CH103 – Physical Chemistry:Introduction to Bonding

Understanding Atomic and Molecular Structure using

Quantum ChemistryInstructor for D3: Prof. Arindam Chowdhury,

Chemistry, Room 215

Phone: x-7154 Email: achowdhury(at)iitb.ac.inarindam(at)chem.iitb.ac.in

Contents (10 Lectures): 3.40-5.00 pm• Schrodinger Equation (Origin or Quantization): Lecture 1-3

Failures of Classical MechanicsPostulates of Quantum MechanicsEnergy Quantization: Particle in a Potential Well

• Atomic Structure: Lecture 4-6Hydrogen Atom and Quantum NumbersAtomic Orbitals and Electron DensitiesMulti-electronic atoms

• Chemical Bonding: Lectures 7-9Molecular Orbital TheoryEnergy and electron densities of diatomic molecules

• Molecular Structure: Lecture 10Bonding in polyatomics using hybridization

Recommended Text (Physical)• Physical Chemistry – I.N. Levine, 5th Ed.• Physical Chemistry – P.W. Atkins 2nd Ed.• Physical Chemistry: A Molecular Approach - McQuarrie and Simon

Websites:www.chem.iitb.ac.in/academics/menu.php(CH103 Course Material for 2008 and 2009); ,

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

http://www.falstad.com/mathphysics.html#qm

http://www.meta-synthesis.com/webbook.html

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

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

Why should we study Chemistry?

Is there a role of Chemistry inreshaping the modern world?

All of Science and Engineering is moving towards interdisciplinary fields of cutting-edge research!!!

Knowing only one subject often not good enough!

It is important to know basics of Chemistry…

1. Nanoelectronics/Nanotechnology: Molecular Electronics2. Energy Science – “Solar Energy” conversion3. BioTechnology – Disease cure, health, medicine4. Atmospheric Science – Have to “Save the World”

Organic Light Emitting Diodes (O-LEDs)LEDs from Conducting Polymers: MEH-PPV

Plastic Electronics and DisplaysConducting-polymers are replacing liquid crystals

Micro & Nano-electronics

1947, Transistor, Bell LabsSilicon Transistor, TI 1954 Intel, 1990s, hundreds of

Transistors in a single chip

Transistors, Intel, 2006, 45 nm separationNext Generation: Molecular Chips

Molecular Electronics

Mechanics of Electrons and Atoms

Nano-scienceAnd Nanotechnology

Mult-electron Atoms(Periodic Table)

Electron Microscopy

Intermolecular ForcesAnd Interactions

Multi-atomicBonding, Molecular

Structure

Biology, Materials Science

Condensed Matter Physics

Chemical ReactionsMolecular Dynamics

Atomic/MolecularSpectroscopy

Atomic Structure - History

Plum-Pudding Model

J.J. Thompson (Discovered Electrons, 1996)

Rutherford’s Gold Foil Experiment

Rutherford beamed alpha particles (++) through gold foil and detectedthem as flashes of light on a screen. The gold foil was only 0.00004 cm thick, meaning on a few hundreds of atoms thick

1871-1937

Planetary model loggerheads with classical electromagnetic theory

Classical EM theory can not explain Blackbody Radiation

Theories based on classical physics unable to explain temperaturedependence of emitted radiation (radiant energy density)

Sun, stars…hot iron rodAll classical theories led to the so called “Ultraviolet Catastrophe”

23

8( ) ; bk T cT d dcν

πρ ν ν ν νλ

= =

Max Planck assumed energies of oscillators are discontinuous

3

/( )1v bv t

aT de

νρ ν =−

Assumption: Energy of electronic oscillators were discrete;Proportional to integral multiple of frequencies

E = Energy of electronic oscillatorsv = frequency of electronic oscillatorsh = Planck’s constant = 6.626 x e-34 joule-secNote: h came in as a fitting parameter

OscE nhν=

1858-1947

Planck never believed his theory was right, since he was a classical physicist

3

3 /8( )

1Bv h k T

h dT dc e νπ ν νρ ν

=−

Line Spectra of Atoms

Rydberg’s formula:

1 22 21 2

1 1 1 ; HR n nc n nνν

λ

= = = − >

RH = 109677.57 cm-1

1854-1919

Bohr’s Phenomenological Model(or…Rutherford-Planck-Bohr Model)

• 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 = n h/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 toanother of energy Eb, it emits or absorbsa photon of light: Ea – Eb = hv

1885-1962

Bohr’s model explains atomic spectra n=1,2,3,...

2(2 )

nhmvr

r nπ

π λ

=

=

4

2 2 2 2

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

ei f

i f

m eE h n nh n n

νε

∆ = − = =

Spectral Transitions: ∆E=hv

Further details in Tutorials as problems!Explains Rydberg’s Formula

4

2 2 20

1.8

en

m eEh nε

= −

Photoelectric Effect

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

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

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

0

20

( )Wave Energy relatedto

of

E E Sin kx t

Intensity EIndependent

ω

ω

= −

∝

Photodetectors, Photovoltaics, Elevator sensor, smoke detectors

Experimental Observations

Einstein: light can behave like particles

2

0

Energy to remove e' from surface

12

0

P M

M

E hv KE mv

KE hv

φ φ

φφ

=

= = + = +

= − ≥

Borrowing Planck’s idea that ∆E=hv, Einstein further proposed that radiation itself existed as small packets of energy (Quanta), now known as PHOTONS

PE hν=

1879-1955; Nobel prizeFor explanation of Photoelectric effect

Energy of photonfrequency dependent

Light is EM Radiation: Waves

Diffraction or Interference Pattern can be possible only if light is a wave

10

( )( )

~ 3 10 / sec

m

m

m

m

E E Sin kx tB B Sin kx tE c cmB

ωω

= −= −

= ×

Wave-Particle Duality of Light

Light has both wave-like andparticle-like propertieseven though

not observed by us (humans)

simultaneously

De Broglie Hypothesis: Matter Waves

h hp mv

λ = =

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

Wave-Packet1892-1987

-3410

-31 6

6.6x10 J s 7 109.1x10 Kg 1x10 m/s

h mmv

λ −= = = ××

Electron-wave moving @ 106 m/s λ too small for MACROSCOPIC

OBJECTS

Electrons are well known “particles” with a negative charge

J.J. Thompson, Nobel Prize in 1906FOR SHOWING ELECTRONS ARE PARTICLES

Essentially how a CRTMonitor or standard TVworks

Electron show interference patterns WavesDavisson and Germer 1927

J.J’s son, George Thompson, Nobel Prize 1937For showing ELECTRONS are WAVES

Light can have Particle or Wave like propertiesElectron/matter can have Particle or Wave-

like properties, but we do not know what they really are!!

Electrons or photons show both wave and particle nature “WAVICLE”:

Best suited to be called some form of “Energy”

Wave-Particle Duality

Uncertainty Principle - Heisenberg

. 4xhx pπ

∆ ∆ ≥

IMPORTANT FOR SUB-MICROSCOPIC SYSTEMS

h = 6.626 x 10-34 J s

1901-1976

Further details in Tutorials as problems!

Uncertainty Principle: ∆X.∆Px~ h

Wave-Packet

Specifying position accurately would require many waves, each having a fixed value of momentum (k),

P becomes completely uncertain.

If only one wave with known k (momentum), the position becomes completely uncertain,since 1 wave spreads indefinitely in space.

