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Q.
The bond dissociation energies of $X_2 , Y_2$ and $XY $ are in the ratio of $1 : 0.5 :1. \Delta H$ for the formation of $XY$ is - $200\, kJ\, mol^{-1}$. The bond dissociation energy of $X_2$ will be
$\Delta H$ for the formation of $XY$ is $-200 kJ mol ^{-1}$.
The bond dissociation energies of $X _{2}, Y _{2}$ and $XY$ are in the ratio of $1: 0.5: 1$.
Let the bond dissociation energy of $X _{2}$ will be $kJ / mol$.
The bond dissociation energy of $Y _{2}$ will be $0.5 a kJ / mol$.
The bond dissociation energy of $XY$ will be $kJ / mol$.
$ \frac{1}{2} X _{2}+\frac{1}{2} Y _{2} \rightarrow XY $
$\Delta H$ (reaction) $=\Sigma \Delta H$ (reactant bonds ) $-\Sigma \Delta H$ ( product bonds)
$\Delta H ($ reaction $)=\frac{1}{2} \Delta H ( X - X )+\frac{1}{2} \Delta H ( Y - Y )-\Delta H ( X - Y )$
$ -200=\frac{1}{2}( a )+\frac{1}{2}(0.5 a )-( a ) $
$ -200=-0.25 \text { (a) } $
$ a =800 $
The bond dissociation energy of $X _{2}$ will be $800 kJ / mol$