2 1 2 12

2 2 2 2

2 1 21 2 12

2 2

2 2

e ee e A B A B

e ee A e B B A

e eB A


m m r r r r r R


m r m r r r r R

e e e eH H H H H H Q Q Q Q

r r r R

ignore

Cannot be Solved


For H2+

[ ]( )ψ ψ φ φ= = +

+1 1 1

1

2 2 A Bbonding s sS

Place the second electron in the bonding orbital to get H2

[ ]( )

[ ]( ) [ ]

ψ ψ ψ

φ φ φ φ α β β α

= ⋅

= + ⋅ + − + +

2 1 2

1 1 2 21 1 1 1

( )

1 1 1(1) (2) (1) (2)

22 2 2 2A B A B

bonding

s s s s

H

S S


[ ]( ) ( ) [ ]

ψ

φ φ φ φ α β β α = + ⋅ + − +

2

1 1 2 21 1 1 1

( )

1 1(1) (2) (1) (2)

2 1 2A B A B

bonding

s s s s

H

S

[ ]

[ ][ ]

ψ φ φ φ φ φ φ φ φ = + + + +

⋅ + ⋅ + ⋅ + ⋅+

1 2 1 2 1 2 1 21 1 1 1 1 1 1 1

12 1

11 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2)

2 1

A A B B A B B Abonding s s s s s s s s

A A B B A B B A

S

s s s s s s s sS

Spatial Part


[ ]( ) ( ) [ ]

ψ

φ φ φ φ α β β α

−

= − ⋅ − − −

2

1 1 2 21 1 1 1

( )

1 1(1) (2) (1) (2)

2 1 2A B A B

anti bonding

s s s s

H

S

[ ]

[ ][ ]

ψ φ φ φ φ φ φ φ φ− = + − − −

⋅ + ⋅ − ⋅ − ⋅−

1 2 1 2 1 2 1 21 1 1 1 1 1 1 1

12 1

11 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2)

2 1

A A B B A B B Aanti bonding s s s s s s s s

A A B B A B B A

S

s s s s s s s sS

Spatial Part


[ ]

[ ][ ]

ψ φ φ φ φ φ φ φ φ = + + + +

⋅ + ⋅ + ⋅ + ⋅+

1 2 1 2 1 2 1 21 1 1 1 1 1 1 1

1

2 1

11 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2)

2 1

A A B B A B B Abonding s s s s s s s s

A A B B A B B A

S

s s s s s s s sS

[ ]

[ ][ ]

ψ φ φ φ φ φ φ φ φ− = + − − −

⋅ + ⋅ − ⋅ − ⋅−

1 2 1 2 1 2 1 21 1 1 1 1 1 1 1

1

2 1

11 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2) 1 (1) 1 (2)

2 1

A A B B A B B Aanti bonding s s s s s s s s

A A B B A B B A

S

s s s s s s s sS


Effective nuclear charge changes the absolute energyLevels and the size of orbitals!

Matching of energies of AOs important for LCAO-MOIf energies are not close to each other, they wouldNot interact to form MOs.

Diatoms of First Row: H2+, H2 ,He2, He2

+


Effective nuclear charge changes the absolute energylevels and the size of orbitals!

Matching of energies of AOs important for LCAO-MO, if the energies of two Aos are not close they will not interact to form MOs.

Matching of AO energies for MO

Due to large difference in energy of 1s(H) and 1s(F), LCAO-MO for both 1S is not feasible in HF. Rather, 2pz(F) and 1s(H) form a sigma bond.

Both symmetry and energyMatching is required for MO.

Valence electrons are most important

Bonding in First-Row Homo-Diatomic Molecules

1s

2s

2p

1s

2s

2p

The orbital energies of the two approaching atoms are identical before they start interacting to form a BOND


1s

2s

2p

1s

2s

2p

1σ

1σ*

2σ

2σ*3σ

3σ*

1π1π*

The interaction between the energy and symmetry matched orbitals leads to various types of BONDs

MO Energies of Dinitrogen

Experiments tell us this picture is incorrect!


