1. Explain the trends of the variation of the metallic radius for the second row transition metal elements (Y to Cd) in terms of metallic bonding.
The metallic radii are:
Metalrmet (pm)Electron configuration
Y182[Kr]5s24d1
Zr160[Kr]5s24d2
Nb147[Kr]5s14d4
Mo140[Kr]5s14d5
Tc135[Kr]5s24d5
Ru134[Kr]5s14d7
Rh134[Kr]5s14d8
Pd137[Kr]4d10
Ag144[Kr]5s14d10
Cd152[Kr]5s24d10
Two competing effects are occurring as the elements change from Y to Cd. The atomic orbitals used to create the band structure are the 5s and 4d (and perhaps the 5p, as well) so there are at least six overlapping bands. As the electrons fill these bands, initally more bonding states are occupied so the metallic radius decreases since there is more bonding character to attract atoms together. Near the middle of the series nonbonding states are occupied, so there should be little change in the metallic radius. Finally, at the end of the transition series the electrons are being used to fill antibonding states so the bonding between atoms decreases and the radius increases. Competing with this is the effect of Z*. As the atomic number increases Z* increases across the series. This causes a reduction in orbital size, which decreases the atomic radius and will reduce the amount of overlap between neighboring orbitals. Thus, the observed trend mainly follows the prediction based on band arguments from Y to Cd but the constancy of the radius from Tc to Pd is caused by the approximate cancelation of the antibonding effects with the Z* effects.
2. When titanium(IV) oxide, a colorless solid, is heated in a substoichiometric amount of hydrogen gas, a blue color develops, indicating light absorption in the red. Write the balanced chemical reaction and use an appropriate bonding theory to explain the color change.
The H2 gas causes reduction of some of the Ti4+ to Ti3+:
Ti4+ + ½ H2 → Ti3+ + H+
Reduction causes a apparent decrease in the band gap, as evidenced by the absorption of the low energy red light (compared to the original titanium(IV) oxide, which absorbs light in the UV. Reduction also introduces extra electrons into the lattice, formally in the form of Ti(III). An excess of electrons is the signature of an n-type semiconductor.
3. Predict the products for the following reactions:
(a) [CuI4]2–(aq) + [CuCl4]3–(aq)
(b) NH2–(aq) + H2O(l)
(a) Use hard-soft acid base theory to approach the problem:
[CuI2]2–(aq)+ [CuCl4]3–(aq) →← [CuCl4]2–(aq)+ [CuI4]3–(aq)
HA-SBSA-HB HA-HBSA-SB
(b) Since the acid portion is the same in each of these, HSAB theory is less useful. Recall that water can either donate or accept a hydrogen ion in water but ammonia is a very poor hydrogen ion donor, meaning that amide ion is a very strong base:
NH2–(aq)+ H2O(l) →← NH3(aq)+ OH–(aq)
4. Predict the order of increasing lattice energy for the following compounds: ZnO (wurtzite structure), NH4F (wurtzite structure), and CaF2 (fluorite structure). Explain your reasoning.
The lattice energy is mostly controlled by the Coulomb interactions, which is dominated by the ionic charges, but slightly modified by the Madelung constant (A). Thus, to a reasonably good approximation Elat ~ Z+Z–A. Thus, for ZnO Elat ~ (+2)(–2)(1.641) = 6.6, for NH4F Elat ~ (+1)(–1)(1.641) = 1.6, and for CaF2 Elat ~ (+2)(–1)(2.519) = 5.0. Therefore, the predicted order of increasing lattice energy is NH4F < CaF2 < ZnO.
5. Complete and balance the following oxidation-reduction reaction and find the standard cell potential:
MnO4–(aq) + HCO2H(aq) → CO2(g) + Mn2+(aq)
MnO4–(aq) + C2O42–(aq) → CO2(g) + Mn2+(aq)
Reduction half-reaction: MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(l) E°red = 1.51 V
Oxidation half-reaction: HCO2H(aq) → CO2(g) + 2 H+(aq) + 2 e– E°red = –0.114 V
Net reaction: 2 MnO4–(aq) + 5 HCO2H(aq) + 6 H+(aq) → 2 Mn2+(aq) + 5 CO2(g) + 8 H2O(l) E° = 1.51 – (–0.114) = 1.62 V