Bond length of ${\mathrm{H}}_2$  is $0.07 \mathrm{nm}$, Bond length of ${\mathrm{Cl}}_2$ is $1.98{ \mathrm{A}}^0$. Bond length of $\mathrm{HCl}$ is

$$\begin{gathered} \text{ Bond length of }{\mathrm{H}}_2=0.074\mathrm{nm} \\ \Rightarrow {\mathrm{r}}_{{\mathrm{H}}_2}=\frac{0.074}{2}\mathrm{nm}=0.37{\mathrm{A}}^{\circ } \\ \text{ Bondlength of }{\mathrm{Cl}}_2=1.98{\mathrm{A}}^{\circ } \\ \Rightarrow {\mathrm{r}}_{{\mathrm{Cl}}_2}=\frac{1.98{\mathrm{A}}^{\circ }}{2}=0.99{\mathrm{A}}^{\circ } \\ \text{ Bondlength of }\mathrm{HCl} \\ \Rightarrow {\mathrm{r}}_{{\mathrm{H}}_2}+{\mathrm{r}}_{{\mathrm{Cl}}_2}=0.37+0.99=1.027\mathrm{nm}. \end{gathered}$$

Hence, the correct option is option (C) 1.025nm.