1. Calculate the theoretical lattice energy of MgO using both theoretical equations and using a Born-Haber cycle. How do the three values compare? MgO crystallizes in the rock salt structure, the same as NaCl. Are the values you found reasonable compared to the lattice energy of NaCl (787.1 kJ/mole) determined in class? Explain why or why not. The necessary data is found below.
For the theoretical calculations: do = 72 + 140 = 212 pm; A = 1.748; and n = 7 (both Mg2+ and O2– have the [Ne] configuration).
Born-Landé = –(6.022×1023)(1.748)(+2)(–2)(1.602×10–19)(1 – 1/7)/ (1.113×10–10)(212×10–12) = 3.93×106 J/mol = 3930 kJ/mol
Born-Mayer = –(6.022×1023)(1.748)(+2)(–2)(1.602×10–19)(1 – 34.5/212)/ (1.113×10–10)(212×10–12) = 3.83×106 J/mol = 3830 kJ/mol
Born-Haber cycle:
Mg(s) → Mg(g) S = 146 kJ/mole
Mg(g) → Mg+(g) + e– IE1 = 737 kJ/mole
Mg+(g) → Mg2+(g) + e– IE2 = 1476 kJ/mole
½ O2(g) → O(g) D = ½(497) kJ/mole = 248.5 kJ/mole
O(g) + e– → O–(g) EA1 = 141 kJ/mole
O–(g) + e– → O2–(g) EA2 = –780 kJ/mole
Mg2+(g) + O2–(g) → MgO(s) Elat
Mg(s) + ½ O2(g) → MgO(s) ΔHf° = –601.6
Then, S + IE1 + IE2 + ½D – EA1 – EA2 – Elat – ΔHf° = 0
Elat = S + IE1 + IE2 + ½D – EA1 – EA2 – ΔHf° = (146) + (737) + (1476) + (248.5) – (141) – (–780) – (–601.6) = 3848 kJ/mole
The agreement between the two theoretical values and the experimental value is excellent. The ratio of lattice energies between NaCl and MgO is expected to be about 4 (Z+Z– = 1 for NaCl compared to Z+Z– = 4 for MgO). The actual ratio is 4.9, which is reasonable given that the additional charge on Mg2+ and O2– scales the ionic radii nonlinearly.
2. Al is a metal while the adjacent element, Si, is a semiconductor. Explain, in terms of band theory, why this change occurs.
Al has an electron configuration of [Ne]3s23p1, which means that the band formed from the s orbital may be filled but the bands formed form the p orbitals is partially filled, which is the requirement for a metal. In the case of Si, the electron configuration is [Ne]3s23p2. Again, the s-band may be filled but the bands formed from the p orbitals would still seem to be unfilled unless the three different p bands do not overlap in energy. This could happen if the orientation of the p orbitals is such that there is no overlap between the different bands so that the three p orbitals lose their degeneracy. This can happen because of the structure of Si and Z*.
3. Predict the products and balance the following reactions:
a. H2SO4(aq) + Na2O(s)
b. F3B-S(CH3)2(solvate) + CH3NH2(g)
c. [CuI4]2–(aq) + [CuCl4]3–(aq)
a. H2SO4(aq) + Na2O(s) → H2O(l) + 2 Na2+(aq) SO42–(aq)
b. F3B-S(CH3)2(solvate) + CH3NH2(g) → F3B-NH2CH3(solvate) + S(CH3)2(solvate)
c. [CuI4]2–(aq) + [CuCl4]3–(aq) → [CuCl4]2–(aq) + [CuI4]3–(aq) (the metathesis occurs because Cu2+ and Cl– are both relatively hard compared to Cu+ and I–)
4. Calculate ΔH° for the reaction in question 3b. Hint: Use a thermochemical cycle.
The Drago-Wayland equation can be used to find ΔH° for addition reactions. Thus, find ΔH° for the two adducts and then subtract the reactions to get the net reaction:
F3B + S(CH3)2 → F3B-S(CH3)2 ΔH°(1)
F3B + CH3NH2 → F3B-NH2CH3 ΔH°(2)
Then, ΔH° = ΔH°(2) – ΔH°(1)
ΔH°(1) = –[(20.2)(0.70) + (3.3)(15.26)] = –64.5 kJ/mole
ΔH°(2) = –[(20.2)(2.66) + (3.3)(12.00)] = –93.3 kJ/mole
ΔH° = ΔH°(2) – ΔH°(1)) = (–93.3) – (–64.5) = –28.8 kJ/mole
5. Balance the reaction for the disproportionation of AgO. Find the standard potential for this reaction. Is the reaction spontaneous or not?
From the Latimer Diagram one possible unbalanced reaction is AgO(s) → Ag2O3(s) + Ag+(aq)
Reduction: [ AgO(s) + 2 H+(aq) + e– → Ag+(aq) + H2O(l) ] × 2
Oxidation: 2 AgO(s) + H2O(l) → Ag2O3(s) + 2 H+(aq) + 2 e–
Net: 4 AgO(s) + 2 H+(aq) → Ag2O3(s) + 2 Ag+(aq) + H2O(l)
E° = +1.77 – +1.57 = +0.20 V. The reaction is spontaneous because E° > 0.
OR
From the Latimer Diagram one possible unbalanced reaction is AgO(s) → Ag2O3(s) + Ag(s)
Reduction: AgO(s) + 2 H+(aq) + 2 e– → Ag(s) + H2O(l)
Oxidation: 2 AgO(s) + H2O(l) → Ag2O3(s) + 2 H+(aq) + 2 e–
Net: 3 AgO(s) → Ag2O3(s) + Ag(s)
E° = (+1.77 + 0.80)/2 – +1.57 = –0.29 V. The reaction is nonspontaneous because E° < 0.