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$IE (I)$ and $IE (Il)$ of $Mg$ are 178 and $348 kcal {mol}^{- 1}$ . The energy required for the reaction $Mg \to {Mg}^{2 +} + 2 e^{-}$ is :

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+ 170 kcal

+ 526 kcal

- 170 kcal

- 526 kcal

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detailed solution

Correct option is B

$Mg (g) \to {Mg}^{+ 2} (g) + 2 e^{-}$ is

$Mg \to {Mg}^{+} \to {Mg}^{+ 2}$ , hence the process requires $178 + 348 = 526 kcal / mol$ of energy.

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IE(I) and IE(Il) of Mg are 178 and 348 kcal mol-1. The energy required for the reaction Mg→Mg2++2e-is :