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Hydrogen Atom
M
m
r
Potential Energy2
( )4 4
e n
o o
q q ZeV r
r r
+
Kinetic Energy
ˆ ˆ ˆn eK K K
R
C2 2
2 2ˆ2 2R rKM m
22ˆ ( )
2 rK rm
M m m
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Hydrogen Atom2 2
2ˆ ˆ ˆ( ) ( )2 4r
o
ZeH K r V r
m r
22 2
2 2 2 2 2
1 1 1sin
sin sinr
d d d d dr
r dr dr r d d r d
22 2
2 2
1 1ˆ sinsin sin
d d dL
d d d
2 2 22
2 2
ˆˆ
2 2 4 o
d d L ZeH r
mr dr dr mr r
Radial Angular Coulombic
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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 22
. .2 2
( 1)( , ) ( ) ( , ) ( )
2 2l m n l m n
d d l lY r R r Y R r
mr dr dr mr
2
. .( , ) ( ) ( ) ( , )4l m n n n l m
o
ZeY R r E R r Y
r
2 2 22
2 2
( 1)( ) 0
2 2 4 n no
d d l l Zer E R r
mr dr dr mr r
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Hydrogen Atom
22
2 2 2 2
21 ( 1)( ) 0
2n
no
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
2
2
40.0529o
oa nmme
Bohr Radius
2
2 2 2
22 ( 1) 2( ) 0n
no
E md d l l ZR r
r dr dr r a r
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Hydrogen Atom
21
2 2 2
22 ( 1) 20r
o
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
o
E ml l Ze
r r a r
2 12 2
21 2 1( 1) 2 0
o
E mZl l
r a r
2 12
22( 1) 0; 2 0; & 0
o
E mZl l
a
It is a ground state as it has no nodes
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Hydrogen Atom
2 2
10; ; &2o
Zl E
a m
2 12
22( 1) 0; 2 0; & 0
o
E mZl l
a
22 2 2 2 2
1 2 202 2 4 o
Z Z meE
a m m
2 4
2 24 2o
Z me
0
1
0,( , , ) ( ) ( , ) ( )
Zr
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
2 2
1 202
ZE
a m
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Hydrogen Atom
In general:
Laguerre Polynomials
11
12
33
1 23
33
55
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
l L x
n l L x x x
l L x x
l L x
0
2Zrx
na
0
32 1
4 3 30 0 0
4 ( 1)! 2 2( )
[( )]
l Zr
nal ln n l
Z n l Zr ZrR r e L
n a n l na na
2 1
0
2ln l
ZrL
na
1S- 0 nodes
2S- 1 node
3S-2 nodes
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Energies of the Hydrogen Atom
In general:
4 2
2 2 2
1
24n
o
me ZE
n
2 2
20
1
24 o
e Z
a n
2
22
Z
n
2
0
27.24H
o
eE eV
a
Hartrees
kJ/mol
627.51 / 2625.5 /kcal mol kJ mol
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Wave functions of the Hydrogen Atom
In general:
0
32 1
4 3 30 0 0
4 ( 1)! 2 2( )
[( )]
l Zr
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
0
1
2C
11
0
21
rL
a
01 0,0( , , ) ( ) ( , )r R r Y
001 3
0
2( )
r
aR r ea
0,0
1( , )
2Y
0 0
3 30 0
2 1 1
2
r r
a ae ea a
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Z=1, n = 2, l = 0, and m = 0:
12
0 0
2! 2r r
La a
0,0
1( , )
2Y
02
300
122 2
r
ae r
aa
0202 3
00
1( ) 1
22
r
a rR r e
aa
02 0,0( , , ) ( ) ( , )r R r Y
Wave functions of the Hydrogen Atom
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Hydrogen AtomZ=1, n = 2, l = 1
022,1,0 3
0 0
1 2( , , ) cos
8
r
arr e
a a
m = 0: m = +1/-1:
022,1, 1 3
0 0
1 1( , , ) sin
8
r
arr e e
a a
022,1, 2,1, 1 2,1, 1 3
0 0
1 1 1( , , ) ( , , ) ( , , ) sin cos
2 8
r
ax
rr r r e
a a
022,1, 2,1, 1 2,1, 1 3
0 0
1 1 1( , , ) ( , , ) ( , , ) sin sin
2 8
r
ay
rr r r e
i a a
+
_
-+-+
+-
-+
+- -
+
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* 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 *
, ,
0 0
( ) ( ) ( ) ( , ) ( , )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
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Radial Distribution Functions0
22
30
4r
are
a
0* 0 21,0,0 1 1( ) ( ) ( )P r R r R r r00
1 30
2( )
r
aR r ea
0202 3
00
1( ) 1
22
r
a rR r e
aa
0* 0 22,0,0 2 2( ) ( ) ( )P r R r R r r
0 3 42
30 0 02 4
r
ae r rr
a a a
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* 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
X
Y
Z
Probability Distributions
0
42
2,1 1,0 50
( ) ( , ) cos sin32
r
arP r P e
a
, ( )n lP r, ( , )l mY
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Atomic Units
Set:2
2
41 . .o
oa a ume
2 4 2
2 22 2 24 2n
o
Z me ZE
nn
2 2 22
2 2
ˆˆ
2 2 4e e o
d d L ZeH r
m r dr dr m r r
2
0
1, 1, & 14
e
em
Hartrees
22
2 2
ˆ1
2 2
d d L Zr
r dr dr r r
2
2
Z
r
a.u.
