CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.
Born-Oppenheimer Approximation
1. What is the main assumption of the Born-Oppenheimer approximation?
The Born-Oppenheimer approximation assumes that the nuclei are stationary, and
electron motion can be treated separately. Fixed bond distances and bond angles are assumed,
and a Hamiltonian operator is written for electronic motion only (Mortimer, p.648).
2. Using the Born-Oppenheimer approximation, write the Hamiltonian for the H2
molecule (2 electrons).
2 2
e2  1
1
1
1
1
1 


Hˆ  
1   22 





2m
40  R AB rA1 rA2 rB1 rB 2 r12 

e1

R12
RA1
e2
RB1
RB2
RA2
A
RAB
B
where rA1 is the distance between nucleus A and electron 1; rA2 is the distance between nucleus
A and electron 2; rB1 is the distance between nucleus B and electron 1; rB2 is the distance
between nucleus B and electron 2; r12 is the distance between electron 1 and electron 2; and RAB
is the distance between nucleus A and nucleus B (and remains constant by the BornOppenheimer approximation).
Variation Method; LCAO-MO
3. Determine the energy for H 2 in terms of Haa, Hab, and S using the variation method.
The secular determinant for this system is
H aa  ES aa
H ab  ES ab
H ba  ES ba
H bb  ES bb
0
In this case, H aa  H bb , H ab  H ba , S ab  S ba  S , S aa  S bb  1 , giving
H aa  E
H ab  ES
H ab  ES
H aa  E
0
H aa  E 2  H ab  ES 2  0
H aa  E   H ab  ES H aa  E   H ab  ES   0
H aa  E1   H ab  E1 S   0
E1  E1 S  H aa  H ab
E1 1  S   H aa  H ab
E1 
H aa  H ab
1 S
or
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CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.
H aa  E 2   H ab  E 2 S   0
E 2  E 2 S  H aa  H ab
E 2 1  S   H aa  H ab
E2 
H aa  H ab
1 S
Normalization; Hybrid Orbitals
4. Show that the sp2 hybrid orbital  sp 2 
s
 2 p is normalized if the s and p orbitals
3
are also normalized.
2
1
1
1
 * d 
s  2 p d 
s 2  2 p 2  2 2sp d  1  2  0  1
3
3
3
2
2
Note: We used  s d  1 ;  p d  1 ; and  spd  0 for normalized, orthogonal s and p





orbitals.
Electronic Configuration for Atoms & Molecules; Pauli Exclusion Principle
5. What are the electron configurations for H, Li+, O2, F, Na+, and Mg2+?
H  : 1s 2
Li  : 1s 2
O 2 : 1s 2 2s 2 2 p 6
F  : 1s  2s  2 p 
2
2
6
Na  : 1s  2s  2 p 
2
2
6
Mg 2 : 1s 2 2s 2 2 p 6
6. Which of the following transitions are allowed in the normal electronic emission
spectrum of an atom
(a) 2s to 1s
(b) 2p to 1s
(c) 3d to 2p
For a single-electron transition, l  1 and n = any integer.
(a) not allowed, l  0
(b) allowed, l  1
(c) allowed, l  1
7. Write the electronic configurations for N2, N2+, N2.
The configurations are
2
CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.
 2 σ u* 1s 2 σ g 2s 2 σ u* 2s 2  u 2 p 4 σ g 2 p 2
2
2
N 2 : σ g 1s 2 σ u* 1s  σ g 2s 2 σ u* 2s   u 2 p 4 σ g 2 p 1
2
2
1
N 2 : σ g 1s 2 σ u* 1s  σ g 2s 2 σ u* 2s   u 2 p 4 σ g 2 p 2  *g 2 p 
N 2 : σ g 1s
Term Symbols for Diatomics; Electronic Transitions
8. Determine the complete term symbol for each of the following electronic
configurations
1
(a) σ g 
(b) σ u 1
(c) σ u 2
(d)  u 1
(e)  u 3
1
(a) σ g  , the single unpaired electron gives a doublet state because S = 1/2 so that
1
2 S  1  2   1  2 . The symbol is 2  g
2
(b) σ u 1 , the term symbol is 2  u
(c) σ u 2 , the electrons must be paired. Therefore, S = 0 and the state is a singlet. The
product of two ungerade functions is gerade. The term symbol is 1  g
(d)  u 1 , the electron could be in either the     1 or the     1 state.
Therefore, there is a degenerate pair of states with   1 and the term symbol is
2
u .
(e)  u  , there are two possible configurations:
3
  2   1 for which
  1   2 for which
  1  1  1  1 and
  1  1  1  1 . Therefore   1 and the state is  . Also, the
function is ungerade because it is a product of three ungerade functions. The term symbol is
therefore 2  u .
9. Write the electronic configuration for Li2, and predict the term symbol for the ground
level.

The electronic configuration is σ g 1s
2 σ u* 1s 2 σ g 2s 2 . For this configuration, M L  0 .
Since   M L , we know that   0 and S = 0, which results in the symbol 1  . Because the
wave function for a σ molecular orbital does not change sign upon reflection across the xz plane,
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CHEM 342. Spring 2002. Problem Set #5. Mortimer Chapters 17, 18. Answers.
the + superscript is used. The parity can be found by multiplying the parities of the orbitals being
used, according to the laws of odd (u) and even (g) multiplication: g  g  g ; u  u  g ; u  g  u .
Therefore, g 4 u 2  g , and the complete term symbol is 1  g .
10. Which of the following electronic transitions are allowed?
(a) 1  u 1  g
(b) 1  u  3  g
(c) 1  g 1  u
The selection rules for electronic transitions in diatomic molecules are
  0,1
S  0
g u
   and    only;    not allowed
(a)
  0; S  0; g  u;  
The transition is allowed.
(b) The transition is forbidden because S  1 .
(c)
  1; S  0; g  u
The transition is allowed.
11. The ground-level term for a heteronuclear diatomic molecule is 3   . Write the term
symbols of the electronic transitions allowed for this molecule.
According to the transition rules (see the previous problem), the transition must be
   ; the superscript must be 3; and   0 or   1 . The allowed transitions are 3    3  
and 3   3  . Note that the notations g and u are not used for heteronuclear diatomic
molecules.
4