Bond enthalpies of \(A_{2}, B_{2}\) and \(A B\) are in the ratio \(2: 1: 2\). If bond enthalpy of formation of \(A B\) is \(-100 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The bond enthalpy of \(B_{2}\) is

Given,
\((\mathrm{BE})_{A_{2}}:(\mathrm{BE})_{B_{2}}:(\mathrm{BE})_{A B}=2: 1: 2\)
where, \(B E=\) Bond enthalpy
\(\left(\Delta H_{f}\right)_{A B}=-100 \mathrm{~kJ}\)
Let the BE of \(A_{2}, B_{2}\) and \(A B\) be \(2 x, x\) and \(2 x\) respectively.
As we know, \(\Delta_{f} H^{\circ}=\Sigma \Delta_{\text {diss }} H_{R}^{\circ}-\Sigma \Delta_{\text {diss }} H_{P}^{\circ}\) where, \(\quad \Lambda_{\text {diss }} H^{\circ}=\) Enthalpy of dissociation
\(\begin{aligned}
\frac{1}{2} A_{2} &+\frac{1}{2} B_{2} \rightarrow A B \\
\therefore \quad-100 &=\left(\frac{1}{2} 2 x+\frac{1}{2} x\right)-x \\
x &=200
\end{aligned}\)
\(\therefore\) Bond enthalpy of \(B_{2}\) is \(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\).