1. (6 points) A beaker is filled with an aqueous solution known to contain comparable quantities of Mg²⁺, HCO₃^-, Pb²⁺, and thiourea, H₂N-C(=S)-NH₂. What adducts are likely to be present in the beaker?

Likely adducts are

Mg2+ with HCO3-, or possibly Mg2+ with CO32-.

Pb2+ with the S atom of thiourea

Water with water.

2. (4 points) A number of metal ion reduction processes are shown below:

a) Hg²⁺ + 2e ® Hg(l)

b) Na⁺ + e ® Na(s)

c) Fe²⁺ + 2e ® Fe(s)

d) La³⁺ + 3e ® La(s)

Many chemists view these as Lewis acid base reactions in which the cation is the acid, the electron is the base, and the metal is the adduct. Assuming the electron to be the ultimate soft base, arrange the four processes in order from most spontaneous (most positive e ^o) to least spontaneous (least positive e ^o).

Soft acids should react more spontaneously with the electron than hard acids. Thus

a > c > b > d 

3. (5 points) In the adduct of Fe³⁺ with dimethylsulfoxide (DMSO), (CH₃)₂SO, the oxygen atom of DMSO serves as the donor atom. Ru²⁺ also forms an adduct with DMSO. Do you think that this will have the same structure as the Fe³⁺ adduct? EXPLAIN.

Ru2+ is a borderline acid, so could bind to either the oxygen or the sulfur donor of DMSO. The sulfur donor is probably more likely. 

4. (5 points) A chemist claims that s/he has synthesized a compound of formula SeOF₂. S/he suggests two possible structures for the compound, shown below. Which is preferable? Why?

Both structures have zero formal charge on all atoms. No basis for preference here.

In the left structure, Se is in the 4+ oxidation state while O and F are in their normal oxidation states. In the right structure, Se is in the 2+ oxidation state, F is in the normal -1 oxidation state, but O is in an unusual zero oxidation state. This provides one reason to prefer the left structure to the right one. The best argument is based on bond energies. Consider conversion of the left structure to the right one. To disrupt the left structure requires breaking one Se=O bond and two Se-F bonds. This is a total input of 440 + 2(351) = 1140 kJ. (The Se=O double bond strength was estimated using the Pauling equation at the top of the bond energy handout). Forming the second structure involves making one O-F bond, one Se-O bond, and one Se-F bond, providing a total of 190 + 230 + 351 = 770 kJ. The left structure is clearly much more stable than the right structure. The left structure is preferred.