1. MgO is found in the rock salt (NaCl) structure with a lattice constant of 421.2 pm. Is this consistent with the ionic radii given in the textbook? Explain.
The rock salt structure has both the cation and the anion in 6-coordinate environments. Thus, the appropriate radius to use for Mg2+ is 86 pm and the appropriate radius to use for O2– is 126 pm. The sum of these two values should give the inter-ion distance, 86 + 126 = 212 pm. In the rock salt structure the lattice constant is measured from O2– to O2–, i.e. two Mg-O distances (another way to see this is that the lattice constant measures the distance across O-Mg-O). Thus, the predicted lattice constant is 2×212 = 424 pm. This is excellent agreement with the experiment (less than 0.7 % error).
2. The band gap in MgO is 7.8 eV. Is MgO likely to be an effective semiconductor? Why or why not? What could you dope MgO with to form a p-type semiconductor? An n-type semiconductor? Explain your reasoning in each case.
The band gap in MgO is very large so it does not act as a semiconductor. Compared to Si, for example, which has a band gap of about 1.1 eV, it is better to characterize MgO as an insulator. For n-type doping more electrons are required, so Na+ might work. By using a monocation the uncompensated charge on the oxide ions can act as the source of extra electrons. Similarly, for a p-type semiconductor extra positive charges are required so Al3+ might work. Even with doping, the effective band gap in MgO is too large to create a useful semiconductor.
3. In terms of Brønsted-Lowry theory, do you expect MgO to be an acid, a base, or a neutral compound? Explain.
MgO is an ionic oxide so is expected to act as a base: MgO(s) + H2O(l) → Mg2+(aq) + 2 OH–(aq). The low solubility of MgO in water, because of the high lattice energy, means that not too much MgO can react under neutral conditions. Under acidic conditions, MgO becomes significantly more soluble.
4. Predict if the following reactions occur. If reaction does occur, balance the reaction.
a. MgO(s) + HF(aq)
b. MgO(s) + H2S(aq)
a. MgO(s) + HF(aq)
Mg2+ is a little bit harder than H+ (because of the hydration sphere around the hydrogen ion) and O2– is a little bit softer than F– (because of the charge) so a metathesis reaction should occur.
MgO(s) + 2 HF(aq) → MgF2(s) + H2O(l)
b. MgO(s) + H2S(aq)
Mg2+ is a little bit harder than H+ and O2– is a little bit harder than S2– (because of size) so no reaction is expected.
5. MgO can be formed by dehydration of Mg(OH)2. Using data from Latimer Diagrams and a Born-Haber (or thermochemical) cycle, find ΔG° for the dehydration reaction of Mg(OH)2. ΔGf° for MgO is –569.4 kJ/mol (this is for the formation of MgO from the elements). ΔG° = –nFE°.
The target reaction is: Mg(OH)2(s) → MgO(s) + H2O(l)
The formation reaction from the elements is: Mg(s) + ½ O2 → MgO(s) (1) ΔGf° = –569.4 kJ/mol
Under basic conditions: Mg(OH)2(s) + 2 e– → Mg(s) + 2 OH–(aq) (2) E°red = –2.687 V
Under basic conditions: O2(g) + 2 H2O(l) + 4 e– → 4 OH–(aq) (3) E°red = +0.401 V
Combining reactions (2)(reversed) and ½(3): Mg(s) + ½ O2(g) + H2O(l) → Mg(OH)2(s) (4) E° = +0.401 – (–2.687) = +3.088 V
Reversing reaction (4): Mg(OH)2(s) → Mg(s) + ½ O2(g) + H2O(l) (5) ΔG° = –[–(2)(96485 Coul/mol)(3.088 V)] = +595.9 kJ/mol
Adding reactions (1) and (5): Mg(OH)2(s) → MgO(s) + H2O(l) (6) ΔG° = +595.9 kJ/mol + (–569.4) = +26.5 kJ/mol
Reaction (6) is the target reaction so ΔG° = +26.5 kJ/mol. This means that magnesium hydroxide must be heated to a fairly high temperature to dehydrate.