If the bond dissociation energies of XY,   ${\mathrm{X}}_2\text{ and }{\mathrm{Y}}_2$   (all diatomic molecules ) are in the ratio 1:1:0.5 and  ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}$ of XY is -200 kJ/mol. The bond dissociation energy of  ${\mathrm{X}}_2$ will be:

$\frac{1}{2}{\mathrm{X}}_2+\frac{1}{2}{\mathrm{Y}}_2\to \mathrm{XY};\mathrm{\Delta H}=-200 \mathrm{kJ}/\mathrm{mol}$

$\text{ Let bond dissociation energy of }{\mathrm{X}}_2,{\mathrm{Y}}_2\text{ and }\mathrm{XY}\text{ be }\mathrm{a}:\frac{\mathrm{a}}{2}:\mathrm{a}\text{ (the ratio given) }$

$$\begin{gathered} {\mathrm{X}}_2\to 2\mathrm{X};\mathrm{\Delta H}=\mathrm{a} \\ {\mathrm{Y}}_2\to 2\mathrm{Y};\mathrm{\Delta H}=\frac{\mathrm{a}}{2}\left(2\right) \\ \mathrm{XY}\to \mathrm{X}+\mathrm{Y};\mathrm{\Delta H}=\mathrm{a} \end{gathered}$$

By (1)+(2)-(3)x(5)

$\begin{array}{l}{\mathrm{X}}_2+{\mathrm{Y}}_2\to 2\mathrm{XY}\\ \mathrm{\Delta H}=\mathrm{a}+\frac{\mathrm{a}}{2}-2\mathrm{a}\end{array}$

or

$\frac{1}{2}{\mathrm{X}}_2+\frac{1}{2}{\mathrm{Y}}_2\to \mathrm{XY}$

$\mathrm{\Delta H}=\frac{\mathrm{a}}{2}+\frac{\mathrm{a}}{4}-\mathrm{a}=-\frac{\mathrm{a}}{4}$

$\therefore -\frac{a}{4}=-200 \therefore a=800 \mathrm{kJ}$