1. Ionic compounds typically have high melting points because the lattice energy is so high. If one were to design an ionic compound with a melting point near room temperature, what properties would be required in the ions and in the lattice? Explain.
The lattice energy depends upon the size of ionic charges, the Madelung constant, and the distance between ions modified by a small nuclear-nuclear repulsive term. To lower the melting point, the charges should be small (i.e., +1, –1), the Madelung constant should be small, and the distance between ions must be large. To create a large distance between ions, the ionic radii should be large so complex ions or organic ions are good candidates for creating low melting ionic compounds. Such ionic liquids exist and are a topic of significant research.
2. K+ has an ionic radius of 1.52 Å for coordination number 6 while Au+ has an ionic radius of 1.51 Å for coordination number 6. Which ion would you expect to be the stronger acid in an aqueous solution? Or would they have the same acid/base reactivity? Explain.
Despite having similar ionic radii, Au+ is the stronger acid. This is due to the poor shielding of the d and f electrons in Au+, which leads to a higher Z*. The high Z* polarizes the solvated water molecules significantly, leading to ionizable hydrogen ions.
3. FeO is typically nonstoichiometric, i.e. a more accurate representation of the chemical formula is FexO, where x ~ 0.95. A consequence of the nonstoichiometry is that FeO is a semiconductor. Is this a p-type or n-type semiconductor? Explain your reasoning.
In order to maintain charge neutrality some of the Fe2+ in FexO must be oxidized to Fe3+. This introduces extra positive charges into the lattice, so that FexO is a p-type semiconductor. For x ~ 0.95, the stoichiometry must be (Fe2+)0.85(Fe3+)0.10O.
4. Which is the better base for boron trifluoride, methylamine or trimethylphosphine? Explain your reasoning.
Use the Drago-Wayland equation to answer the question:
BF3 + N(CH3)H2 → F3B–N(CH3)H2
–ΔH = EAEB + EBCB = (20.2)(2.66) + (3.31)(12.00) = 93.5 kJ/mol
BF3 + P(CH3)3 → F3B–P(CH3)3
–ΔH = EAEB + EBCB = (20.2)(17.2) + (3.31)(13.40) = 391.8 kJ/mol
Trimethylphosphine is much better Lewis base for boron trifluoride. This is consistent with the electron donating properties of methyl groups and the larger size of the P atom.
5. Consider the Latimer diagram for cobalt in acidic solution, given below. From this information, find the potential for the disproportionation of Co3+ at neutral pH. Assume that all cobalt ions are 0.1 M concentration.
CoO2 → 1.4 Co3+ → 1.92 Co2+ → –0.282 Co
The disproportionation reaction under acidic conditions is:
2 Co3+(aq) + 2 H2O(l) → CoO2(s) + Co2+(aq) + 4 H+(aq)
The potential at standard conditions is E° = 1.92 – 1.4 = 0.5 V
At pH = 7 (neutral conditions), the Nernst equation must be used to find the potential:
E = E° – RT — nF lnQ where Q = [Co2+][H+]4 ———— [Co3+]2 = [10–1][10–7]4 ————— [10–1]2 = 10–27
E = 0.5 – (8.314)(298) ————— (1)(96500) ln(10–27) = 2.1 V