Let the bond dissociation energies of $X_2$, $Y_2$, and $XY$ be $x$, $0.5x$, and $x$ respectively. The reaction for the formation of $XY$ is: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightarrow XY$ The enthalpy of formation ($\Delta_f H$) is given by: $\Delta_f H = \sum \text{B.E. (reactants)} - \sum \text{B.E. (products)}$ $\Delta_f H = \frac{1}{2}\text{B.E.}(X_2) + \frac{1}{2}\text{B.E.}(Y_2) - \text{B.E.}(XY)$ Given that $\Delta_f H = -200 \text{ kJ mol}^{-1}$: $-200 = \frac{1}{2}(x) + \frac{1}{2}(0.5x) - x$ $-200 = 0.5x + 0.25x - x$ $-200 = -0.25x$ $x = \frac{200}{0.25} = 800 \text{ kJ mol}^{-1}$ Therefore, the bond dissociation energy of $X_2$ is $800 \text{ kJ mol}^{-1}$.