Bond dissociation enthalpy of $H_2,C{l}_2$ and $HCl$ are 434, 242 and 431 kJ $mo{l}^{-1}$ respectively.

Enthalpy of formation of $HCl$ is:

The correct option is D $-93kJmo{l}^{-1}$

The reaction between hydrogen and chlorine molecules to form hydrogen chloride. This reaction can be broken down into these simple steps:

$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$

Step 1: Dissociation of Hydrogen and Chlorine molecules into their respective atoms

$H_2\left(g\right)\to 2H\left(g\right)$

$C{l}_2\left(g\right)\to 2Cl\left(g\right)$

Step 2: Combination of these atoms to form hydrogen chloride

$2H\left(g\right)+2Cl\left(g\right)\to 2HCl\left(g\right)$

Since in the first step, one-mole of each $H-H$ and $Cl-Cl$ bonds are broken, it is an endothermic process ($\Delta H_1>0$) while in the second step two moles of $HCl$ are being formed which releases energy making $\Delta H_2<0$

Now, $\Delta H_1=\Delta H_{Cl-Cl}+\Delta H_{H-H}$

$=434+242$

$=676kJmo{l}^{-1}$

Now, $\Delta H_2=-2\Delta H_{H-Cl}$

$=-862kJmo{l}^{-1}$

Applying Hess Law, i.e, $\Delta H^{{}^{\prime }}=\Delta H_1+\Delta H_2$, we get

$\Delta H^{{}^{\prime }}=-186kJmo{l}^{-1}$

This is the enthalpy of formation of two moles of $HCl$. Hence the enthalpy of formation of one mole of $HCl$ will be

$\Delta H=-93kJmo{l}^{-1}$