If the enthalpy of formation of a diatomic molecule $\mathrm{AB}$ is $-400 \mathrm{~kJ} / \mathrm{mol}$ and the bond dissociation energies of $\mathrm{A}_2$ and $\mathrm{B}_2$ and $\mathrm{AB}$ are in the ratio $2: 1: 2$, what is the bond dissociation enthalpy of $\mathrm{B}_2$ ?

The formation reaction for the diatomic molecule $\mathrm{AB}$ is given by: $\dfrac{1}{2} \mathrm{A}_2(g) + \dfrac{1}{2} \mathrm{B}_2(g) \rightarrow \mathrm{AB}(g)$.

The enthalpy of formation $\Delta H_f^{\circ}$ is related to the bond dissociation energies (BDE) by the equation: $\Delta H_f^{\circ} = \sum \text{BDE}(\text{reactants}) - \sum \text{BDE}(\text{products})$.

Let the bond dissociation energies of $\mathrm{A}_2$, $\mathrm{B}_2$, and $\mathrm{AB}$ be $2x$, $x$, and $2x$ respectively, based on the given ratio $2:1:2$.

Substituting these into the enthalpy equation: $-400 = \left( \dfrac{1}{2}(2x) + \dfrac{1}{2}(x) \right) - (2x)$.

Simplifying the expression: $-400 = x + 0.5x - 2x$.

$-400 = -0.5x$.

Solving for $x$: $x = \dfrac{400}{0.5} = 800 \text{ kJ/mol}$.

Since the bond dissociation energy of $\mathrm{B}_2$ is $x$, the value is $800 \text{ kJ/mol}$.

Answer: 800 kJ/mol