The nature of the world within the submicroscopic atoms is not directly observable –

we deduce it on the basis of highly indirect evidence, and then proceed to describe it……in terms/quantities which we know from our

experience of the macroscopic world

However, the analogies may not always be good ones

Classical quantities such as position and momentum have decreased suitability for describing the subatomic world.

Inherently, “nature” at the atomic scaleis not deterministic, and their exact behaviors

can not be predicted – Probabilistic

Bohr Model with a De Broglie Touch

Constructive InterferenceOf the electron-waves Can result in stationaryStates (Bohr orbits); If wavelength don’t match,There can not be any Energy Level (state)

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

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

r nh

mvnhmvr

π λ

λ

π

=

=

=

Need a new theory for dynamics of electrons, atoms or molecules

• Probabilistic, not deterministic (non-newtonian)• Wavelike equation for describing sub/atomic systems

Schrodinger 1925!!!

Mathematical Tools Required

1

0

4, , , , , , /( ) ( ) : dSQR CUBE Sin new functions numbersdx y

Af x g x ∂⇒

∂= ∫

1. Operators: A symbol that tells you to do something to whatever follows it. Operators can be real or complex, and can be represented as matrices. An operatorEssentially perturbs a function or system, as if a measurement is being done!

2. Solve 2nd order partial differential equations with multiple variables:

2 2

2 2( ) ( ) ( ) ( ) ( ) ( ) 0f f f f mx y x y f x nf y

x yx y+ +

∂ ∂ ∂ ∂+ + + =∂ ∂∂ ∂

2

2( ) ( )fd x kf x C

dx+ =

2 2

2 2( ) ( ) ( ) ( )( ) ( )f f f fmd x d x d y d yf x nf y const

dx dydx dy+ + = + + =

[ ] 1 1 2 2 1 1 2 2

3

:

( ) ( ) ( ) ( ) : ;

but or , : Operators

Linear OperatorsdA c f x c f x c Af x c Af x Lineardx

SQR or Sin Cos Nonlinear

+ = + ⇒

Separation of Variables…

2 2 2

1 1: ( ) ( ) and ( ) ( )

n n

j j j j j jj j

Mean E x x x P x E x x x P x= =

= = = =∑ ∑

4. Expectation Values or Average or Mean values

, ( ), ( ) ( ) : Given A find the function x such that A x a x Eigenvalue Equationφ φ φ=

3. Eigenfunctions and Eigenvalues:

Operator operating on a function results in re-generating the same function (which is called eigenfunction) multiplied by a number (eigenvalue)

; ? ?( ) n

xn

d Can or Sin x Eigenvaluesdx

A x be eα αφ=

If P(x) is the probability

(#) for , , but (#) for , n

n

d Sin x Sin x n even it is Sin x n odd not eigenfucntiondx

α α α= ⊗ = ≠ ⊗ =

! :n

x n x nn

d Yes EVdx

e eα αα α= ⇒

LAWS of Quantum Mechanics The state of a system is completely specified by thewavefunction ψ(x,y,z,t). It’s square gives probability density…

To every observable in classical mechanics, there corresponds a linear operator in quantum mechanics

In measurement of observable associated with operator A, only values that will be observed are the eigenvalues of A

The average value of the observable corresponding to A is

Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger Eqn:

ˆ*a A dυ= Ψ Ψ∫

1. Born Interpretation of Wavefunction1. The state of a QM system is completely specified by a wave function Ψ(x,y,z,t) & all possible information can be derived from Ψ

Ψ can be real or complex

2. Ψ*. Ψ dv is the probability that the particle lies in volume element dv, so Ψ∗Ψ is the probability density at that point

1882-1970

y

x

z

xa

ya

za

dxdy

dz

*

2

1 3 : ( , , , ')Pr( , , )

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

( , , , ')

a a a a a a

a a a a a a

a a a

Particle System in D x y z tx x x dx y y y dy z z z dz

x y z t x y z t dxdydz

x y z t dv

Ψ≤ ≤ + ≤ ≤ + ≤ ≤ +

= Ψ Ψ

= Ψ

*

2* 2 2

:

( )( )

can be complex f ig and f ig

f ig f ig f g

Ψ Ψ = + Ψ = −

∴ Ψ Ψ = − + = + = Ψ

1a. Corollary: Normalization

* *

( , , ). ( , , ) 1all space all space

x y z x y z dxdydz dτΨ Ψ = Ψ Ψ = Ψ Ψ =∫∫∫ ∫

Normalization of Wave function: Since Ψ∗Ψdv is the probability, the total probability of finding the particle somewhere in space has to be unity.

If integration diverges, i.e infinity: Ψ can not be normalized, and therefore can NEVER be an acceptable wave function. But if it is a constant A ≠1, then it is OK

Work on problems related to normalization of wave functions

Automatically, a condition has been imposed on the nature of wave function Ψ:Ψ must vanish at ± infinity, or rather at the boundaries! Also Ψ has to be finite!!!

X

Ψ

X

Ψ

The state of a QM system is completely specified by a wavefunctionwhich is ψ(x,y,z,t). Square gives probability density…

To every observable in classical mechanics, there correspondsa linear operator in quantum mechanics

In measurement of observable associated with operator A, only values that will be observed are the eigenvalues of A

The average value of the observable corresponding to A is

Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger Eqn:

ˆ*a A dυ= Ψ Ψ∫

LAWS of Quantum Mechanics

2. Operator FormalismTo every observable in classical mechanics, there corresponds

a linear operator (real or complex) in quantum mechanics

Classical quantity Quantum Mechanical operator

Position, x x̂

Momentum, px dxdi−

Kinetic energy in 1-D, 2

2x

xpKm

= 2

22

2 dxd

m

−

Kinetic energy in 3-D,

( )2 2 212 x y zK p p p

m= + +

∂∂

+∂∂

+

∂∂

− 2

2

2

2

2

22

2 zyxm

Potential Energy, V(x) Multiply by V(x)

Total Energy, E= K+V

∂∂

+∂∂

+

∂∂

− 2

2

2

2

2

22

2 zyxm

+ V(x)

Find out QM operators corresponding to angular momentum, and KE for rotating systems

The state of a QM system is completely specified by a wavefunctionwhich is ψ(x,y,z,t). Square gives probability density…

To every observable in classical mechanics, there corresponds a linear operator in quantum mechanics

In measurement of observable associated with operator A, the only values that will be experimentally measured are the eigenvalues of A, (which can not be imaginary)

The average value of the observable corresponding to A is

Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger Eqn:

ˆ*a A dυ= Ψ Ψ∫

LAWS of Quantum Mechanics

3. Real (observable) eigenvalues

ˆ n n nA aΨ = Ψ

•In any measurement of abservable associated with operator A, the only values thatwill be ever observed are the eigenvalues an, which satisfy the eigenvalue equation:

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

All the eigenfunctions of Quantum Mechanical operators are “Orthogonal”

* ( ) ( ) 0 for m n m nx x dx m nψ ψ ψ ψ+∞

−∞

= = ≠∫

Ψn corresponds to the eigenfunctions or eigenstates of the system.

Eigenvalues represented by an are observables

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

There may be, and typically are, many Eigen-f(n)s for the same QM operator!

