Calculate the EA of O atom to O^2- ion from the following data:
i. $Δ_{f} H^{⊖} [MgO (s)] = - 600 kJ {mol}^{- 1}$
ii. $Δ_{u} H^{⊖} [MgO (s)] = - 3860 kJ {mol}^{- 1}$
iii. ${IE}_{1} + {IE}_{2} of Mg (g) = 2170 kJ {mol}^{- 1}$
iv. $Δ_{diss} H^{⊖} [O_{2} (g)] = + 494 kJ {mol}^{- 1}$
v. $Δ_{sub} H^{⊖} of Mg (s) = + 150 kJ {mol}^{- 1}$

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$- 69.3 kJ {mol}^{- 1}$

$- 693 kJ {mol}^{- 1}$

$+ 693 kJ {mol}^{- 1}$

$+ 69.3 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

Using Born-Haber cycle for MgO, calculation of EA of O atom ot O^2- ion (All values in kJ mol^-1)

$\begin{array}{l} ∴ Δ_{f} H^{⊖} = Δ_{sub} H^{⊖} + ({IE}_{1} + {IE}_{2}) + \frac{1}{2} Δ_{diss} + ({EA}_{1} + {EA}_{2}) + Δ_{U} H^{⊖} \\ - 600 = 150 + 2170 + \frac{1}{2} \times 494 + ({EA}_{1} + {EA}_{2}) - 3860 \\ ∴ ({EA}_{1} + {EA}_{2}) = + 693 kJ {mol}^{- 1} \end{array}$