1. (9 points) Provide the formula and a guessed structure for the oxide of each species below:

a) S⁴⁺

SO2. Radius ratio is 51/126 = 0.40, very close to the border between CN 4 and 6. Try CN 4 for S4+. It has one lone pair, so the S will bond to 4-1 = 3 oxygen atoms. CN of O is 3*1/2 = 3/2. Thus one of the oxygens bridges and the other does not. Click for structure.

b) N³⁺

N2O3. Radius ratio is 30/126 = 0.24 (close to the 3/4 cutoff), implying total coordination number of 3 or 4 for N. Try CN 4 first. With one lone pair, this means that each N will bond to 3 oxygens. CN(oxygen = 3*2/3 = 2, meaning all oxygens are bridging. The structure is like that of As2O3, an infinite 2-dimensional sheet polymer.

Assuming total CN of 3 for N means that each N binds 2 oxygens. CN(O) = 2*2/3 = 4/3, so one O is bridging and two are terminal. The structure has each N single bonded to a bridging O, double bonded two a terminal O, with one lone pair.

c) Br⁵⁺.

Br2O5. Radius ratio is 55/126 (55 is estimated from the Br3+ and Br7+ values) = 0.44. This is close to the 4/6 cutoff but on the 6 side, so go with that. Total CN(Br) = 6. Subtract one lone pair to show that each Br is bonded to 5 oxygens. CN(O) is therefore 5*2/5 = 2, meaning that all oxygen atoms are bridging. Click for structure.

2. (6 points) A chemist claims that he has discovered a previously unknown silicate with formula SiO_2.5^-1. He proposes the structure shown below (in shaded triangles, the Si and fourth O atoms are below the plane of the paper; in unshaded triangles, they are above the plane). Does his proposed structure match his proposed formula? (NOTE: Answering YES or NO without embellishment will earn you no points).

The indicated structure was not intended to be an infinite sheet; instead it was proposed to be as shown (triangle-tipped star of David).

As such, it does NOT agree with the formula. A peripheral SiO4 tetrahedron involves 3 terminal and one bridging O, for a formula of SiO3.5. An inner SiO4 tetrahedron involves 1 terminal and three bridging O, for a formula of SiO2.5. There are 6 of each type of triangle, so the formula for the structure is the average of these 2 types, or SiO3.

3. (5 points) Combine the two half reactions below to obtain a spontaneous overall reaction, and balance the overall reaction.

ClO₃^- + 3H⁺ + 2e ® HClO₂ + H₂O, e ^o = 1.21 V

O₂ + 4H⁺ + 4e ® 2H₂O, e ^o = 1.23 V

The second half reaction occurs more spontaneously, so will occur as written. The first half reaction will be forced to occur in reverse. The first half reaction must be taken twice to insure equal numbers of electrons:

O₂ + 2HClO₂ ® 2H⁺ + 2ClO₃^-