The state of a QM system is completely specified by a wavefunctionwhich is ψ(x,y,z,t). Square gives probability density…

To every observable in classical mechanics, there corresponds a linear operator in quantum mechanics

In measurement of observable associated with operator A, only values that will be observed are the eigenvalues of A

The average value of the observable corresponding to A is

Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger Eqn:

ˆ*a A dυ= Ψ Ψ∫

LAWS of Quantum Mechanics

4. Average or Expectation value

2 *( ) ~ . ( 1 )P x dx dx in DΨ = Ψ Ψ

Prescription for obtaining the average or mean value of a classical observable

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

ˆ ˆ*all space

a A dv A= Ψ Ψ = Ψ Ψ∫

nd 2 2( ) ( ) and 2 moment ( )E x x xP x dx x x P x dx+∞ +∞

−∞ −∞

= = =∫ ∫

2 *

ˆ 1 . ( ) = . ~all space

In D a A P x dx A dx A dx+∞ +∞

−∞ −∞

= Ψ Ψ Ψ∫ ∫ ∫

The state of a QM system is completely specified by a wavefunctionwhich is ψ(x,y,z,t). Square gives probability density…

To every observable in classical mechanics, there corresponds a linear operator in quantum mechanics

In measurement of observable associated with operator A, only values that will be observed are the eigenvalues of A

The average value of the observable corresponding to A is

Ψ(x,y,z,t) of a system evolves according to time-dependent Schrodinger Equation:

ˆ*a A dυ= Ψ Ψ∫

( , ) ( , )i r t H r tt

∂Ψ = Ψ

∂

LAWS of Quantum Mechanics

5: Schrodinger Equation (SE)

1887-1961

The SE equation is somewhat equivalent to Newton’s laws, but applicable to atomic/molecular and subatomic systems

2 2 2

2

(1 )

2 2

x

x

P D ix

PKm m x

∂→ −

∂

∂= = −

∂

Predicts the probability of finding the particle at a specified volume element at future time.. counterpart to Newton’s Equations of motion.

2 2

For ( , ),

( , ) ( , ) ( , )2

x

x t Schrodinger Equation

i x t V x t x tt m

Ψ

∂Ψ = − ∇ + Ψ ∂

2 2

22

2

( )

( , )2

, 1

x

x

Hamiltonian Energy Operator

H V x tm

where in Dx

⇒ = − ∇ +

∂∇ =

∂

2nd order partial differential equation

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

Classical quantity Quantum Mechanical operator

Position, x x̂

Momentum, px dxdi−

Kinetic energy in 1-D, 2

2x

xpKm

= 2

22

2 dxd

m

−

Kinetic energy in 3-D,

( )2 2 212 x y zK p p p

m= + +

∂∂

+∂∂

+

∂∂

− 2

2

2

2

2

22

2 zyxm

Potential Energy, V(x) Multiply by V(x)

Total Energy, E= K+V

∂∂

+∂∂

+

∂∂

− 2

2

2

2

2

22

2 zyxm

+ V(x)

Time-Dependent Schrödinger Equation

1 2 1 2( , ,... , ) ( ). ( , ,... )n nx x x t f t x x xψΨ =Very often, V(x,t) = V(x): solutions to the TDSE has the form:

i.e. The space part and the time part can be separated out in the solutions!

Maths: Separation of variables to solve linear 2nd order differential equations

22

1 2

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

( , ) ( , ,... , )

x

n

i x t x t V x t x tt m

x t x x x t

∂Ψ = − ∇ Ψ + Ψ

∂Ψ = Ψ

2 22 2

2

( )

( , ) , 12

x x

Hamiltonian Energy Operator

H V x t where in Dm x

∂⇒ = − ∇ + ∇ =

∂

Separation of space and time variables

:1 ( ) ( ) ( ) ( )

dt

TIME DEPENDENT PART of solutiondf t df t iEE

i f t dt f t

−

− = ⇒ = −

[ ] [ ]

2 2

2

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

LHS F(x) = RHS F(t) :

f tV x E constm x x i f t t

E independent of x and t

ψψ

∂ ∂− + = − =

∂ ∂

2 2

2 2

Let us consider the solutions for TDSE that have the form:

( ) ( )( ) ( )

( , ) ( ). ( )f t xx and f t

t t x x

x t f t xψψ

ψ∂Ψ ∂ ∂ Ψ ∂

= =∂ ∂ ∂ ∂

Ψ =

For a one-particle, 1-D, If V(x,t)=V(x):2 2

2: ( )2

TDSE V xm x i t

∂ Ψ ∂Ψ− + Ψ = −

∂ ∂

Substitute & divide by ( ) ( )f t xψΨ =

/ ( ) ~ iEtf t e−

Time Independent Schrödinger Equation 2 2 2 2

2 2

Rearranging ( ) :

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

time INDEPENDENT SPACE part

d x dV x E V x x E xm x dx m dx

ψ ψ ψψ

−

− + = ⇒ − + =

2 2

2

1 ( ) ( ) dimension (Energy), 2 ( )

( !!!)

d x V x E E has same as Vm x dx

this const is the ENERGY of the system energy comes automatically

ψψ

− + =

2 2 2 2

2 2 21

1 (3 ) :

( , , )2

N

i i i i i

For particle D system solve TISE

V x y z Em x y z

ψ ψ ψ=

∂ ∂ ∂⇒ − + + + = ∂ ∂ ∂

∑

2 2

2

: ( )

( )2

( ) ( )Time Independent Schrodinger Equation Stationaty States

dwhere H V x T Vm dx

H x E xψ ψ−

≡ − + = +=

We have suddenly come up with another EIGENVALUE PROBLEM to solve!!!

N

HΨn=EnΨn : Energies and Wavefunctions

( ) ( )( )( )

2 2 *

/ * /

20 *

. ( )

iEt iEt

f f f

e e

eF t

ψ ψ ψ

ψ ψ

ψ ψ ψ

−

Ψ = =

=

= =

≠

For Stationary states (i.e. for a particular solution): Ψ2 and Energy Const in time

2 2

2, ( ) 2

( ) ( )n n n where H V x Total Energym x

H x E xψ ψ ∂≡ − + ⇒

∂=

MANY solutions to TISE possible different Ψ with different energies

All energies are allowed unless there are some special conditions (just like a wave when restricted at the ends – Stretched string)

Does not mean that the system is at rest (recap Bohr’s stationary state), but energies and the probability do not change with time.

Well-Behaved or valid wavefunctions1. Ψ must be normalizable: Ψ must be finite and 0 at boundaries/ ±infinity2. Ψ must be a solution of the Schrodinger equation:

A. Must be a continuous function of x,y,z.B. The slope of the function i.e. dψ/dq must be continuous in x,y,z.

3. Ψ must be single-valued

Quadratically integrable

The essence of Quantum Mechanics• Not deterministic: Can not precisely determine many parameters in the

system, but Ψ can provide all the information (spatio-temporal)

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

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

• Final outputs tally very well with experimental results –that’s why QM is still out there! Does not violate Classical mechanics for large value of mass.

