One mole of magnesium in the vapor state absored `1200 kJ mol^(-1)` of enegry. If the first and second ionization energies of `Mg` are `750` and `1450 kJ mol^(-1)`, respectively, the final composition of the mixture is

Options

A. `86%Mg^(+)+14%Mg^(2+)`
B. `69%Mg^(+) +31%Mg^(2+)`
C. `14%Mg^(+)+86%Mg^(2+)`
D. `31%Mg^(+)+69%Mg^(2+)`

Answer

B