If the bond dissociation energies of $$XY$$, $$X_2$$ and $$Y_2$$ (all diatomic molecules) are in the ratio of $$1 : 1 : 0.5$$ and $$\Delta_f H$$ for the formation of $$XY$$ is $$-200$$ kJ mole$$^{-1}$$. The bond dissociation energy of $$X_2$$ will be

The standard enthalpy of formation of a gaseous diatomic molecule $$XY(g)$$ is defined for the reaction

$$\tfrac12 X_2(g) + \tfrac12 Y_2(g) \rightarrow XY(g)$$

To apply bond‐energy data, remember:

• Breaking a bond requires energy (positive).
• Forming a bond releases the same amount of energy (negative sign).

Hence, for the above reaction,

$$\Delta_f H^\circ = \Bigl[\tfrac12 D(X_2) + \tfrac12 D(Y_2)\Bigr] - D(XY)$$ $$-(1)$$

The question states the bond dissociation energies are in the ratio

$$D(XY) : D(X_2) : D(Y_2) = 1 : 1 : 0.5$$

Let $$k$$ be the common factor. Then

$$D(XY)=k,\; D(X_2)=k,\; D(Y_2)=0.5k$$ $$-(2)$$

The given standard enthalpy of formation is $$\Delta_f H^\circ = -200\;{\rm kJ\,mol^{-1}}$$. Substitute (2) in (1):

$$-200 = \bigl[\tfrac12 k + \tfrac12(0.5k)\bigr] - k$$

$$\;\;\; = (0.5k + 0.25k) - k = 0.75k - k = -0.25k$$

Thus

$$-200 = -0.25k \;\;\Longrightarrow\;\; k = 800\;{\rm kJ\,mol^{-1}}$$

From (2), the bond dissociation energy of $$X_2$$ is $$D(X_2)=k=800\;{\rm kJ\,mol^{-1}}$$.

This value is not listed among the options A, B, or C, so the correct choice is

Option D which is: None of these