Question

$I{E}_1$ and $I{E}_2$ of $Ca$ are $141$ and $274\text{kcal}$ respectively. The energy required for the following reaction is:
$Ca\to C{a}^{2+}+2e^{-}$
(Where $IE$ = ionization energy)

• A. $-415\text{kcal}$
• B. $+415\text{kcal}$
• C. $-175\text{kcal}$
• D. $+175\text{kcal}$

Answer 1

The correct option is B $+415\text{kcal}$

According to given conditions
$Ca\to C{a}^{+}+e^{-}I{E}_1=141\text{kcal}$
$C{a}^{+}\to C{a}^{2+}+e^{-}I{E}_2=274\text{kcal}$
Therefore,
$Ca\to C{a}^{2+}+2e^{-}IE=I{E}_1+I{E}_2$
Hence, ionization energy of the given reaction is:
$IE=141\text{kcal}+274\text{kcal}$
$=415\text{kcal}$