Much simpler forms.
0
3 3
1 30
Zr
a Zrs
Z Ze e
a
023 3 2
2 300
1 12 22 22 2
r ra
s
Z e r Z e r
aa
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AtomsPotential Energy
2
( )4 4
e nen i
o i o i
q q ZeV r
r r
Kinetic Energy2 2
2 2ˆ2 2 iR r
i e
KM m
22ˆ ( )
2 ii ri i e
K rm
C
me
me
2
( )4 4
i jee
o ij o ij
q q eV r
r r
=r12
M
2
2i
i
i
Z
r
1
ijr
1
1( ) ( )en i ee ij
i i j i iji ij
ZV V r V r
r r
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Helium Atom
C
me
me
=r12
M
2 1ˆ ˆ ˆ2i
i i j ii ij
ZH K V
r r
2 1
2i
i i ji ij
Z
r r
,
1ˆi
i i i j ij
Hr
1 212
1ˆ ˆH Hr
Cannot be separated!!!
2
2i
ii
ZH
r
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?
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Ground State Energy of Helium Atom
Eo
E1
E2
I1 = 24.587 ev
Eo
E1
E2
I2 = 54.416 ev
Ionization Energy of He
EFree
Eo=- 24.587 - 54.416 ev
=- 79.003 ev =- 2.9033 Hartrees
Perturbation Theory1 2
12
1ˆ ˆ ˆH H Hr
01 2
ˆ ˆ ˆH H H 1
12
1H
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
1
20 11
12r Z
Hr
2 20 11 2 2
1
22
2 2 1
ZE
n
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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
1
22
ZE E
n
Not even close.Off by 1.1 H, or3000 kJ/mol
Therefore e’-e’ correlation, Vee, is very significant
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Ground State Energy of Helium Atom
0
12
1ˆ ˆH Hr
1
12
1H
r
0 0 01 2 1 2( , ) ( ) ( )r r r r
0 01 1 1 1 1 1
ˆ ( ) ( )H r E r
01 2
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
1 2
0 1 0 0* 01 2 1 2 1 2 1 2 1 2
12
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
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Ground State Energy of Helium Atom
30 (1 ) iZri
Zs e
1 2
1 0 1 0 0* 01 2 1 2 1 2 1 2 1 2
12
1ˆ, , , ,S S
E r r H r r r r r r dV dVr
1 2
0* 0* 0 01 2 1 2 1 2
12
1
S S
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
r
12
1 5 51 (1)1 (2) 1 (1)1 (2)
8 4
Zs s s s
r
0 1 54 2.75H
4E E E Closer but still far off!!!
1
0
1.2531.5%
4
E
E
Perturbation is too large for PT to be accurate, much higher corrections would be required
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Variational Method
0ˆ
exact exactH E
i ii
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:
Then
0
ˆt t
t t
HE
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
t
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Variational Method
0 0ˆ ˆ
t t i i j ji j
H E c H E c
*0
ˆi j i j
i j
c c H E *
0ˆ
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
i
c c E E *
00 & 0i i ic c E E
Since
0
ˆt t
t t
HE
Variational Energy
var
ˆ( ) ( )( )
( ) ( )t t
t t
HE
E0
Evar()
var ( ) 0d
Ed
min
2
var2( ) 0
dE
d
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Variational Method For He Atom3
1,0,01 ( ) ( ) iZri
Zs i e
r
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( , ) ( , )
( , ) ( , )
HE
r r r r
r r r r
33
1,0,01 ( ) ( ) eff ieff Z r
i
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
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Variational Method For He Atom
var 1 212
1ˆ ˆ1 (1)1 (2) 1 (1)1 (2)E s s H H s sr
1 2
12
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
1 2
12
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
1 212
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
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Variational Method For He Atom
31
3
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
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)effeff
Zs K s Z Z s s
r r
1 11
ˆ ˆ effZH K
r
2
1ˆ 1 (1) 1 (1)
2effZ
H s s 2
22eff
n
ZE
n
2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
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Variational Method For He Atom2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
3 31 1
3 322
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
r r
31
3 22
1 1 1 1 1
0 0 0
sineffZ reffZre dr d d
3123
1 1
0
4 effZ r
effZ re dr
200
1( 1)au auue du au e
a
23 0
2
14 2 1 2 0 1
2
effZ aeff eff eff
eff
Z Z e Z eZ
32
14 1
4eff effeff
Z ZZ
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Variational Method For He Atom2
11 1
1ˆ1 (1) 1 (1) 1 (1) 1 (1)2eff
eff
Zs H s Z Z s s
r
2
2eff
eff eff
ZZ Z Z
2
2ˆ1 (2) 1 (2)
2eff
eff eff
Zs H s Z Z Z
12
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
5
2 2 8eff eff
eff eff eff eff eff
Z ZZ Z Z Z Z Z Z
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Variational Method For He Atom
2 2var
52
8eff eff eff effE Z Z ZZ Z
2 52
8eff eff effZ ZZ Z
var
5 52 2 0
8 16eff effeff
dE Z Z Z Z
dZ
5 272
16 16effZ
2
var
27 27 5 272(2) 2.8479H
16 16 8 16E
Much closer to -2.9033 H
(E= 0.055 H = kJ/mol error)
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Variational Method For He Atom
271.69
16effZ 3 27
161 271 ( )
16
ir
s i e
