1. Consider the hypothetical compound solid AlCl. Estimate the lattice energy using the Born-Mayer equation. To find the cationic radius examine the per cent reduction of the radius for several metals from the neutral to the +1 state and use this as a guide to estimate the cationic radius. Assume the NaCl lattice and that d* = 34.5 pm. Use your estimate to calculate the standard heat of formation for AlCl. Does your calculated heat of formation account for the nonexistence of solid AlCl? Why or why not? Cite the source for any data you need to look up.
Answer
Looking at the ratio of r+ (Table 1.4 from the textbook) to ratomic (Table 1.3 from the textbook) for the group 1 atoms finds values in the range of 0.48 to 0.62 for 6 a 6-coordinate cation. Using a middle value of 0.55 for Al estimates r+(Al+) to be ~ 0.55×125 = 69 pm. This seems reasonable since the cation radius for Al3+ (6-coordinate) is 53 pm (Table 1.4). The anion radius for Cl– is 181 pm (Table 1.4) so do = 69 + 181 = 250 pm.
Using A = 1.748 (Table 4.8) with the Born-Mayer equation gives:
Elat = –(NAAe2Z+Z– /4πεodo)×(1 – d*/do)
= –[(6.022×1023)(1.748)(1.602×10–19)2(+1)(–1)/ (1.113×10–10)(250×10–12)]×[1 – 34.5/250]
= 836900 J/mol = 836.9 kJ/mol
Born-Haber cycle
Sublimation: Al(s) → Al(g) S = 330.0 kJ/mol (NIST web site)
Ionization: Al(g) → Al+(g) + e– IE = 577 kJ/mol (textbook, Table 1.5)
Bond dissociation: ½ Cl2(g → Cl(g) BDE = 242 kJ/mol (textbook, Table 2.7)
Electron attachment: Cl(g) + e– → Cl–(g) EA = 349 kJ/mol (textbook, Table 1.6)
Lattice energy: Al+(g) + Cl– → AlCl(s) Elat = 836.9 kJ/mol (from above)
Formation from elements: Al(s) + ½Cl2(g) → AlCl(s) ΔfH°
The cycle is S + IE + ½BDE – EA – Elat – ΔfH° = 0
ΔfH° = S + IE + ½BDE – EA – Elat = 330.0 + 577 + ½(242) – 349 – 836.9 = –158 kJ/mol
Based on this estimate, AlCl has a favorable heat of formation and might be expected to be stable. AlCl has been detected in the gas phase at high temperatures and low pressure. It does not form a solid. Rather, it disproportionates into Al and AlCl3.
2. MoS2 is currently a heavily studied material because it can act as a semiconductor. Explain why this compound is semiconducting?
Answer
The band structure for this compound is complicated but looking at the oxidation numbers can help. If the S exists as sulfide, S2–, then the Mo must be Mo4+. However, if the S exists as disulfide, S22–, then the Mo must be Mo2+. A mixture of these two states creates charge carriers that can be responsible for the semiconducting behavior. Another way to think of this is that the mixed oxidation states leads to self-doping.
3. Consider H2CrO4 and HMnO4. Write the Brønsted-Lowry reaction of each of these acids with water. Which of these is the stronger acid? Explain your reasoning.
Answer
H2CrO4(aq) + H2O(l) → H3O+(aq) + HCrO4–(aq)
HMnO4(aq) + H2O(l) → H3O+(aq) + MnO4–(aq)
The oxidation number of Cr is +6 while that of Mn is +7. The larger central charge suggests that the stronger acid is HMnO4. Another approach is to use Pauling's rules, which estimate the pKa of H2CrO4 to be ~ –2 while for HMnO4 pKa ~ –7.