Applications- 1. Free Particle

2 2

2

:

( ) ( ) ( ) ( )2

( ) ( )

Time Independent Schrodinger Equation

x V x x E xm x

H x E x ψ ψ ψψ ψ

−

∂− + =∂

= ⇒

2 2 2 22

2 : : ( ) sin cos( ) cos sin

2( ) ( sin cos ) ( )

ndSolve the Order Linear DE Trial Solution x A kx B kxD x A kx B kx

mED x k A kx B kx k x k

ψψ

ψ ψ

= += −

= − + = − =

2 2 22 22 2

2 2 2

:( ) 0 :" "

( ) 2( ) ( ) ( ) 0 2

For a free particleV x as there are no external potential forces free

x mEE x D x k x where D and km x x

ψ ψ ψ ψ

=

∂ ∂⇒ − = ⇒ + = = =

∂ ∂

2 2

; 2

2 2 ( ) sin cos

kE No restrictions on k Energies of free particle continuousm

mE mEAnd x A x B xψ

∴ = ⇒

= +

No Quantization:All energies allowed

2. Particle in 1-D Potential Well

2 22

2 : ( ) ( ) ( ) 0 ( ) ( )

: ( ) 0

mFor regions I and III D x E V x D x x

Normalization condition not satisfied Only solution x

ψ ψ ψ ψ

ψ

+ − = ⇒ = ∞

⇒ ⇒ =

2 2 2 22 2

2 2 : ( ) ( ) 0 ( ) ( ) 0

Similar to Free Particle solution,

: ( ) sin cos

mE mEFor regions II D x x D x k x k

but now boundary conditions present

Trial Solution x A kx B kx

ψ ψ ψ ψ

ψ

+ = ⇒ + = =

⇒

= +

Boundary ConditionsΨ(0)=0 and Ψ(L)=0

V=0V=∞ V=∞

Region IRegion II

Region III

XX=0 X=L

: 0, ( ) 0 0 ( ) sin , ( ) 0 sin 0 0 sin 0

0, ( ) 0 !! sin 0 1, 2,3, 4,...

Boundary Conditions x x B x A kxAlso at x L x A kL either A or kLIf A x everywhere or kL kL n n

ψ ψψ

ψ π

= = ⇒ = ⇒ == = ⇒ = ⇒ = =

= = = ⇒ = =

x

V x x L

x L

0

( ) 0 0

∞ <= ≤ ≤∞ >

Particle in Box (PIB) Energies are Quantized: Certain Levels Allowed

2 2 2 2 2 22

2 2 2

2

:sin 0 for 1, 2,3, 4,...

22 8

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

n

n

From Boundary ConditionskL kL n n

m n n n hE k EL mL mL

or E nAutomatically energy is no longer continuousbut discrete or quantize

π

π π

β

= ⇒ = =

∴ = = ⇒ = =

=

!!!d

Note: minf. energies mergeLlarge, energies merge

22 2

2

1 :

( )8

Larger ,

i f

f i f i

Spectroscopy of PIB in D n n

hh E E E n nmL

the box smaller the energy of hv

ν

→

= ∆ = − = −

Zero Point Energy

Concept!

PIB Wave Functions and Probability

* 2 2

0 0

: ( ) sin sin

: ( ) ( ) 1 sin 1

2 2 : ( ) sin

L L

n

nWavefunction x A kx A xL

nNormalization x x dx A x dxL

n xSolve for A Normalized Wavefunctions xL L L

πψ

πψ ψ

πψ

= =

= ⇒ =

= ⇒ =

∫ ∫

No. of Nodes(zero crossings)= n-1

Symmetric(even function)And Anti-Symmetric(odd function)

Expectation Values and Probability

ψ ψ

π π

π

∗= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅

∫

∫

∫

0

2

0

2 2sin sin

2sin

L

L

x x dx

n nx x x dx

L L L Ln

x x dxL L

ψ ψ

π π

π π π

∗ ∂ = ⋅ − ⋅ ⋅ ∂

∂= − ⋅ ⋅ ⋅

∂−

= ⋅ ⋅

∫

∫

∫

0

2 0

2 2sin sin

2sin cos

x

L

L

p i dxx

n ni x x dx

L L x L Li n n n

x x dxL L L

Particle in a 2-D Well: Approach

Hamiltonian x yH H H

m x m y

2 2

2 22 2∂ ∂

= + = − −∂ ∂

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

E is sum of the eigenvalues of the parts of Ĥ

nH x y E x y( , ) ( , )ψ ψ⋅ = ⋅

yx

x x y y

nnx y x y x y

L L L L2 2

( , ) ( ) ( ) sin sinππ

ψ ψ ψ= ⋅ = ⋅

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

V=0

Lx

Ly

Square Box ⇒ Lx = Ly = L

Particle in a 2-D Well: Energies

2D Well: 2 Quantum Numbers are required to describe the system

V=0 222

, 2 2 , 8

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

x y

yxn n

x y

x y

nnhEm L L

n n

= +

=

Degeneracy is manifestation of symmetry

Ener

gy

2-D Potential Well - Wavefunctions

What Quantum Numbers in x and ydo this wavefunction correspond to?

Number of nodes = ni-1

Particle in a 3D-Box:Wavefunctions

3-D Box: 3 Quantum NumbersCubic Box: Wavefunctions

Importance in Chemistry/Spectroscopy• Electronic spectra of conjugated molecules (loose p-electrons): 1D-PIB

V=0V=∞ V=∞

Region IRegion II

Region III

XX=0 X=L

22 2

2

1 :

( )8

,

i f

f i f i

Spectroscopy of PIB in D n n

hh E E E n nmL

Longer the box smaller the energy of hv

ν

→

= ∆ = − = −Increasing length of the box(i.e. the conjugation length)reduces the energy gaps andhence lower energy photons

are absorbed or emitted

Nanoscience: Quantum Confinement

Band gap changes due to confinement,

and so will the color of emitted light

CB

VB

CB

VB

CB

VB

Quantum Dots – Particle (excitons) in a Sphere!

Quantum Dots have a huge applicationIn chemistry, biology, and materials science

For photoemission imaging purpose, As well as light harvesting/energy science

- + - + - +

The Hydrogen Atom

A Completely Solvable problem!!

(kind of rare, in QM!)

CH103 Lecture 4 AC

What we learnt from solving PIB?Formulate a correct Hamiltonian

(energy) Operator H

Impose boundary conditionsfor eigenfunctions (restriction)and obtain Quantum Numbers

Probability andAverage Values

Solve HΨ=EΨ (2nd order PDE)by separation of variable and

intelligent trial/guess solutions

Energies etc. of statesCorresponding toQuantum Numbers

Eigenstates or Wavefunctions: Should be “well behaved” -

Normalization of WavefunctionQuantum Numbersthat specify the

“state” of the system

H-Atom: Constructing H=T+V

2 2 2 2 2

2 2 21

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

( , , , )

N

i i i i i

P x y zH KE PE V x y z t V x y z tm m x y z

P i and V x y z t Potential energyx y z

=

∂ ∂ ∂= + = + = − + + + ∂ ∂ ∂

∂ ∂ ∂= − + + = ∂ ∂ ∂

∑

2 2

2 2 2 2

Nucleus Electron

Electron NucleusNucleus Electronm mH V −+∇ ∇= − −

Hydrogenic Atoms: 2-Particle System1 electron moving around a

(massive) central nucleus (+ve)

2 2 2

2 2 2

22 2

1, ( ) ;

2 i i i

N

i ii i

where Laplacian i particlesx y z

H Vm=

∂ ∂ ∂= + + →

∂ ∂ ∂= − ∇ + ∇∑

r

2 2

2 2 2 2

Electron Nucleus

Nucleus ElectronNucleus Electronm m

H V −+∇ ∇= − −

( ) ( ) ( )

