20_01fig_PChem.jpg Hydrogen Atom M m r Potential Energy + Kinetic Energy R C.

transcript

20_01fig_PChem.jpg

Hydrogen Atom

Potential Energy2

( )4 4

q q ZeV r

Kinetic Energy

ˆ ˆ ˆn eK K K

2 2ˆ2 2R rKM m

22ˆ ( )

2 rK rm

20_01fig_PChem.jpg

Hydrogen Atom2 2

2ˆ ˆ ˆ( ) ( )2 4r

ZeH K r V r

2 2 2 2 2

1 1 1sin

sin sinr

d d d d dr

r dr dr r d d r d

1 1ˆ sinsin sin

d d dL

2 2 22

2 2 4 o

d d L ZeH r

mr dr dr mr r

Radial Angular Coulombic

20_01fig_PChem.jpg

Hydrogen Atom( , , )r will be an eigenfunction of 2ˆ ˆ ˆ, & zH L L

.( , , ) ( ) ( , )n l mr R r Y Separable

ˆ ( , , ) ( , , )H r E r

2 2 22

. .2 2

ˆ( ) ( , ) ( ) ( , )

2 2 4 n l m n n l mo

d d L Zer R r Y E R r Y

mr dr dr mr r

. .2 2

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

2 2l m n l m n

d d l lY r R r Y R r

mr dr dr mr

. .( , ) ( ) ( ) ( , )4l m n n n l m

ZeY R r E R r Y

2 2 22

( 1)( ) 0

2 2 4 n no

d d l l Zer E R r

mr dr dr mr r

20_01fig_PChem.jpg

Hydrogen Atom

2 2 2 2

21 ( 1)( ) 0

E md d l l Zmer R r

r dr dr r r

2 22 2

2 2 2 2

1 1 22

d d d d d dr r r

r dr dr r dr dr r dr dr

Recall

40.0529o

oa nmme

Bohr Radius

22 ( 1) 2( ) 0n

E md d l l ZR r

r dr dr r a r

20_01fig_PChem.jpg

Hydrogen Atom

22 ( 1) 20r

E md d l l Ze

r dr dr r a r

Assume1( ) 0 asR r r

1( ) rR r e Let’s try

2 12 2

22 ( 1) 20r

E ml l Ze

r r a r

2 12 2

21 2 1( 1) 2 0

E mZl l

22( 1) 0; 2 0; & 0

E mZl l

It is a ground state as it has no nodes

20_01fig_PChem.jpg

Hydrogen Atom

10; ; &2o

22( 1) 0; 2 0; & 0

E mZl l

22 2 2 2 2

1 2 202 2 4 o

Z Z meE

2 24 2o

0,( , , ) ( ) ( , ) ( )

aln l mr R r Y CR r Ce

The ground state as it has no nodes n=1, and since l=0 and m = 0, the wavefunction will have no angular dependence

20_01fig_PChem.jpg

Hydrogen Atom

In general:

Laguerre Polynomials

1 0 ( ) 1

2 0 ( ) (2!)(2 )

1 ( ) (3!)

3 0 ( ) (3!) 3 3 0.5

1 ( ) (4!)(4 )

2 ( ) (5!)

n l L x

n l L x x

n l L x x x

l L x x

4 3 30 0 0

4 ( 1)! 2 2( )

nal ln n l

Z n l Zr ZrR r e L

n a n l na na

1S- 0 nodes

2S- 1 node

3S-2 nodes

Energies of the Hydrogen Atom

In general:

27.24H

Hartrees

kJ/mol

627.51 / 2625.5 /kcal mol kJ mol

Wave functions of the Hydrogen Atom

In general:

4 3 30 0 0

4 ( 1)! 2 2( )

nal ln n l

Z n l Zr ZrR r e L

n a n l na na

1( , ) (cos( ))

2mm im

l m l lY C P e

,( , , ) ( ) ( , )ln l mr R r Y

Z=1, n = 1, l = 0, and m = 0:

00 (cos( )) 1P 0

01 0,0( , , ) ( ) ( , )r R r Y

aR r ea

1( , )

3 30 0

a ae ea a

Z=1, n = 2, l = 0, and m = 0:

2! 2r r

1( , )

0202 3

1( ) 1

a rR r e

02 0,0( , , ) ( ) ( , )r R r Y

Wave functions of the Hydrogen Atom

Hydrogen AtomZ=1, n = 2, l = 1

022,1,0 3

1 2( , , ) cos

m = 0: m = +1/-1:

