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Q.
Bond enthalpies of $A_2$, $B_2$ and $AB$ are in the ratio $2 : 1 : 2$. If bond enthalpy of formation of $AB$ is $-100 \,KJ \,mol ^{-1} $.The bond enthalpy of $B _{2}$ is
Let bond enthalpy of $A _{2}$ be $x , B _{2}$ be $\frac{ x }{2},AB$ be $x$
$A _{2}+ B _{2} \rightarrow 2 AB$
$x +\frac{ x }{2}\,\,\,2x$
$\Delta H _{\text {form }}=\frac{1}{2} A _{2}+\frac{1}{2} B _{2} \rightarrow AB$
$-100=\left(\frac{ x }{2}+\frac{ x }{4}\right)- x$
$x =400$
$B _{2}=\frac{ x }{2}=200 \,kJ$