2 2 2 2 2 2 2 2

2 2 2 2 2 2

2 2 2

2

2 2

N N N N e e e e

e N e N e N

m x y z m x y z

where r x x y y z z

ZerH ∂ ∂ ∂ ∂ ∂ ∂

+ + + + ∂ ∂ ∂ ∂ ∂ ∂

= − + − + −

−= − −

2 2 22 2

,

,2 2

( , , , , ) e e e N N N Total e N

N e

Total

eN Total Total Total Total TotalZe

m m rwhere x y z x y z and E E E

E

=

− =

Ψ Ψ = +

∇ Ψ ∇ Ψ Ψ Ψ− −

2 2

0

: ( ) ( ) ~4

Coulomb PotentialZe ZePotential Energy V U r

r rπε= = − −

r

Potential Energy: Coulomb Potential

: TotalIf Complete Wavefunction for H Atom TISE becomesΨ = −

Reduced Form of TISE for H-Atom:Separation of Variables

Reduced Center of Mass:

e e N Ne N CM

e e N Ne N CM

e e N Ne N CM

e N

m x m xx x x xM

m y m yy y y yM

m z m zz z z zM

m mM

µ

+= − =

+= − =

+= − =

=

2

2 2

2

2

2

2

in terms of CM and electronic coordinates( , ) ( ) ( )

, , , , , , , ,

N N N

Total e e e N N N e e e e N N N N

CM

e e e e

EM

Zer

Separate H

and

E

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

µ

=

− =

Ψ = Ψ • Ψ

∇ Ψ Ψ

∇ Ψ Ψ

−

−

Free Particle: movement ofThe whole atom: You solved it! Relative motion of the electron and

With respect to the Nucleus

2 2 2 2 2

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

2 e e e ei i i

e eZex y z x y z x y z x y z x y z

x y z x y zE

µ=

∂ ∂ ∂− Ψ + Ψ + Ψ − Ψ ∂ ∂ ∂ + +

Ψ

Problem: 2nd order PDE with 3 variables - can not be separated!

Relative motion of electron wrt nucleus:

: .

Movement of electron much faster than heavy nucleusSeparate translational motion relative frame⇒

Spherical Polar Coordinates

Hamiltonian (Energy) Operator in 3D Spherical Polar Coordinates

To solve this PDE, need to separate variables, which is POSSIBLE

Recap: Particle in a 3-D Box

TISE for H-Atom in spherical-polar coordinates

s ( ) ( ) ( ) : ( , , ) ( ) ( ) ( )Special olution if H H r H H r R rθ φ θ φ θ φ= + + Ψ = Θ Φ

[ ] [ ]

[ ]

2

2 2

2 2

2

2

1( ) ( ). ( ) sin ( ) ( ). ( )sin1

1 ( ) ( ). ( )sin

2

d d d dr R r R rdr dr d d

r d R rd

ZeE Rr

θ φ θ θ φθ θ θ

φ θθ φ

µ

Θ Φ + Θ Φ

+ Φ Θ

+ +

( ) ( ) ( ) 0r θ φΘ Φ =

22

2 2 2

2

2

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

2 ( , , ) 0r r r r

r r r

ZeE rr

θ φ θ θ φ θ φθ θ θ θ φ

µ θ φ

∂ ∂ ∂ ∂ ∂ Ψ + Ψ + Ψ ∂ ∂ ∂ ∂ ∂

+ + Ψ

=

2nd order Partial Differential Equation with three variables

2 2 22 2 2

2 2

sin sin 2 1sin sind dR d d Ze dr E rR dr dr d d r d

θ θ µθ θθ θ φ

Θ Φ + + + = − Θ Φ

2( , ) ( ) ( , ) ( ) . ( )LHS F r G RHS F r G Const m sayθ φ θ φ= = = ⇒ = = =

2 22

2 2 2 2

1 2sin 0sin sin

d dR R d d R d Zer E Rr dr dr d d d r

µθθ θ θ θ φ

Φ Θ Θ Φ ΘΦ + + + + ΘΦ =

22

2

1 0mφ

∂ Φ∴ + =

Φ ∂

2 2sin :. .

rMultiply by and rearrangeR

θΘ Φ

Separation of Variables r, θ, φ( , , ) ( ). ( ). ( ) . .r R r Rθ φ θ φΨ = Θ Φ = Θ Φ

Solve 2nd order DE to obtain functional form of Φ(φ)

: ( )Solution imAe φφ ±Φ =

Solving φ-part is relatively simple!

Another Quantum Number “popped out” out of Boundary Conditions: Quantization of Angular Momentum!

Magnetic Quatnum Number: Theoretically can take any integral value includingZero, but restricted by another quantum number, l: m<l. Related to orbital

Angular momentum and splitting of energy levels in presence of magnetic field.

Boundary Condition

22

2

1 ( ) 0( )

mφφ φ

∂∴ Φ + =

Φ ∂ 22

2

&

im

m

φ

φ

∂Φ= ± Φ

∂

∂ Φ= − Φ

∂

Solving R(r) and Θ(θ) part not so simple, but can be done!

2 2 22 2

2

2

2sin sin 2sin sin 0

sin :

d dR d r Zer ER dr dr d r

Divide by and rearrange

mθ θ µθ θθ θ

θ

∂ Θ + + + = Θ ∂ −

Need mathematical skills to solve the Differential Equations for R(r) and Θ(θ)

2 2 22

2 2

1 2 1 sin ( ) .sin sin

d dR r Ze d d mr E say constR dr dr r d d

µ θ βθ θ θ θ

Θ + + = − + = = Θ

Only r : Can be solvedto obtain R(r) part

Only θ: Can be solvedto obtain Θ(θ) part

Boundaryconditions

applied!

Θ(θ)Φ(φ) are Spherical Harmonics Ylm(θ,φ)

l=Azimuthal Quantum Numberor orbital quantum number l≤n-1

Easier to solve if written differently: Rigid-Rotor Problem already have solved the angular (q, f) part: Related to Angular Momentum!

L2 Square of angular momentum: Eigenfunctions are “Spherical Harmonics”

n= Principal Quantum Number=1,2,3,…

Radial Wavefunction depends on n and l:

Solve for R(r): Quantized Energies

Only n dependence of E: Note that V term in the H is needed for providing energies as eigenvalues. Angular parts does not have it!!!

Essentially sameEquation as Bohr’s,With slight changes

Energies: En:

Particle in 3D-Box:Three Quantum Numbers

3 Quantum Numbers needed toDescribe the system completely

Normalization Conditions (each dimension)

Meaning of Quantum Numbers

L=0 s-orbitalL=1 p-orbitalL=2 d-orbitalL=3 f-orbital

How to obtain normalized Ψn,l,m(r,θ,π)?

Radial Solutions depend on n and l (l=n-1)

Additional restrictions on l: n ≥ l+1

2 1

0

02( , ). . .l lnl n l

Zr naZr

naR A n l r L e

+ +

− =

f(r,n,l,Z) g(r,Z)Exponential Decay g(r)

Note: Rnl0 as rinfinity

H-Atom Complete Ψ(r,θ,φ) for n=1,2σr/a0 F(r) only

Linear combinationOf two solutions is

Also a solution(linear operators!)

