Bond dissociation energy XY, X₂ and Y₂, of are in the ratio 1:1:0.5 and $∆{\mathrm{H}}_{\mathrm{f}}$ of XY is $-200\mathrm{KJ}.{\mathrm{mol}}^{-1}$. The bond dissociation energy of X₂ will be

$\begin{array}{l}\frac{1}{2}{\mathrm{X}}_2+\frac{1}{2}{\mathrm{Y}}_2\to \mathrm{XY}.{\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}=-200,\mathrm{XY}:{\mathrm{X}}_2:{\mathrm{Y}}_2=2:2:1\\ \mathrm{\Delta H}=\sum \mathrm{B}\cdot {\mathrm{E}}_{\left(\text{Reactents }\right)}-\sum \mathrm{B}\cdot {\mathrm{E}}_{\left(\text{products }\right)}\\ -200=\frac{1}{2}[2\mathrm{x}+\mathrm{x}]-2\mathrm{x}\\ -200=\frac{1}{2}\left(3\mathrm{x}\right)-2\mathrm{x}, \mathrm{x}=400\\ \\ \text{ B.E of }{\mathrm{X}}_2=2\mathrm{x}=2\mathrm{x}400=800\mathrm{KJ}{\mathrm{mol}}^{-1}\end{array}$