Question

$I{P}_1$ and $I{P}_2$ of $Mg$ are 178 and $348{\text{K cal mol}}^{-1}$ respectively. The energy required for the reaction
$Mg\to M{g}^{2+}+2e^{-}$ is:

• A. $+170{\text{K cal mol}}^{-1}$
• B. $+526{\text{K cal mol}}^{-1}$
• C. $-170{\text{K cal mol}}^{-1}$
• D. $-526{\text{K cal mol}}^{-1}$

Answer 1

The correct option is B $+526{\text{K cal mol}}^{-1}$

$Mg\to M{g}^{2+}+2e^{-}$
The energy required in ${\text{kcal mol}}^{-1}$ is the sum of the first ionization energy and the second ionization energy.
$I{p}_1+I{p}_2=178+348=+526{\text{K cal mol}}^{-1}$