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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
Explanation
$\Delta H$ = bonds broken – bonds formed.
Detailed Solution
The reaction for $\Delta_fH^\circ(XY)$: $\frac{1}{2}X_2(g) + \frac{1}{2}Y_2(g) \rightarrow XY(g)$
Bond energies of $X_2$, $Y_2$ and XY are X, $\frac{X}{2}$ and X respectively.
$\Delta H = \frac{X}{2} + \frac{X}{4} - X = -200$
On solving, $-\frac{X}{4} = -200$
X = 800 kJ/mol