022,1, 1 3

1 1( , , ) sin

arr e e

022,1, 2,1, 1 2,1, 1 3

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

rr r r e

022,1, 2,1, 1 2,1, 1 3

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

rr r r e

20_06fig_PChem.jpg

* 2, , , ,( ) ( , , ) ( , , ) sinn l m n l mP R r r r drd d

* * 2, ,( ) ( , ) ( ) ( , ) sinl l

n l m n l mR r Y R r Y r drd d * 2 *

, ,( ) ( ) ( , ) ( , )sinl ln n l m l mR r R r r dr Y Y d d

For radial distribution functions we integrate over all angles only

2* 2 *

( ) ( ) ( ) ( , ) ( , )sinl ln n l m l mP r R r R r r Y Y d d

* 2( ) ( )l ln nR r R r r Prob. density as a function of r.

Radial Distribution Functions

20_09fig_PChem.jpg

Radial Distribution Functions0

0* 0 21,0,0 1 1( ) ( ) ( )P r R r R r r00

aR r ea

0202 3

1( ) 1

a rR r e

0* 0 22,0,0 2 2( ) ( ) ( )P r R r R r r

0 3 42

30 0 02 4

ae r rr

20_08fig_PChem.jpg

* 2, , , ,( , , ) ( , , ) ( , , ) sinn l m n l mP r r r r

* 2 *, , , ,( ) ( ) ( , ) ( , )sinl l

n n l m l mR r R r r Y Y

, ,( ) ( , )n l l mP r P

Probability Distributions

2,1 1,0 50

( ) ( , ) cos sin32

arP r P e

, ( )n lP r, ( , )l mY

20_12fig_PChem.jpg

Atomic Units

41 . .o

oa a ume

2 22 2 24 2n

Z me ZE

2 2 22

2 2 4e e o

d d L ZeH r

m r dr dr m r r

1, 1, & 14

Hartrees

d d L Zr

r dr dr r r

Much simpler forms.

Z Ze e

023 3 2

1 12 22 22 2

Z e r Z e r

AtomsPotential Energy

( )4 4

e nen i

o i o i

q q ZeV r

Kinetic Energy2 2

2 2ˆ2 2 iR r

22ˆ ( )

2 ii ri i e

( )4 4

o ij o ij

q q eV r

1( ) ( )en i ee ij

i i j i iji ij

ZV V r V r

Helium Atom

2 1ˆ ˆ ˆ2i

i i j ii ij

ZH K V

i i ji ij

i i i j ij

1ˆ ˆH Hr

Cannot be separated!!!

Hydrogen like 1 e’ Hamiltonian

i.e. r12 cannot be expressed as a function of just r1 or just r2

What kind of approximations can be made?

