To find the bond enthalpy of $$\mathrm{A}_{2}$$, we can use the given information about the reaction and the bond enthalpies' ratio. The reaction is:

$$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$$

The enthalpy change for the reaction, $$\Delta H_{f}^{0}$$, is given as:

$$\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

The bond enthalpies of $$\mathrm{A}_{2}$$, $$\mathrm{B}_{2}$$, and $$\mathrm{AB}$$ are in the ratio of $$1: 0.5: 1$$.

Let's denote the bond enthalpies of $$\mathrm{A}_{2}$$, $$\mathrm{B}_{2}$$, and $$\mathrm{AB}$$ as $$x$$, $$0.5x$$, and $$x$$, respectively.

The enthalpy change for the reaction can be calculated using the bond enthalpies:

$$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$$

For the given reaction:

$$-200 = (x + 0.5x) - 2x$$

Now we can solve for $$x$$, which represents the bond enthalpy of $$\mathrm{A}_{2}$$:

$$-200 = 1.5x - 2x$$

$$-200 = -0.5x$$

$$x = \frac{-200}{-0.5}$$

$$x = 400$$

The bond enthalpy of $$\mathrm{A}_{2}$$ is $$400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$$.