1S

2S

2pz

2px,y

F(r,θ)

F(r,θ,φ)

F(r) only

S-Orbitals (l=0,ml=0)” Rnl and Rnl2

Seems that maximum probability of finding the electron on the nucleus itself

100 ' / 21S N e ρ−Ψ = 200 '' / 2

2 (2 )S N e ρρ −Ψ = − 300 ''' 2 /33 (27 18 2 )S N e ρρ ρ −Ψ = − +

Surface plot of Ψ and Ψ2 for S

1S 2S

(1S)2 (2S)2

R2(r) predicts maximum probability at the center of the atom (for s)!!!

2 2

2

2 2 2

Probability

4 ( ) ( )

increasing 4 ( ) 0 4 0

nl

nl

of finding the electronanywhere in a shell of thicknessdr at radius r is r R r dr for Sr funtion

r R r dr as r dr

π

π π

→

→ →

Productof increasingfunction and decreasing function:MAXIMUM

Radial Distribution Functions: 4pr2Rnl2(r)

Number of Radial Nodesis always n-l-1

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

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

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

N=1, l=0

N=2, l=0

N=3, l=0

nS nSr r= Ψ Ψ

Shapes of orbitals dependon Orbital quantum number land Magnetic quantum no. ml

No f dependence: symmetric around z axis

Angular part of Wave Functions

/ 2210 (2 ) .coszp N e ρρ θ−Ψ =

/ 2

/ 2

(2 ) .sin .cos

(2 ) .sin .sinx

y

p N ep N e

ρ

ρ

ρ θ φ

ρ θ φ

−

−

Ψ =

Ψ =

Angular + Radial Electron Densities

Surface plot of Ψ and Ψ2 for 2p

2

2 2

2 2 /31

2 /32

2 /33

2 2 /34

2 2 /35

3 (3cos 1)

3 sin cos cos

3 sin cos sin

3 sin cos 2

3 sin sin 2

z

xz

yz

x y

xy

d N e

d N ed N e

d N e

d N e

ρ

ρ

ρ

ρ

ρ

ρ θ

ρ θ θ φ

ρ θ θ φ

ρ θ φ

ρ θ φ

−

−

−

−−

−

= −

=

=

=

=

d-orbitals: n=3, l=2, ml=-2,-1,0,1,2

Angular Part

Angular + Radial

Blue: -veYellow: +ve

Surface plot of Ψ and Ψ2 for 3d

f-orbitals: n=4,l=3,ml=-3,-2,-1,0,1,2,3

Cross Sections of Orbitals: Highly Symmetric Visual Treats

Radial parts not to scale!

How do they look from outside in 3D?

Radial parts not to scale!

How do they look from inside in 3D?

Radial parts not to scale!

Can you guess what orbital this is? What are the quantum numbers?

Probably a tough question to ask, but look into it carefully…

Why did we talk about all these complex orbitals when there is only one electron in H-atom?

Excited Electronic States!

Basis for all kinds of spectroscopy!

Multi electron systems: Orbitals are not that different!

Multi-Electronic Atoms

More than 1e: Can not be solved exactly!

He-atom (2e): 3-particle system!

+

1

2

r1

r2

r12=r1-r2

In atomic units, for convenience, often constants me,e’, 4πε0, h/2π are omitted

1 21 2 1 2 1 2 1 2

12

1( , ) ( , ) ( , ) ( , )HeH r r H r r H r r r rr

⇒ Ψ = Ψ + Ψ + Ψ

1 2

12

1electronicHeH H H

r= + +

Multi-electron atoms

Hamiltonian is no longer spherically symmetric

due to Σ (1/r12) term and therefore numerical

methods must be used to solve the TISE

Hi1 electron hydrogenic Hamiltonians for each particleSuch that HiΨi=EiΨi, where Ei are 1-electron energies

1 1 1 2 2 2

1 2

1( , , , , , ,...... , , ) ( , , ) .N N

N N N i i i ii i j ij

H r r r H r constr

θ φ θ φ θ φ θ φ= < =

= +∑ ∑

Inter-electronic interaction terms mess things up!

Orbital Approximation for N electrons

1 electron wavefunctions are termed orbitals!

Implies that e’-e’ interaction is neglected, as first approximation!!

TISE for He: 2 electron system

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

1 11 21 1 1 2 2 2

12

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

He He He He

e eHe

H r r E r r

H H r H rr

θ φ θ φ θ φ θ φ

θ φ θ φ

Ψ = Ψ

= + +

1 1 1 2 2 2 1 1 1 1 2 2 2 2( , , , , , ) ( , , ) ( , , ) ( 1 )He r r r r product of e wavefunctionsθ φ θ φ θ φ θ φΨ ≈ Ψ Ψ

Orbital Approximation (for He)

3 32 2

0 0

1 20 01 1 1 (1)1 (2).He

Zr Zra aZ Z s s

a ae e

π π

− −

Ψ ≈ ⇒ Ψ ≈

1 11 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2( , , ) ( , , ) ( , , ) ( , , )e e

HeH r r r rε θ φ θ φ θ φ ε θ φ Ψ = Ψ Ψ + Ψ Ψ

Hydrogenic orbitals since both electrons will be in the 1s orbital!

1 11 21 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 2 2

12

1 ( ) :

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

He

In order to solve for e orbital energies NEGLECT Electron Electron REPULSION

H H r r r H r r rr

θ φ θ φ θ φ θ φ

−

Ψ = Ψ Ψ + Ψ Ψ + Ψ Ψ

[ ] [ ]1 11 1 1 1 1 2 2 2 2 2 1 1 1 1 2 2 2 2( , , ) ( , , ) ( , , ) ( , , )e e

HeH r r r rε θ φ θ φ ε θ φ θ φΨ = Ψ Ψ + Ψ Ψ

( )[ ]1 1 1 11 2 1 1 1 1 2 2 2 2 1 2( , , ) ( , , )e e e e

He He HeH r r E Eε ε θ φ θ φ ε εΨ = + Ψ Ψ = Ψ ⇒ = +

1 11 21 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 2 2

12

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

HeH H r r r H r r rr

θ φ θ φ θ φ θ φ Ψ = Ψ Ψ + Ψ Ψ + Ψ Ψ

2 4 2

1 2 1 22 2 21 1

1 2

( ) 13.6 4( 13.6) 108.8

. : 24.6 54.4IE IE

Z e ZE Theory eV eV E E E E eVn n

Exp Ionization Energies E eV E eV

µ= − = − = − = ⇔ = + = −

= − = −

Most serious assumption of “orbital approximation” is drastic ! Electrons stay out of each other’s way by undergoing “correlated”

Motion to ensure electronic repulsion is minimum

Need to use other (numerical) approximate methods such as Variational Methodor Perturbation Theory without using orbital approximation

Effective Nuclear ChargeLet us find out what the ionization energies come out to be:

For He: Z=2, n=1 (GS), say a=a0

Electrons must be shielding each other!!!Outer electrons feel that the net charge on the nucleus

is less than what is expected: Screening of Z“Effective Nuclear Charge”

1 1 2 212

1 ( ) ( ) , !r r not small cannot be neglectedr

Ψ Ψ ⇒

+2

Electronic Shielding: Effective Nuclear Charge

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

several chemical properties of multi-electron systems

+2

3 32 2

0 0

1 20 01 (1)1 (2) .eff eff

He

eff effZ r Z ra aZ Z

s sa a

e e− −

Ψ ≈ ≈

σ

σ=

= −

−= ⋅

∑

1

eff

n

Hatomi i

Z Z

ZE E

n

See slides of Prof. Naresh Patwari (details provided)

σ= − =

= − ×

2

, , , , ...

