Step 1: {Using Given Data}
Let the bond dissociation energy of X\(_2\) be \(a\) kJ/mol.
Thus, the bond dissociation energy of Y\(_2\) is \(0.5a\) kJ/mol and the bond dissociation energy of XY is \(a\) kJ/mol.
The formation reaction is:
\[
\frac{1}{2} {X}_2 + \frac{1}{2} {Y}_2 \rightarrow {XY}, \, \Delta H = -200 \, {kJ/mol}
\]
Step 2: {Bond Energy Formula}
The bond dissociation energy can be calculated using the formula for enthalpy change:
\[
\Delta H = BE({Reactants}) - BE({Products})
\]
\[
\Delta H = \left[\frac{1}{2} \, BE({X}_2) + \frac{1}{2} \, BE({Y}_2)\right] - BE({XY})
\]
\[
\Delta H = \left[\frac{a}{2} + \frac{0.5a}{2}\right] - a
\]
\[
-200 = \frac{a + 0.5a}{2} - a
\]
\[
-200 = \frac{1.5a}{2} - a
\]
\[
-200 = 0.75a - a
\]
\[
-200 = -0.25a
\]
\[
a = 800 \, {kJ/mol}
\]
Thus, the correct answer is (D).