The bond dissociation energies of ${\mathrm{X}}_2,{\mathrm{Y}}_2\text{ and }\mathrm{XY}$ are in the ratio of $1:0.5:1.\mathrm{\Delta H}$ for the formation of $\mathrm{XY}\text{ is }-200 \mathrm{kJ}{ \mathrm{mol}}^{-1}$. The bond dissociation energy of ${\mathrm{X}}_2$ will be

$\mathrm{LetB}.\mathrm{E}.\mathrm{of} {\mathrm{X}}_2,{\mathrm{Y}}_2 \mathrm{and} \mathrm{XY} \mathrm{are} \mathrm{x} \mathrm{kJ}{ \mathrm{mol}}^{-1},0.5\mathrm{x} \mathrm{kJ}{ \mathrm{mol}}^{-1}\mathrm{and} \mathrm{x} \mathrm{kJ}{ \mathrm{mol}}^{-1}\mathrm{respectively}.$

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

$\mathrm{\Delta H}=\mathrm{\Sigma }( \mathrm{B}.\mathrm{E}. {)}_{\mathrm{Reactant} }-\mathrm{\Sigma }( \mathrm{B}.\mathrm{E}. {)}_{\mathrm{Product} }$

$\therefore -200=\left[\frac{1}{2}\times \left(\mathrm{x}\right)+\frac{1}{2}\times (0.5\mathrm{x})\right]-[1\times (\mathrm{x}\left)\right]$

$\text{ B.E. of }{\mathrm{X}}_2=\mathrm{x}=800 \mathrm{kJ}{ \mathrm{mol}}^{-1}$