( , ') ( 13.6 )

i

ieff

Hi

Z where i s p d f

Zavg per e IE eV

n

Multi-electronics: Consequences of Zeff

• How do we get 2p energy higher than 2S?• How does Radial distributions change? • How does Zeff affect atomic properties?

Read up on electronic

configuration of multi-

electronic atoms: Hund’s Rule

& Aufbau Principle

Quantization of Spatial (Orbital) Angular Momentum of Electrons in the atom

L=2ml+1

Na/Ag atoms: ns1: l=0: Zero Orbital Momentum For 1S electrons

If electrons are classical, "spinning" particles, then distribution of their spin angular momentum vectors is taken to be truly random and each particle wouldbe deflected up or down by a different amount, producing an even distribution.But electrons are deflected either up or down by a specific amount.

Intrinsic angular momentum or “Spin” of an electron is quantized

This can only mean that spin angular momentum is quantized (i.e., it can only take ondiscrete values). There is not a continuous distribution of possible angular momenta.

Electrons are spin-1⁄2 particles. These have only two possible spin angular momentum values, called “spin-up” (or α) and “spin-down” (or β)

The exact value in the z direction is ms= +ħ/2 or −ħ/2. If this value arises as a result of the rotating particles, then they would have to be spinning impossibly fast (GREATER THAN THE SPEED OF LIGHT).

Spin S(ω) where w is unknown (internal, if exists) coordinate (NON-CLASSICAL)

Implication of “Spin” on Hydrogen 1-e wavefunctions: Spin Orbitals

Introduction of spin results in the 4th quantum numberFor each electron: 3 spatial, and one for spin (4th coordinate)

Introduce a “spin” component to each of the 1e WFs along with spatial components. This doubles the degeneracy of each level little effect on energy.

1e WFs are now called SPIN ORBITALS.

2e Spin orbitals and Exclusion Principle

The complete wavefunction (both spatial and spin coordinates) of system of identical fermions (i.e. electrons) must be anti-symmetric with respect to interchange of all the coordinates (spatial and spin) of any two particles.

(Pauli’s?) Exclusion Principle: (by Dirac and Heisenberg)

Wavefunctions must reflect the indistinguishibility of electrons: No known experiments can distinguish between the last two functions, which clearly distinguishes between electrons – CAN NOT BE ALLOWED

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

Indistinguishibility

Exchange OperatorΨ = ±Ψ Symmetric:

Anti-symmetric:

Postulate 6: Complete wavefunctions describing a many electron system must be antisymmetric under the exchange of any two electrons.

For He atom ground state (both electron is 1s):

Space part is SYMMETRIC: 1s(1).1s(2) since both e in same orbital

We must therefore multiply this with an ANTI-SYMMETRIC Spin function

Spin Part:(1,2)(2,1)

6th LAW of Quantum Mechanics

Total approximate Ψ of He (GS): 1s(1)1s(2) * 2-1/2 [α(1)β(2)-β (1)α(2)]

Which is also anti-symmetric w.r.t exchange of electronic coordinates

Also, two electrons is 1s orbitals have OPPOSITTE Spins!!!

Ground and excited states of HeGround state: Both electrons are in same orbital (1s): 1s(1)1s(2)Symmetric

If they are in different orbitals: spatial part: 1s(1)2s(2) or 1s(2)2s(1)Both these are neither symmetric nor anti-symmetric.

Spatial Part can only be: 1s(1)2s(2) + 1s(2)2s(1) (SYMMETRIC)

or 1s(1)2s(2) - 1s(2)2s(1) (ANTI-SYMMETRIC)

Complete (spatial & spin) wavefunctions for an excited state of He(Note: Spatial*Spin part is antisymmetric)

Molecular Electronic Structureand Bonding in Diatomic and

Polyatomic Molecules

Lec 7 CH103 AC

Simplest Molecules: H2+ and H2

For H2 Molecule (2 Nuclei + 2 Electrons):

R

r1Ar1B

r2A r2B

r12

( ) ( )2 2 2 21 2

1 2 1 2 122H

1 1 1 1 1 1~ A B e eA A B B AB

Hr r r r R r

−∇ − ∇ + −∇ − ∇ + − − − − + +

( ) ( )2 2 21

1 12H

1 1 1~ A B eA B AB

Hr r R

+

−∇ − ∇ + −∇ + − − +

H+2 Molecule Ion (2 Nuclei + 1 Electron)

Born-Oppenheimer Approximation

+2

2H (1 )

1 1

1 1 1 1 1~ H e

e A AB e B AB eB

H Hr R r R r

−∇ − + − = + −

Nuclei are stationary wrt electronic motion

( ) ( )2 2 21

1 12H

1 1 1~ A B eA B AB

Hr r R

+

−∇ − ∇ + −∇ + − − +

CONSTANT

Model 1: Valence Bond Theory (VBT)1 electron in H1 2 electron in H2

Inclusion of ionic terms “Resonance”

Model 2: Molecular Orbital Theory

Both VBT and MOT are used frequently – each ofthem work good for certain systems, respectively,and often are not so good for other systems!

Molecular Orbital Theory: LCAO Linear Combination of Atomic Orbitals (LCAO)

1 1 2 1 1 1 2 2

2 2 2 2 21 1 2 2 1 2 1 22

A BMO Approx S S

MO

c c c c

c c c c

ϕ ϕ ψ ψ ψ ψ

ϕ ψ ψ ψ ψ

= = + = +

= + +2 2 2

1 2 1 2 1 22 2 2

2 1 1 2 1 2

, ( ) ( ) ~ 1 1

, ( ) ( ) ~ 1 1b b b b b A B

a a a a a A B

for c c c c c s sfor c c c c c s s

ϕ ψ ψ ϕ ψ ψ

ϕ ψ ψ ϕ ψ ψ

= = = + ⇒ = ± + +

= − = − = − ⇒ = ± − −

2 21 2 1 2: , symmetry c c c c= ⇒ = ±

Importance of Overlap Integral2 2

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

? !

1 ( ) ( )

b a

b b b b

How to find out c and c Normalization

c cϕ ϕ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ = = + + = + + +

( ) ( )1 2 1 21 1;

2 2 2 2b aS Sϕ ψ ψ ϕ ψ ψ= + = −

+ −

S can be positive, zero or negativeS is a qualitative measure of bond strength

1 2 2 11 1 where

2 2 2 2b ac and c S Overlap IntegralS S

ψ ψ ψ ψ= = = = ⇔+ −

1 1

( ) ( )1 1 1 1/

1 12 2 2 2A B A Bb b S S S Sb

E H HS S

ϕ ϕ ψ ψ ψ ψ+

= = + ++ +

1 1 1 1

/ /

Let

and 1 1

S S S Si i jiii ij

ii ij ii ijb a

ij ij

H H and H H

H H H HE E

S S

ψ ψ ψ ψ

+ −

= =

+ −= =

+ −

H2+Bonding and Anti-bonding Orbitals

( )

( )

1 1

1 1

12 2

12 2

A B

A B

a S S

b S S

S

S

ϕ ψ ψ

ϕ ψ ψ

= −−

= ++

1 1 1 1 1 1 1 1/

1 =2(1 ) A A B B A B B AS S S S S S S Sb

E H H H HS

ψ ψ ψ ψ ψ ψ ψ ψ+

+ + + +

(1 )1 1 1 1 1 1 1 1.

