IE $_{1}$ and IE $_{2}$ of $M g$ metal are 178 and 348 $k c a l / m o l$ respectively. The energy required for the given reaction is:
$M g (g) ⟶ M g^{+ 2} (g) + 2 e^{-}$

+ 170 k c a l / m o l

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+ 526 k c a l / m o l

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- 170 k c a l / m o l

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- 526 k c a l / m o l

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Solution

The correct option is D $+ 526 k c a l / m o l$
$M g (g) ⟶ M g^{+ 2} (g) + 2 e^{-}$ is
$M g \to M g^{+} \to M g^{+ 2}$ so energy required for the reaction is $178 + 348 = 526$ kcal/mol.

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