CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
Valence Bond Method; Slater Determinants
1. Use the following spin functions
1  12
 2  12
2
12  12
2
2
12  12
4 
2
to write the four complete wave functions for H2 (including the spin component) using the
1
following Heitler-London valence bond wave functions   
 a 1 b 2   a 2 b 1 and
2
1
 a 1 b 2   a 2 b 1. What are the degeneracies of the energy levels described
 
2
by these wave functions? Hint: Your wave functions for H2 should be in terms of
 a 1;  a 2;  b 1;  b 2; 1; 2; 1; and/or 2.
1
The wave function   
 a 1 b 2   a 2 b 1 is symmetric with respect to the
2
exchange of electrons as shown by operating on it with the permutator operator as follows:
1
 a 1 b 2   a 2 b 1  1  a 2 b 1   a 1 b 2   
Pˆ 1,2   Pˆ 1,2
2
2
According to the Pauli exclusion principle, this symmetric wave function must be used
with an anti-symmetric spin function, which results in the following wave function.
1
 a 1 b 2   a 2 b 1 2 12  12
1 
2
2
1
1   a 1 b 2   a 2 b 112  12
2
1
Similarly, the wave function   
 a 1 b 2   a 2 b 1 can be shown to be
2
antisymmetric and must be used with a symmetric spin function as follows:
1
 a 1 b 2   a 2 b 112
2 
2
1
 a 1 b 2   a 2 b 112
3 
2
1
 4   a 1 b 2   a 2 b 112  12
2
There is only one spin state for   , resulting in a degeneracy of 1; while there are 3 spin
states for   , resulting in a degeneracy of 3.
3 
1
CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
2. Express as Slater determinants the 4 valence bond wave functions for H2 (found in the
previous problem).
1  a 11  a 2 2 
1  a 11  a 2 2 
1 

2  b 11  b 2 2 
2  b 11  b 2 2 
2 
3 
4 
1  a 11  a 2 2 
2  b 11  b 2 2 
1  a 11  a 22 
2  b 11  b 2 2
1  a 11  a 2 2 
1  a 11  a 2 2 

2  b 11  b 22
2  b 11  b 2 2 
Dipole Moments
3. Consider the ClF molecule to consist of two ions of opposite charge separated by the
bond length of 0.163 nm. Calculate the dipole moment for this model.
  Qr



  1.6022 10 19 C 1.63 10 10 m  2.6110  29 C m
Electronegativity
4. The Cl-F, Cl-Cl, and F-F bond energies are 255 kJ/mol, 243 kJ/mol, 159 kJ/mol,
respectively. Using the Pauling definition of electronegativity (Motimer, Chapter 18), calculate
the difference in electronegativities of Cl and F.
1
E ClF  E ClF  E ClCl  E FF 
2
1
E ClF  255 kJ mol 1  243 kJ mol 1  159 kJ mol 1
2
1
E ClF  54 kJ mol



EN F  EN Cl  0.102 mol 1 / 2 kJ 1 / 2

54 kJ mol 1
EN F  EN Cl  0.75
5. An alternative method for calculating electronegativities is the Mulliken definition
1
given by EN i  IE i  EAi  , where EN is the electronegativity, IE is the ionization energy, and
2
EA is the electron affinity. The ionization of Cl is 1251.1 kJ/mol and its electron affinity is
 348.7 kJ/mol. Using the Mulliken definition, calculate the electronegativity of Cl.
1
EN i  1251.1 kJ mol 1   348.7 kJ mol 1
2
EN i  451.2 kJ mol 1



Huckel Method; Secular Determinants
6. What are the 3 approximations of the Huckel method?
2
CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
(a) The Hamiltonian for the  electrons is assumed to be separate from that for the σ
electrons. Therefore, the treatment of  electrons is independent of the treatment of σ
electrons. (See Mortimer, p. 702).
(b) All overlap integrals equal zero.
(c) All resonance integrals between non-adjacent atoms equal zero. All resonance
integrals between adjacent atoms equal .
7. Using the Huckel approximations, write the secular determinant for hexatriene. If we
E
were to expand this determinant, we would obtain x 6  5x 4  6 x 2  1 where x 
. The

 j 
 where j
general solutions for x for this type of determinant are given by x j  2 cos
 nC  1 
labels different energy states with j  1,2,...nC ; while nC is the number of carbons in the
molecule. Find expressions for each of the six energy states of hexatriene in terms of  and  .
The secular determinant for hexatriene is
E

