Question

One mole of magnesium in the vapour state absorbed 1200 kJ ${mole}^{-1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ ${mole}^{-1}$ respectively, the final composition of the mixture is

• A. 31% ${Mg}^{+}$ + 69 % ${Mg}^{+}$
• B. 69% ${Mg}^{+}$ + 31 % ${Mg}^{+}$
• C. 86% ${Mg}^{+}$ + 14% ${Mg}^{+2}$
• D. 14% ${Mg}^{+}$ + 86% ${Mg}^{+2}$

Answer 1

The correct option is B

69% ${Mg}^{+}$ + 31 % ${Mg}^{+}$

Energy absorbed for converting $M{g}_{\left(g\right)}$ $\to$ ${M{g}_{\left(g\right)}}^{+}$ = 750 kJ

Energy left unconsumed = 1200 = 750 = 450 kJ.

This energy will be required to convert ${M{g}_{\left(g\right)}}^{+}$ to ${M{g}_{\left(g\right)}}^{+2}$

Thus % of ${M{g}_{\left(g\right)}}^{+2}$ = $\frac{450}{1450}$ $\times$ 100 = 31%

% of ${M{g}_{\left(g\right)}}^{+}$ = 100 - 31 = 69%