Ground State Energy of Helium Atom

I1 = 24.587 ev

I2 = 54.416 ev

Ionization Energy of He

Eo=- 24.587 - 54.416 ev

=- 79.003 ev =- 2.9033 Hartrees

Perturbation Theory1 2

1ˆ ˆ ˆH H Hr

ˆ ˆ ˆH H H 1

r 0 0 0

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

0 0 01 1 1 1 1 1

ˆ ( ) ( )H r E r 101

1(1 ) rs e

2 20 11 2 2

Ground State Energy of Helium Atom0

1 2ˆ ˆ ˆH H H 0 0 0

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

0 0 0 01 2 1 2 1 2

ˆ ˆ ˆ( , ) ( ) ( )H r r H H r r 0 0 0 0

1 1 2 2 1 2ˆ ˆ( ) ( ) ( ) ( )H r r H r r

0 0 0 02 1 1 1 2 1

ˆ ˆ( ) ( ) ( ) ( )r H r r H r

0 0 0 02 1 1 1 2 1( ) ( ) ( ) ( )r E r r E r

1 2 0 00 0 2 1( ) ( )E E r r

0 01 2( , )E r r

0 0 01 2 2 2 4E E E H

20 011 22

Not even close.Off by 1.1 H, or3000 kJ/mol

Therefore e’-e’ correlation, Vee, is very significant

1ˆ ˆH Hr

0 0 01 2 1 2( , ) ( ) ( )r r r r

0 01 1 1 1 1 1

ˆ ( ) ( )H r E r

ˆ ˆ ˆH H H

0 1 0 0 0 1 01 2 1 2 1 2

ˆ, ,E E E E E r r H r r

0 1 0 0* 01 2 1 2 1 2 1 2 1 2

1ˆ, , , ,S S

r r H r r r r r r dV dVr

1 2 1 2ˆ ( , ) ( , )H r r E r r 0 1

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

30 (1 ) iZri

1 0 1 0 0* 01 2 1 2 1 2 1 2 1 2

1ˆ, , , ,S S

E r r H r r r r r r dV dVr

0* 0* 0 01 2 1 2 1 2

r r r r dV dVr

1 2 1 2

2 22 2

1 2 1 2 1 2 1 2 1 2120 0 0 0 0 0

2 2 2 2 1 2 2 2 2sin sinZr Zr Zr Zre e e e r r dr dr d d d d

1 5 51 (1)1 (2) 1 (1)1 (2)

Zs s s s

0 1 54 2.75H

4E E E Closer but still far off!!!

1.2531.5%

Perturbation is too large for PT to be accurate, much higher corrections would be required

Variational Method

exact exactH E

c i i iH E

The wavefunction can be optimized to the system to make it more suitable

Consider a trail wavefunction t and exact

Is the true wavefunction, where:

The exact energy is a lower bound

,n exactis a complete set

Assume the trial function can be expressed in terms of the exact functions

0 0ˆ ˆ 0t t t t t tH E H E

We need to show that

texact

Variational Method

0 0ˆ ˆ

t t i i j ji j

H E c H E c

ˆi j i j

c c H E *

i j i j i ji j

c c H E *

0i j i ij iji j

c c E E

*0 0i i i

c c E E *

00 & 0i i ic c E E

Variational Energy

ˆ( ) ( )( )

( ) ( )t t

Evar()

var ( ) 0d

var2( ) 0

Variational Method For He Atom3

1,0,01 ( ) ( ) iZri

Zs i e

Let’s optimize the value of Z, since the presence of a second electrons shields the nucleus, effectively lowering its charge.

1 2 1 2var

1 2 1 2

ˆ( , ) ( , )

( , ) ( , )

r r r r

1,0,01 ( ) ( ) eff ieff Z r

Zs i e

r 1 2( , ) 1 (1)1 (2)s s r r

1 2 1 2( , ) ( , ) 1 (1) 1 (1) 1 (2) 1 (2) 1s s s s r r r r

var 1 212

1ˆ ˆ1 (1)1 (2) 1 (1)1 (2)E s s H H s sr

Variational Method For He Atom

var 1 212

1ˆ ˆ1 (1)1 (2) 1 (1)1 (2)E s s H H s sr

ˆ ˆ1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2)

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

s s H s s s s H s s

s s s sr

ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (2) 1 (2) 1 (1) 1 (1)

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

s H s s s s H s s s

s s s sr

1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)s H s s H s s s s sr

1 (1) effeff Z rZs e

var 1 212

1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)E s H s s H s s s s sr

1 1 11 1 1 1

ˆ ˆ ˆ1 (1) 1 (1) 1 (1) 1 (1) 1 (1) 1 (1)eff effZ ZZ Zs H s s K s s K s

r r r r

1ˆ1 (1) 1 (1) 1 (1) 1 (1)effeff

Zs K s Z Z s s

ˆ ˆ effZH K

1ˆ 1 (1) 1 (1)

H s s 2

1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff

Zs H s Z Z s s

1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff

Zs H s Z Z s s

3 31 1

1 1 1 1 11 10 0 0

1 11 (1) 1 (1) sineff effeff effZ r Z rZ Zs s e e r drd d

1 1 1 1 1

sineffZ reffZre dr d d

4 effZ r

effZ re dr

1( 1)au auue du au e

14 2 1 2 0 1

effZ aeff eff eff

Z Z e Z eZ

4eff effeff

1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff

Zs H s Z Z s s

eff eff

ZZ Z Z

2ˆ1 (2) 1 (2)

eff eff

Zs H s Z Z Z

1 51 (1)2 (2) 1 (1)2 (2)

