Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Interatomic distance in hydrogen$\left({\mathrm{H}}_2\right)$molecule$\left({\mathrm{L}}_{{\mathrm{H}}_2}\right)=74$pm and interatomic distance in chlorine$\left(\mathrm{C}{\mathrm{l}}_2\right)$molecule$\left({\mathrm{L}}_{\mathrm{C}{\mathrm{l}}_2}\right)=198\mathrm{ }\mathrm{p}\mathrm{m}$. Atoms linked by one$\mathrm{\sigma }$-bond, covalent radius$\left(\mathrm{r}\right)=\frac{\mathrm{I}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{i}\mathrm{c}\mathrm{ }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{2}$. Therefore covalent radius of hydrogen$\left({\mathrm{r}}_{\mathrm{H}}\right)=\frac{{\mathrm{L}}_{{\mathrm{H}}_2}}{2}=\frac{74}{2}=37$pm and that of chlorine$\left({\mathrm{r}}_{\mathrm{C}\mathrm{l}}\right)=\frac{{\mathrm{L}}_{\mathrm{C}{\mathrm{l}}_2}}{2}=\frac{198}{2}=99$pm

As${\text{H}}_2\left(\text{g}\right)+{\text{Cl}}_2\left(\text{g}\right)\to 2\text{HCL}\left(\text{g}\right)$

Therefore bond length of HCl molecule$\left({\mathrm{L}}_{\mathrm{H}\mathrm{C}\mathrm{l}}\right)$

$={\mathrm{r}}_{\mathrm{H}}+{\mathrm{r}}_{\mathrm{C}\mathrm{l}}=37+99=136$pm.