1s

2s2p

1s

2s2p

The 2s and 2p orbitals are degenerate in Hydrogen. However in the many electron atoms these two sets of orbitals are no longer degenerate.


1s

2s

2p

1s

2s

2p

The difference in the energies of the 2s and 2p orbitals increases along the period. Its is minimum for Li and maximum for Ne

MO Energies of Dinitrogen

Mixing of 2s and 2p orbital occur because of small energy gap between them2s and 2p electrons feels not so different nuclear charge.

Note how the MO of 2s→σ have p-type looks, while π-levels are clean

s-p Mixing: Hybridization of MO

Mixing of 2s and 2p orbital occur because of small energy gap between them 2s and 2p electrons feels not so different nuclear charge

s-p Mixing: Hybridization of MO

B2 is paramagnetic. This can only happen if the two electrons with parallel spin are placed in the degenerate π-orbitals and if π orbitals are energetically lower than the σ orbital

Incorrect!

MO diagram of F2: No s-p Mixing

No Mixing of s and p orbital because of higher energyGap between 2s and 2p levels in Oxygen and Fluorine!2s and 2p electrons feels very different nuclear charge

MO Energy Level Diagram for Homo-Diatomics

Upto N2 Beyond N2

Bond-Order and Other Properties

N2 : (1σg)2 (1σ*u)

2 (2σg)2 (2σ*u)

2

(1πux)2 (1πuy)

2 (3σg)2

BO = 3 All spins paired: diamagnetic

O2 : (1σg)2 (1σ*u)

2 (2σg)2 (2σ*u)

2

(3σg)2 (1πux)

2 (1πuy)2 (1πux)

1

(1π*uy)1

BO = 2 2 spins unpaired: paramagnetic

MO Contours and Electron Density

Hetero-Diatomics: HF

Due to higher electronegativityof F than H, the electron distribution is lopsided

Hetero-Diatomics: CO

Hybridization

Linear combination of atomic orbitals within an atom leading to more effective bonding

2s

2pz 2px 2py 2px 2py

αααα 2s-ββββ 2pz

αααα 2s+ββββ 2pz

The coefficients αααα and ββββ depend on field strength

Hybridization is close to VBT approach. Use of experimental information

All hybridized orbitals are equivalent and are ortho-normal to each other

Contribution from s=0.5; contribution from p=0.5Have to normalize each hybridized orbital

= − 1

1

2s pψ ψ ψ

= + 2

1

2s pψ ψ ψ

2 equivalent hybrid orbitalsof the same energy andshape (directions different)

Linear geometry with Hybridized atom at the center

s and p orbital of the Same atom! Not same as S (overlap)

s+p (sp)Hybridization

The other p-orbitals are available for π bonding


1

2

3

1 20

33

1 1 1

3 2 6

1 1 1

3 2 6

s px py

s px py

s px py

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ

= + ⋅ +

= + −

= − −

The other p-orbital are available for π bonding

s+2p (sp2)Hybridization

1

2

3

4

1 1 1 1

2 2 2 21 1 1 1

2 2 2 21 1 1 1

2 2 2 21 1 1 12 2 2 2

s px py pz

s px py pz

s px py pz

s px py pz

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

= + + +

= − − +

= + − −

= − + −

How to calculate the coefficients?

Use orthogonality of hybrid orbitalsand normalization conditions

There is no unique solution

1

2

3

4

1 30 0

2 2

1 2 10

2 3 2 3

1 1 1 12 6 2 2 3

1 1 1 1

2 6 2 2 3

s px py pz

s px py pz

s px py pz

s px py pz

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

= + ⋅ + ⋅ +

= + + ⋅ −

= − + −

= − − −



No other p-orbital is available for π bonding

s-p3-d2 & s-p3-d Hybridization

Sp3d2

Octahedral

Sp3dTrignoal bipyramidal

Hybridization and Geometry

Do Orbitals Really Exist?

Tomographic image of HOMO of N2

Nature; Volume 342; Year 2004; 867-871

Collaboration between Chemists and Engineers

The tensile strength of spider silk is greater than the same weight of steel and has much greater elasticity

Gnp ch103-lecture notes

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