$(\mathrm{IE}{)}_1$ and $(\mathrm{IE}{)}_2$ of ${\mathrm{Mg}}_{\left(\mathrm{g}\right)}$ are $740,1540 \mathrm{kJ}{ \mathrm{mol}}^{-1}$. Calculate percentage of ${{\mathrm{Mg}}^{+}}_{\left(\mathrm{g}\right)}$ and ${{\mathrm{Mg}}^{2+}}_{\left(\mathrm{g}\right)}$ if $1 \mathrm{g}$ of ${\mathrm{Mg}}_{\left(\mathrm{g}\right)}$ absorbs $50.0 \mathrm{kJ}$ of energy.

We know that No. of moles=Given mass/ Molar mass

then,No. of moles of $\mathrm{Mg}=\frac{1}{24}=0.0417$

Now, Ist ionisation Energy $=740 \mathrm{kJ}/\mathrm{mol}$(Given)

Energy required for conversion of $0.0417\mathrm{Mg}$ to ${\mathrm{Mg}}^{+}$ is $=0.0417\times 747=30.83\mathrm{KJ}$.

So, energy unused $=(50-30.83)=19.17 \mathrm{kJ}$

So, $19.17 \mathrm{kJ}$ will be used in ionisation ${\mathrm{Mg}}^{+}\to {\mathrm{Mg}}^{+}$

So, number of moles of ${\mathrm{Mg}}^{+}$converted into ${\mathrm{Mg}}^{++}$  is $\frac{19.17}{1450}=0.0132$

Therefore, No. of moles of ${\mathrm{Mg}}^{\top }=0.0417-0.0132=0.0285$

% of ${\mathrm{Mg}}^{+}=\left(\frac{0.0285}{0.0417}\times 100\right)\%=68.65\%$

% of $\left.{\mathrm{Mg}}^{2+}=\frac{0.0417}{0.0132}\times \frac{0.0417}{0.00}\right)\%=31.65\%$

Then, $\%{\mathrm{Mg}}^{+}=68.65$ and $\%{\mathrm{Mg}}^{+2}=31.65$

Hence, option(d) is the answer.