0
0
0
0
xj 

0
0
0
E

0
0

E

0
0

E

0
0

E
0
0
0

0
0
0
0

E
E

E j    x j
 j 

 E j    2 cos
 j 

n

1
x j  2 cos
 C

 nC  1 
For hexatriene, nC  6 and we obtain
3
CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
 1 
E1    2 cos 
    1.802
 6  1
 2 
E 2    2 cos 
    1.247
 6  1
 3 
E 3    2 cos 
    0.445
 6  1
 4 
E 4    2 cos 
    0.445
 6  1
 5 
E 5    2 cos 
    1.247
 6  1
 6 
E 6    2 cos 
    1.802
 6  1
8. Trial wave functions for hexatriene are constructed by taking a linear combination of
the basis functions  as follows  r  c1,r 1  c 2,r  2  c 3,r  3  c 4,r  4  c 5,r  5  c 6,r  6 where r =
1,2,3,4,5,6. The coefficients c j , r can be determined from the equation
c j ,r
 2 

 
 nC  1 
1/ 2
sin
jr
nC  1
where j = 1,2,3,4,5,6.
Write the Huckel molecular orbital wave functions 1 and  4 for hexatriene in terms of
1 ,  2 ,  3 ,  4 ,  5 ,  6 .
For 1 , we have
c jr
 2 

 
 nC  1 
 2 
c1,1  

 6 1
1/ 2
sin
sin
11  0.232
sin
21  0.418
1/ 2
 2 
c 2,1  

 6 1
jr
nC  1
1/ 2
6 1
6 1
31  0.521
 2 
c 3,1  
 sin
6 1
 6 1
Similarly, we find that c 4,1  0.521; c 5,1  0.418; c 6,1  0.232; which gives the wave
1/ 2
function 1  0.2321  0.481 2  0.521 3  0.521 4  0.481 5  0.232 6 .
For  4 , we have
4
CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
 2 
c1, 4  

 6  1
sin
14  0.521
sin
24  0.232
1/ 2
 2 
c 2, 4  

 6  1
1/ 2
6 1
6 1
34  0.418
 2 
c 3, 4  
 sin
6 1
 6  1
Similarly, we find that c 4, 4  0.418; c5, 4  0.232; c 6, 4  0.521; which gives the wave
1/ 2
function  4  0.5211  0.232 2  0.418 3  0.418 4  0.232 5  0.521 6 .
Although this problem does not ask you to find all the wave functions, all 6 trial wave
functions of hexatriene can be found by this method. They are as follows:
 2  0.4181  0.521 2  0.232 3  0.232 4  0.521 5  0.418 6
 3  0.5211  0.232 2  0.418 3  0.418 4  0.232 5  0.521 6
 5  0.4181  0.521 2  0.232 3  0.232 4  0.521 5  0.418 6
 6  0.2321  0.481 2  0.521 3  0.521 4  0.481 5  0.232 6
9. The 6  electrons of hexatriene are placed in the bonding orbitals 1 ,  2 , and  3 (the
first three energy states). Determine the total   electron energy and the delocalization energy
for hexatriene. Express both answers in terms of  and/or  . Hint: The delocalization energy is
the energy by which the molecule is stabilized relative to isolated double bonds. The total
  electron energy for one molecule of ethylene is 2  2 .
E   2E1  2E 2  2 E3  2  1.802  2  1.247  2  0.445  6  6.988
DE  E  hexatriene   3E  ethylene 
DE  6  6.988  32  2
DE  0.988
Note: Delocalization energy was abreviated DE.
10. Using the Huckel approximations, write the secular determinant for butadiene, and
determine the energy states.
The secular determinant for butadiene is
E

0

E

0

E
0
0

0
0

E
By expanding, we obtain the following
5
CHEM 342. Spring 2002. Problem Set #6. Mortimer Chapter 18; Appendix H. Answers.
E

0
0
E

0


0

E

0
   E  
E
  0   E
 
0

E

0

E
0

E
0
0

E
   E 2
E



E

0

   E 
 2
 2


E
0 E

E
0 E
   E 4    E 2  2    E 2  2    E 2  2   4 
   E 4  3  E 2  2   4  0
For x 
  E 2
, the expanded determinant has the form x 2  3x  1  0 . The roots are

x  2.62 and 0.38. Therefore,
2

  E
x
2
2
x 2    E 
2
E    x 2
E    x
The energies of the four LCAO-MOs are
E1    1.62
E 2    1.62
E 3    0.62
E 4    0.62
11. Using the Huckel approximations, write the secular determinant for cyclobutadiene.
(Please note that there is no need to determine the energy states for this problem.)
For four electrons, the secular determinant is
E

0

E

0

E

0


0

E
Note the two extra  terms in this secular determinant compared to those written for
noncyclic butadiene (previous problem).
6