1 2
3 27 27
16 161 2
1 27( , ) 1 (1)1 (2)
16
r rs s e e
r r
Optimized wavefunction
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Variational Method For He Atom
27
16effZ 3 27
161 271 ( )
16
ir
s i e
1 2
3 27 27
16 161 2
1 27( , ) 1 (1)1 (2)
16
r rs s e e
r r
Optimized wavefunction
1 2 1 2
3
2
1 2( , ) Z r Z r Z r Z rZ Ze e e e
r r
1.19 & 2.18Z Z var 2.8757HE
Other Trail Functions
(E= 0.027 H = kJ/mol error)
Optimizes both nuclear charges simultaneously
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Variational Method For He Atom
1 2( )1 2 12
1( , ) (1 )Z r re br
N r r
1.849 & 0.364Z b var 2.8920HE
Other Trail Functions
(E= 0.011 H = kJ/mol error)
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.
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Variational Method For He Atom
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
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The H2+ Molecule
2
( )4
i ine i
o i i
Z e ZV r
r 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.
2
( )4
A B A Bnn AB
o AB AB
Z Z e Z ZV R
R R
2 2 2 21 1 1 1
2 2 2 2A B A BR R r rA BM M
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The H2+ Molecule
1 1 1
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
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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
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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
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Linear Variational WFctns.
( ) ( )i ii
c r r
Suppose the trial wavefunction can be expressed in terms of an expansion of an appropriate set of functions, not necessarily othonormal
2 1ii
c 1
i j ijij
i jS
S i j
var
ˆ( ) ( )( )
( ) ( )
HE
r r
rr r
ˆ ˆi i j j i j i j
i j i j
i j i ji i j j i j
i j
c H c c c H
c cc c
i j iji j
i j iji j
c c H
c c S
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Linear Variational WFctns.
var i j ij i j iji j i j
E c c S c c H
var 0i
dE
dc
2i i j ij i j ijj ji i
d dE c c S c c H
dc dc
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
1 1
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Linear Variational WFctns.
11 12 12 1 1 1 1 1
21 21 22 2 2 2 2 2
1 1 2 2
1 1 2 2
i i j i j n i n i
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
0
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
1i j
ij
i j
S i j
S
Need to diagonalize matrix, to find eigenvalues and eigen vectors:
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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
A B
Hr r
ˆi i iH E
( ) 1 ( ) 1 ( )i iA A iB Bc s c s r r r
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Linear Combination of Atomic Orbitals.
2 21 1 1 1ˆ ˆ1 1 1 12 2A Bk l k r r l kl
A B
s H s s s Hr r
H
0i iE H S
* *i iA iBc c
11 1k l
lk
l ks s
S l k
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
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Linear Combination of Atomic Orbitals.
2 21 1 1 1ˆ1 1 1 1 12 2A Bkk k k k r r k
A B
H s H s s sr r
2
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
2Re
R R
AA BBH H
AB BAH H
AB BAS S
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Linear Combination of Atomic Orbitals.
0AA i AB i AB
AB i AB AA i
H E H E S
H E S H E
2 2
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
2( ) 4( ) 4(1 )
2(1 )
AA AB AB AA AB AB AB AA AB
AB
H H S H H S S H HE
S
2 2 22( ) 2 2
2(1 )(1 )AA AB AB AB AB AA AB AA AB
AB AB
H H S H H H S H S
S S
( ) ( )
(1 )(1 ) (1 )AA ABAA AB AB AB AA AB
AB AB AB
H HH H S H H S
S S S
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Prediction of the Bond
(1 )AA AB
AB
H HE
S
2
2
1 ( 1) ( 1) 1
21 1
3
R R
g
R
R e R R eE
RR R e
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Bonding and Antibonding Orbitals of H2+
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23_09fig_PChem.jpg
Density Difference Between MO’s and 1s O’s
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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)
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-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
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23_13fig_PChem.jpg
MO’s for the Diatomics
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23_02tbl_PChem.jpg
Energy Level Diagram For the Diatomics
Electron Configuration for H2 &He2
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Electron Configuration of N2
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23_16fig_PChem.jpg
Electron Configuration of F2
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23_18fig_PChem.jpg
Electron Configurations of the Diatomics
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23_20fig_PChem.jpg
Bonding in HF