1 1( ) +( )B A B A B A B A

H eij BA S S S S S S S SAB eBconst

H or H H HR r

ψ ψ ψ ψ ψ ψ ψ ψ

= = −

(1 )1 1 1 1 1 1 1 1

.

1 1( ) +( )A A A A A A A A

H eii AA S S S S S S S SAB eBconst

H or H H HR r

ψ ψ ψ ψ ψ ψ ψ ψ

= = −

2(1 )

1 1 1 1 1~ H e

eA AB eB AB eB

H Hr R r R r

−∇ − + − = + −

1 (1 ) 1 11 1( ) where

A Aii AA BB S e S SAB eB

H or H H E J J Coulomb IntegralR r

ψ ψ= = + − = ⇒

1 1 (1 ) 1 1 1 1 11 1

B A B A B Aij S S e S S S S SAB eB

H ER r

ψ ψ ψ ψ ψ ψ= + −

Stabilization Energy due to Bond Formation

1 (1 ) 1 11, where

Re : .

B Aij S e S SAB eB

SH E S K KR r

K Exchange Integral or sonance IntegralLowering of energy that leads to bond formation

ψ ψ= + − =

⇒

K is purely quantum-mechanical concept

1

S

Interaction energy viewed as a -ve charge cloud on atoms with +ve charged nucleus

E1s(1e)

E1s(1e)*S

Bonding/antibonding energy depends on RAB

H2+

Represents the energies ofbonding and anti-bonding levels at equilibrium RAB

Electron Densities and Energy: f(RAB)

H2+ : MO, Electron Density and Energy

σ Bond formation with 1S-orbitals!

φ+

φ − φ−2

φ+2

Electron Density of Sigma Bonds

Note the signs, symmetries and nodes

LCAO-MO of p-orbitals

σ -bonding using p-orbitals

π-bonding involving p-orbitals

( ) ( ) ( )( ) ( ) ( )

:

( , , ) ( , , )

:

:

Inversion Operation I

I x y z x y z

r r r Symmetric

r r r Anti sym

⇒ + + + → − − −

ΙΨ = Ψ − = +Ψ

ΙΨ = Ψ − = −Ψ −

Symmetry and Nomenclature of MOsGerade (g) (symmetric) and Ungerade (u) (antisymmetric) wrt inversion operation!

Hydrogen molecule ion:

“b” Bonding using s: symmetricσg

“a” antibonding using s: antisymmetric σu*

Types of Covalent Bonds

•What are the symmetries?•Where are the nodes?•What are the relative energies?•How S changes with RAB? •What are the Ψ(MO)?

δ bonds

π bonds

σ bonds

Dihydrogen Molecule (H2)

R

r1Ar1B

r2A r2Br12

Place 2e with opp. spin in bonding orbital of H2+ [ ]1 1 1 1

2(1 )e

b A Bs sS

ϕ = ++

Can not be solved exactly Approximate

[ ]2

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

MOH A A B B A B B As s s s s s s s

Sϕ = + + +

+

[ ][ ]2

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

MOH A B A Bs s s s

Sϕ = + +

+Spatial part!

[ ] [ ]1(1) (2) (1) (2) (2) (1)2bonding b bϕ ψ ψ α β α β= • −

Space+Spin

2

2

2 2H 1 2

1 2 1 2 12

H (1 ) (1 )

1 2 12

1 1 1 1 1 1~

1 1 1 1A B B A AB

H e H e

B A AB

Hr r r r R r

H H Hr r R r

−∇ − + −∇ − − − + +

= + − − + +

MOT overemphasizes ionic terms in a covalent bond! VBT better?

Dihydrogen Molecule: One more electron goes to bonding orbital

Bond strength increases: Bond order=1

Effective nuclear charge changes the absolute Energy levels and the orbitals!

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

*

Energies of H2+, H2, He2

+, He2

He2+

He2

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

Electron Density Maps/Contours

H2 Li2: core 1σ

Li2: core 1σ∗ Li2: Valence 2σ

Li2: Total

MO Contours show electron density maps

O2 molecule

Total

2σ and 2σ∗

3σ and 1π

1π∗

Total Electron Density Mapsof 1st row Homonuclear Diatomics

Expected MO and Energies for DinitrogenAre these MO and correct energy level diagram for N2?

There is a problem! Spectroscopy says NO!

Actual MO and Energy Diagram for N2

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

Note how the MO of 2sσ have π-type looks, while π-levels are clean

Nature 2004 vol 432 867

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 effective nuclear charge.

S-P Mixing: B2 confirms it!

Boron is paramagnetic. This can only happen if the two electrons with parallel spin are in the π-orbitals π-bonding energies lower than σ*?

Incorrect!

MO Energy Diagram for F2

Less 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 Diagram (Homo- Diatomics)

Bond Order = ½ (# of bonding electrons - # of antibonding electrons)Bond order = 0 molecular can not exist – No bond formationBond order higher stronger bond, shorter bond length

Bond order, strength, length......magnetic properties

Paired Spins: DiamagneticUnpaired spins: Paramagnetic

Hetero-nuclear Diatomics: HF

Electronegativity of F much more than H (as Zeff more than H):

Electrostatic potentialscan be computed which gives a realistic picture of the electron densitiesin HF: “Egg” Shaped

CO: Cases of S-P Mixing of two atoms having different Zeff

Hybrids: Linear Combination of S and P leads to lowering of energy

•Hybridization is close to VBT approach. Use of experimental information•All hybridized orbitals are equevalent and are ortho-normal to each other

Linus Pauling, ~1930

Linear Environment: s & 1-p mix: sp

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

S and P orbital of the Same atom! This is notThe same as S (overlap)

2 equivalent hybrid orbitalsof the same energy and

shape (directions different)

Linear geometry with Hybridized atom at the center 2S- and 2P- (similar energy)

Mixes to form hybrid orbitalWhich forms a MO with H (1S)

2 more p-orbital available for bonding

Contours & bonding of sp-hybridization

Trigonal Environment: Mixing s & 2-p - sp2

Contribution from s=0.33; from p=0.66

1

2

2

1 2 0.33

1 1 13 2 6

1 1 13 2 6

x y

x y

x y

s p ptr

s p ptr

s p ptr

ψ ψ ψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ

= + +

= + −

= − −

30

30

• How do we calculate the coefficients?Use orthogonality of hybrid orbitals and normalization conditions

• There is no unique combination/solution

Hybridization of s & 3-p:sp3: TetrahedralContribution from s=0.25; from p=0.75

No p-orbital available

1

2

2

2

1 3 0. 0.2 21 2 1 0.2 3 2 31 1 1 12 6 2 2 31 1 1 12 6 2 2 3

y

tr

tr p

tr

tr

x y z

x z

x y z

x y z

s p p p

s p p

s p p p

s p p p

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ ψ ψ ψ

= + + +

= + + −

= − + −

= − − −

Hybridization of s,p,d: sp3d2 and sp3d

Similarly, sp3d trigonal bipyramidal

Sp3d2: octahedral

Visualizing Pentacene with AFM

AFM Image of Pentacene: Science August 30, 2009

Collaboration of engineers and scientistshas led to such fantastic discoveries