8 effs s s s Zr

Similarly

Recall from PT

var 1 212

1ˆ ˆ1 (1) 1 (1) 1 (2) 1 (2) 1 (1)1 (2) 1 (1)1 (2)E s H s s H s s s s sr

2 2 8eff eff

eff eff eff eff eff

Z ZZ Z Z Z Z Z Z

2 2var

8eff eff eff effE Z Z ZZ Z

8eff eff effZ ZZ Z

5 52 2 0

8 16eff effeff

dE Z Z Z Z

16 16effZ

27 27 5 272(2) 2.8479H

16 16 8 16E

Much closer to -2.9033 H

(E= 0.055 H = kJ/mol error)

271.69

16effZ 3 27

161 271 ( )

3 27 27

16 161 2

1 27( , ) 1 (1)1 (2)

r rs s e e

Optimized wavefunction

16effZ 3 27

161 271 ( )

3 27 27

16 161 2

1 27( , ) 1 (1)1 (2)

r rs s e e

Optimized wavefunction

1 2 1 2

1 2( , ) Z r Z r Z r Z rZ Ze e e e

1.19 & 2.18Z Z var 2.8757HE

Other Trail Functions

Optimizes both nuclear charges simultaneously

1 2( )1 2 12

1( , ) (1 )Z r re br

1.849 & 0.364Z b var 2.8920HE

Other Trail Functions

Z’, b are optimized. Accounts for dependence on r12.

In M.O. calculations the wavefunction used are designed to give the most accurate energies for the least computational effort required.

The more accurate the energy the more parameters that must be optimizedthe more demanding the calculation.

In M.O. calculations the wavefunction used are designed to give the most accurate energies for the least computational effort required.

The more accurate the energy the more parameters that must be optimizedthe more demanding the calculation.

-2.862879 H

-2.862871 H

-2.84885 H

Experimental -79.003 ev -2.9003 H

The H2+ Molecule

i ine i

Z e ZV r

2 2 2 22 2 2 2ˆ ˆ ˆ

2 2 2 2A B A Bnuclear electronic R R r rA B e e

K K KM M m m

One electron problem

Two nuclei

Define electron position, i..e. internal coordinates, w.r.t. nuclear positions.

A B A Bnn AB

o AB AB

Z Z e Z ZV R

2 2 2 21 1 1 1

2 2 2 2A B A BR R r rA BM M

The H2+ Molecule

AB A BR r r

2 2 2 21 1 1 1

2 2 2 2A B A BR R r rA BM M

ˆ ˆ ˆH K V

ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( )N A N B e A e BK K R K R K r K r

ˆ ˆ ˆ ˆ( ) ( ) ( )nn AB en A en BV V R V r V r

Since ZA=1 and ZB=1

2 2 2 21 1 1 1 1 1 1ˆ2 2 2 2A B A BR R r r

A B AB A B

HM M R r r

The H2+ Molecule

2 2 2 21 1 1 1 1 1 1ˆ2 2 2 2A B A BR R r r

A B AB A B

HM M R r r

Nuclear Electronic

( , ) ( ; ) ( )R r R r R

The nuclear positions determine the electronic wavefunction

Assume electronic motion is much faster than nuclear motion, implies that the nuclear positions are essentially static

ˆ ˆ ˆ( ) ( ; )N eH H H R R rThe electronic part is determined by the nuclear positions

Separable??

ˆ ( , ) ( , )H R r W R r W- Total Energy

The H2+ Molecule

ˆ ˆ ˆ( , ) ( ) ( ; ) ( ; ) ( ) ( ; ) ( )N e TH R r H H R R E R R r r R r R

ˆ ˆ ˆ ˆ ( ; ) ( ) ( ; ) ( )N nn e ne TK V K V R E R r R r R

ˆ ˆ ( ) ( )N nn NK V E R R

ˆ ˆ ( ; ) ( ; )e ne eK V R E R r r

Potential energy surface.

Of primary interest

Nuclear

Electronic

ˆ ˆ ( ) ( )N nn e TK V E E R R

Linear Variational WFctns.

