1. Cr crystalizes in a body centered cubic lattice with a cell constant of 287.2 pm. The atomic mass of Cr is 52.00 g·mol–1. Calculate the density in units of g·cm–3.
The volume of the cubic unit cell is (287.2 pm)3 = 2.369×107 pm3
The body centered cubic has 2 Cr atoms per unit cell so the mass in the unit cell is 2×(52.00 g·mol–1)/(6.022×1023 mol–1) = 1.727×10–22 g
Density = mass/volume = (1.727×10–22 g)/(2.369×107 pm3) = 7.290×10–30 g·pm–3 = 7.290 g·cm–3
2. For each of the following pairs of acids, indicate the stronger acid and briefly explain your reasoning. a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+; b) HClO3 or HClO4.
a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+: [Fe(OH2)6]3+ because of the greater positive charge.
b) HClO3 or HClO4: HClO4 – the estimated pKa of HOClO2 is ~ –2 and the estimated pKa of HOClO3 is ~ –7
3. Will Ga doped into Ge be an n– or p–type semiconductor? Explain your reasoning.
Ga has one less valence electron than Ge so this will be a p-type semiconductor.
4. Calculate the lattice energy for MgS (which crystallizes in the rock salt lattice) using the Born-Mayer equation and a Born-Haber cycle using the data given below. Do the two values agree, within experimental expectations? If not, provide a reason.
Born-Mayer: Elat = –[(6.022×1023)(1.748)(+2)(–2)(1.60s×10–19)2/ (1.113×10–10)((72 + 184)×10–12)][1 – 34.5/(72 + 184)] = 3.28×106 J·mol–1 = 3280 kJ·mol–1
Born-Haber cycle:
Mg(s) → Mg(g) S(Mg) = 148 kJ·mol–1
Mg(g) → Mg+(g) + e– IE1(Mg) = 737 kJ·mol–1
Mg+(g) → Mg2+(g) + e– IE2(Mg) = 1476 kJ·mol–1
S(s) → S(g) S(S) = 277 kJ·mol–1
S(g + e– → S–(g EA1(S) = 200. kJ·mol–1
S–(g) + e– → S2–(g) EA2(S) = –492 kJ·mol–1
Mg2+(g) + S2–(g) → MgS(s) Elat
Mg(s) + S(s) → MgS(s) ΔfHo = –347 kJ·mol–1
S(Mg) + S(S) + IE1 + IE2 – EA1 – EA2 – Elat – ΔfHo = 0
Elat = S(Mg) + S(S) + IE1 + IE2 – EA1 – EA2 – ΔfHo
Elat = (148) + (277) + (737) + (1476) – (200) – (–492) – (–347) = 3277 kJ·mol–1
The agreement is excellent.
5. Consider methanol (CH3OH), acting as a Lewis acid, reacting with acetone (CH3(C=O)CH3) and DMSO (CH3(S=O)CH3). Find the enthalpy of each reaction. Based on these results, comment on the solubility of these liquids with each other.
CH3OH + CH3(C=O)CH3 → CH3OH–(O=C)(CH3)3
ΔrHo = –[(6.98)(2.02) + (0.296)(4.67)] = –15.5 kJ·mol–1
CH3OH + CH3(S=O)CH3 → CH3OH–(O=S)(CH3)3
ΔrHo = –[(6.98)(2.76) + (0.296)(5.83)] = –21.0 kJ·mol–1
The enthalpies for the two reactions are similar and relatively small. This might suggest that neither acetone nor DMSO is very soluble in methanol. However, the solvents are all miscible, meaning that the solvation for these liquids is not governed by Lewis acid/base interactions.