To solve this problem, we need to use the concept of bond dissociation energies and the enthalpy change for the formation of a compound.$\\$Given:$\\$• Bond dissociation energies of $X_2, Y_2,$ and $XY$ are in the ratio $1:0.5:1. \\$• $\Delta H$ for the formation of $XY$ is $-200$ kJ mol$^{-1}. \\$Let's denote:$\\$• Bond dissociation energy of $X_2$ as $E_{X_2}. \\$• Bond dissociation energy of $Y_2$ as $E_{Y_2}. \\$• Bond dissociation energy of $XY$ as $E_{XY}. \\$According to the given ratio:$\\ E_{X_2} : E_{Y_2} : E_{XY} = 1 : 0.5 : 1 \\$This implies:$\\ E_{Y_2} = 0.5 \cdot E_{X_2} \\ E_{XY} = E_{X_2} \\$The reaction for the formation of $XY$ is:$\\ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightarrow XY \\$The enthalpy change $\Delta H$ for this reaction is given by:$\\ \Delta H = \left( \frac{1}{2} E_{X_2} + \frac{1}{2} E_{Y_2} \right) - E_{XY} \\$Substitute the known values:$\\ -200 = \left( \frac{1}{2} E_{X_2} + \frac{1}{2} \cdot 0.5 \cdot E_{X_2} \right) - E_{X_2} \\$Simplify:$\\ -200 = \left( \frac{1}{2} E_{X_2} + \frac{1}{4} E_{X_2} \right) - E_{X_2} \\ -200 = \frac{3}{4} E_{X_2} - E_{X_2} \\ -200 = \frac{3}{4} E_{X_2} - \frac{4}{4} E_{X_2} \\ -200 = -\frac{1}{4} E_{X_2} \\$Solve for $E_{X_2}: \\ E_{X_2} = 200 \times 4 \\ E_{X_2} = 800$ kJ mol$^{-1} \\$Therefore, the bond dissociation energy of $X_2$ is $800$ kJ mol$^{-1}. \\$This corresponds to Option 1.