( ) ( )i ii

Suppose the trial wavefunction can be expressed in terms of an expansion of an appropriate set of functions, not necessarily othonormal

i j ijij

ˆ( ) ( )( )

( ) ( )

ˆ ˆi i j j i j i j

i j i j

i j i ji i j j i j

c H c c c H

c cc c

i j iji j

var i j ij i j iji j i j

E c c S c c H

var 0i

2i i j ij i j ijj ji i

d dE c c S c c H

2i i j ij i j ijj j

E c c S c c H

For each ci

Find the optimum coefficients, that minimize Evar.

ii j ij i i j ij i j ij

j j ji i i

dE d dc c S E c c S c c H

dc dc dc

0j ij j ij ij

c H c S E

11 12 12 1 1 1 1 1

21 21 22 2 2 2 2 2

1 1 2 2

i i j i j n i n i

j i j j i j jj i jn i jn ij

n i n n i n nj i nj nn i in

H E H E S H E S H E S c

H E S H E H E S H E S c

H E S H E S H E H E S c

H E S H E S H E S H E c

ˆ ˆi jH H

* * * *1 2

ˆ wherei i i i i i ij inH E c c c c

0j ij ij ij

c H S E 0i iE H S

Need to diagonalize matrix, to find eigenvalues and eigen vectors:

Linear Combination of Atomic Orbitals.Lets use the 1s Hydrogen like orbitals to be a basis for a trial function and apply variational theory to find the best approximate wavefunction

( ) 1 ( ) 1 ( )i iA A iB Bc s c s r r r

1 ( ) & 1 ( )A Bs sr rWhere are Hydrogen like wavefunction with n=1, l=0, centred in nucleus a and b, resp.

2 21 1 1 1ˆ2 2A Br r

ˆi i iH E

( ) 1 ( ) 1 ( )i iA A iB Bc s c s r r r

Linear Combination of Atomic Orbitals.

2 21 1 1 1ˆ ˆ1 1 1 12 2A Bk l k r r l kl

s H s s s Hr r

0i iE H S

* *i iA iBc c

11 1k l

l ks s

0AA i AB i AB iA

BA i BA BB i iB

H E H E S c

H E S H E c

ˆ 1 1 1 1iA A iB B i iA A iB BH c s c s E c s c s

2 21 1 1 1ˆ1 1 1 1 12 2A Bkk k k k r r k

H s H s s sr r

1 1 1 1 13

Rkl k l l k lk

RS s s s s S e R

0AA i AB i AB iA

AB i AB AA i iB

H E H E S c

H E S H E c

1ˆ ˆ1 1 1 1 12

Rkl k l kl l kH s H s S R e s H s

21 1 11

AA BBH H

AB BAH H

AB BAS S

0AA i AB i AB

AB i AB AA i

H E H E S

H E S H E

0AA i AB i ABH E H E S

2 2 2 2(1 ) 2 ( ) 0i AB i AB AB AA AA ABE S E H S H H H

2 2 2 2

2( ) 4( ) 4(1 )

AA AB AB AA AB AB AB AA AB

H H S H H S S H HE

2 2 22( ) 2 2

2(1 )(1 )AA AB AB AB AB AA AB AA AB

H H S H H H S H S

( ) ( )

(1 )(1 ) (1 )AA ABAA AB AB AB AA AB

AB AB AB

H HH H S H H S

Prediction of the Bond

(1 )AA AB

1 ( 1) ( 1) 1

R e R R eE

RR R e

Bonding and Antibonding Orbitals of H2+

23_09fig_PChem.jpg

Density Difference Between MO’s and 1s O’s

23_11fig_PChem.jpg

Electron Densities of Sigma and Pi M.O’.s

1 1( ) 1 ( ) 1 ( )

2 2g A Bs s r r r* 1 1( ) 1 ( ) 1 ( )

2 2u A Bs s r r r

1 1( ) 2 ( ) 2 ( )

2 2g x A x Bp p r r r , ,

1 1( ) 2 ( ) 2 ( )

2 2g x A x Bp p r r r

Bonding

BondingAntibonding

Antibonding

g=gerade (same) u=ungerade

(opposite)

-13.6 e.v.

-19.6 e.v.

-18.6 e.v.

Electron population on F is larger, ie. bond in polarized to F, ie. shows the F is more electronegative.

(0.345)1 (0.840)2H zFs p

Other Types of M.O.’s

23_13fig_PChem.jpg

MO’s for the Diatomics

23_02tbl_PChem.jpg

Energy Level Diagram For the Diatomics

Electron Configuration for H2 &He2

23_17fig_PChem.jpg

Electron Configuration of N2

23_16fig_PChem.jpg

Electron Configuration of F2

23_18fig_PChem.jpg

Electron Configurations of the Diatomics

23_20fig_PChem.jpg

Bonding in HF

20_01fig_PChem.jpg Hydrogen Atom M m r Potential Energy + Kinetic Energy R C.

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