1. The point group assignments are:

a) MoO2detc2 is C2

b) MoOCl2detc2 is Cs

c) MoOdetc2 is C2v

2. The point symmetry of [Ni(CN)5]3- is C4v. What is the geometry of the ion?

This symmetry only fits a square pyramidal geometry.

3. (a) Create five complexes having symmetries of these point groups: 
D4h,   C3v,   C2v,   D3,   Cs 
by exchanging water ligands (i.e., making ligand substitutions) on [Ni(H2O)6]2+ using any of the following ligands:
ethylenediamine (en), chloride,  acetylacetonate (acac), thiocyanate,  ammonia. 
Draw structures to illustrate the symmetries of your complexes.

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4. (answer) The thiomolybdate ion [MoO3S]2- has symmetry C3v. The three Mo=O bonds will generate two observed stretch modes of A1 and E which are both IR active.

5. The five d-orbitals for a metal M are degerenate in 'free' atoms . However, when bonds are made from M to other atoms or ligands htis 5-fold degeneracy disappears.

(a) Why does the 5-fold degeneracy disappear?

Because in any geometry, in any symmetry, there is no longer any symmetry 'code' type that is 5-fold degenerate. For example, in Oh, the highest degeneracy in the symmetry codes is 'T' type, triply degenerate.

(b) What happens to the five d-orbitals of M when 6 identical ligands bond to M in an octhedral geometry?

The 5 d's split into two sets, doubly degenerate Eg adn triply degenerate T2g.

(c) Can you suggest specific situations that would make the metal's d's all non-degenerate?

The 5 d orbitals will all be non-degenerate when the symmetry is low. Or, as the symmetry of Oh is lowered to D4h then to C4v or C3v, the degeneracies split. A point group symmetry having only non-degenerate symmetry types like A1, B1, B2 etc as in C2v would accomplish this.

6. (answer) The two hybrid expressions are symmetrical (have the same % atomic orbital composition) but point in different directions. Because the hybrids have a larger % p character than % s (which is only ~20%) the angle must be narrower than for the ideal 120 deg angle of an sp2 hybrid (which has 33% s character). So the second AB2 structure with the smallest angle is the best choice.

7. The hybrid orbital expressions are given below for the central atom (S) in SF2. 

                                     hybrid(a) =  0.33 3s + 0.63 3pz + 0.71 3py
         hybrid                       hybrid(b) =  0.33 3s + 0.63 3pz - 0.71 3py
         orbitals                     hybrid(c) =  0.90 3s - 0.44 3pz
         on sulfur:                  hybrid(d) =  1.0 3px

(a) Which hybrid orbitals will be used for S-F bonds? hybrid (a) and hybrid (b)

      (b) Which hybrid orbitals will be used for non-bonding electron. hybrid (c) and hybrid (d)

(c) Explain your choice: Because the two S-F bonds have to be made from two equivalent hybrids on S; only hybrids (b) and (c) are a symmetry equivalent pair.

8. Using the table of Valence Orbital Ionization Energies below, predict which member of  
   Group 16 will make the most covalent bonds to Zn in a compound ZnX. 
  Which atomic orbital(s) of the chosen Group 16 element will mainly be involved in bonding
   to Zn?

                        principle quantum #                        orbital energies (eV)

O                                                  2                                                                     -32.3                        -15.8                       
S                                                   3                                                                     -20.7                        -11.6
Se                                                4                                                                     -20.8                        -10.8

(answer) The most covalent bond will involve orbitals close in energy on each atom of the bond. Using the Zn 4s orbital at -9.4 eV, the orbital closest in energy is the Se 4p at -10.8 eV. ZnSe is the most covalent compound.

9. (answers)

(a) there is a formal -1 charge on the C atom and a formal +1 charge on the O atom.This is odd since the O atom is the more electronegative element.

(b) MOT and VSEPR both describe a triple bond; MOT specifies a sigma and two pi bonds

MOT and VSEPR both describe two lone pairs; in VSEPR these look equivalent on the C and O but in MOT specifies the O lone pair is the 1sigma MO (mainly O 2s a.o.) at lowest energy while the 3sigma MO is the lone pair on C at much higher energy and is the HOMO.

(c) the metal orbitals will be at higher energies similar to C ( metals tend to be electropositive) and will favor a reaction at the HOMO of CO which is localized on C and looks much like the "lone air" from VSEPR.

10. (answer) In this MO of [Pt(CN)4]2-, you can identify an antibonding pi interaction betweeen the C and N of the CN ligand ( in othe rwords, the cyanide pi* MO) and a pi bonding interaction between the Pt and the C, where the Pt is using a d orbital like dxy with lobes in the plane inbetweeen the bond axes.