CHEM 342. Spring 2002.  PS#7 Answers   PS#5 Answers    PS#6 Questions    PS#6 Answers.doc (Word 97)  

  

Problem Set #6 Answers

 

Note: Relevant Sections in Mortimer are Chapter 18 and Appendix H.

Valence Bond Method; Slater Determinants

1. Use the following spin functions

to write the four complete wave functions for H2 (including the spin component) using the following Heitler-London valence bond wave functions and . What are the degeneracies of the energy levels described by these wave functions? Hint: Your wave functions for H2 should be in terms of

The wave function is symmetric with respect to the exchange of electrons as shown by operating on it with the permutator operator as follows:

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.

Similarly, the wave function can be shown to be antisymmetric and must be used with a symmetric spin function as follows:

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.

2. Express as Slater determinants the 4 valence bond wave functions for H2 (found in the previous problem).

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.

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.

5. An alternative method for calculating electronegativities is the Mulliken definition given by , where EN is the electronegativity, IE is the ionization energy, and EA is the electron affinity. The ionization of Cl is 1251.1 kJ/mol and its electron affinity is kJ/mol. Using the Mulliken definition, calculate the electronegativity of Cl.

Huckel Method; Secular Determinants

6. What are the 3 approximations of the Huckel method?

    1. The Hamiltonian for the electrons is assumed to be separate from that for the electrons. Therefore, the treatment of electrons is independent of the treatment of electrons. (See Mortimer, p. 702).
    2. All overlap integrals equal zero.
    3. All resonance integrals between non-adjacent atoms equal zero. All resonance integrals between adjacent atoms equal b .

7. Using the Huckel approximations, write the secular determinant for hexatriene. If we were to expand this determinant, we would obtain where . The general solutions for x for this type of determinant are given by where j labels different energy states with ; while 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

For hexatriene, and we obtain

8. Trial wave functions for hexatriene are constructed by taking a linear combination of the basis functions as follows where r = 1,2,3,4,5,6. The coefficients can be determined from the equation

where j = 1,2,3,4,5,6.

Write the Huckel molecular orbital wave functions for hexatriene in terms of

For , we have

Similarly, we find that which gives the wave function .

For , we have

Similarly, we find that which gives the wave function .

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:

9. The 6 electrons of hexatriene are placed in the bonding orbitals (the first three energy states). Determine the total 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 energy for one molecule of ethylene is .

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

By expanding, we obtain the following

For , the expanded determinant has the form . The roots are and 0.38. Therefore,

The energies of the four LCAO-MOs are

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

Note the two extra terms in this secular determinant compared to those written for noncyclic butadiene (previous problem).