${I P}_{1}$ and ${I P}_{2}$ of $M g$ are $178$ and $348 k c a l {m o l e}^{- 1}$ . The energy required for the reaction $M g \to {M g}^{2 +} + {2 e}^{-}$ is:

+ 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 B $+ 526 k c a l / m o l$
$M g ⟶ {M g}^{+} + e^{-} . . . . . . . {I P}_{1} = + 178 k c a l {m o l e}^{- 1}$
${M g}^{+} ⟶ {M g}^{2 +} + e^{-} . . . . . . . . {I P}_{2} = + 348 k c a l {m o l e}^{- 1}$
Overall: $M g ⟶ {M g}^{2 +} + {2 e}^{-} . . . . . . I P = I P_{1} + I P_{2}$
$= 178 + 348$
$= + 526 k c a l {m o l e}^{- 1}$
$∴ I P = 526 k c a l {m o l e}^{- 1}$

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