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Chemistry Question on Classification of elements & periodicity in properties

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

A

3131% Mg^++ 69% Mg^{2+}

B

6969% Mg^+ + 31% Mg^2{+}

C

8686% Mg^+ + 14% Mg^2{+}

D

1414% Mg^+ + 86% Mg^2{+}

Answer

6969% Mg^+ + 31% Mg^2{+}

Explanation

Solution

Energy absorbed in the ionisation of 1 mole of Mg (g) to Mg+4(g)=750kJMg^+4 (g) = 750 kJ Energy left unused = 1200 - 750 = 450 kJ % of Mg+Mg^+ (g) converted into Mg2+Mg^{2+} (g) =4501450×100\frac{450}{1450} \times 100 = 31% . Thus the percentage of Mg+Mg^+ (g) = 100 - 31 = 69%