Question

Calculate second electron affinity of oxygen for the process: O\(^-\)(g) + e\(^-\)(g) \(\rightarrow\) O\(^{2-}\)(g) using the following data:
(i) Heat of sublimation of Mg(s) = +147.7 kJ mol\(^{-1}\)
(ii) Ionisation energy of Mg(g) to form Mg\(^{2+}\)(g) = +2189.0 kJ mol\(^{-1}\)
(iii) Bond dissociation energy for O\(_2\) = +498.4 kJ mol\(^{-1}\)
(iv) First electron affinity of O(g) = -141.0 kJ mol\(^{-1}\)
(v) Heat of formation of MgO = -601.7 kJ mol\(^{-1}\)
(vi) Lattice energy of MgO = -3791.0 kJ mol\(^{-1}\)

• A. 235.6 kJ mol\(^{-1}\)
• B. 468.7 kJ mol\(^{-1}\)
• C. 544.4 kJ mol\(^{-1}\)
• D. 744.4 kJ mol\(^{-1}\)

Hint:

Second EA is endothermic (positive) because adding electron to O\(^-\) is energetically unfavourable.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use Born-Haber cycle for MgO formation.
Step 2: Detailed Explanation:
For Mg(s) + \(\frac{1}{2}\)O\(_2\)(g) \(\rightarrow\) MgO(s): \(\Delta H_f = -601.7\)
Born-Haber cycle:
Sublimation of Mg: +147.7
IE\(_1\) + IE\(_2\) of Mg: +2189.0
Dissociation of \(\frac{1}{2}\)O\(_2\): +498.4/2 = +249.2
First EA of O: -141.0
Second EA of O: x (unknown)
Lattice energy: -3791.0
Sum = 147.7 + 2189.0 + 249.2 - 141.0 + x - 3791.0 = -601.7
(147.7 + 2189.0 = 2336.7; +249.2 = 2585.9; -141 = 2444.9)
2444.9 + x - 3791.0 = -601.7 \(\Rightarrow\) x - 1346.1 = -601.7 \(\Rightarrow\) x = 744.4 kJ mol\(^{-1}\).
Step 3: Final Answer:
Thus, second electron affinity of oxygen = +744.4 kJ mol\(^{-1}\).