Testing the ionic model for magnesium oxide with a Born–Haber cycle
A class is asked to evaluate the ionic model for magnesium oxide. The following thermodynamic data are provided in kJ mol−1: enthalpy of formation of MgO(s) = −602; enthalpy of atomisation of Mg(s) = +148; first ionisation energy of Mg = +738; second ionisation energy of Mg = +1451; enthalpy of atomisation of O2(g) to 2O(g) per mole of O atoms = +249; first electron affinity of O = −141; second electron affinity of O = +798. A theoretical calculation using a simple ionic model gives the lattice enthalpy of MgO as −3920 kJ mol−1. The experimental value is obtained from a Born–Haber cycle using the data above. MgO has a melting point of 2852 °C and is used as a refractory material. A student says the close agreement between the two lattice enthalpies proves that MgO is purely ionic. Another student points out that the second electron affinity is positive, which seems to make MgO formation unfavourable. The teacher wants a quantitative check and a critical comment.
1. Construct the Born–Haber cycle in words and calculate the lattice enthalpy of MgO from the data. Show the sign convention clearly.
Read the worked answer
The cycle starts from the elements in their standard states and goes through gaseous ions:
Mg(s) → Mg(g) → Mg+(g) → Mg2+(g) 1/2O2(g) → O(g) → O−(g) → O2−(g) Mg2+(g) + O2−(g) → MgO(s)
Using Hess’s law:
ΔHf(MgO) = ΔHatom(Mg) + IE1(Mg) + IE2(Mg) + ΔHatom(O) + EA1(O) + EA2(O) + ΔHlatt(MgO)
Substitute the values, taking care that electron affinities are enthalpy changes for electron gain:
−602 = 148 + 738 + 1451 + 249 + (−141) + 798 + ΔHlatt
−602 = 3243 + ΔHlatt
ΔHlatt = −602 − 3243 = −3845 kJ mol−1
The lattice enthalpy is defined here as the enthalpy change when one mole of solid ionic lattice forms from its gaseous ions; the negative sign means the process is exothermic.
2. Compare your calculated lattice enthalpy with the theoretical value and comment on the student’s statement that MgO is purely ionic. Refer to the size and direction of the difference.
Read the worked answer
The Born–Haber value is −3845 kJ mol−1. The simple ionic model gives −3920 kJ mol−1. The difference is:
−3845 − (−3920) = +75 kJ mol−1
This is about 2% of the lattice enthalpy. The agreement is good, so the ionic model is a useful first approximation. However, the experimental lattice enthalpy is less exothermic than the theoretical value.
A purely ionic model assumes point charges and no sharing of electron density. If the real compound has some covalent character, the electron density between Mg2+ and O2− is partly shared, and the simple electrostatic model overestimates the attraction. Polarisation of the oxide ion by the small, highly charged Mg2+ ion can also reduce the purely ionic interaction. The discrepancy is small, so MgO is predominantly ionic, but the statement that it is “purely ionic” is too strong. The calculated value also carries uncertainties from the experimental data, so a 2% difference should not be over-interpreted.
3. Explain why the second electron affinity of oxygen is positive and why MgO is nevertheless thermodynamically stable with respect to its elements.
Read the worked answer
The second electron affinity is the enthalpy change for O−(g) + e− → O2−(g). The first electron is attracted to a neutral oxygen atom, so energy is released and EA1 is negative. The second electron is attracted to an already negative O− ion, but it also experiences strong electron–electron repulsion. The repulsion outweighs the nuclear attraction, so energy must be supplied and EA2 is positive (+798 kJ mol−1).
MgO is still stable because the overall formation enthalpy is negative. The very exothermic lattice enthalpy (−3845 kJ mol−1) and the formation of the small, highly charged Mg2+ ion more than compensate for the positive second electron affinity. The cycle gives ΔHf = −602 kJ mol−1, so MgO(s) is lower in enthalpy than Mg(s) + 1/2O2(g) under standard conditions. Thermodynamic stability also depends on entropy, but the large negative enthalpy is the dominant reason why MgO forms.