The internuclear distances in H_2 and Cl_2 mo | vedclass

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(C) The bond length of a molecule $AB$ is given by the sum of the covalent radii of atoms $A$ and $B$,i.e.,$d_{AB} = r_A + r_B$. For $H_2$,the internu

Vedclass · Accepted Answer

$136$

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(C) The bond length of a molecule $AB$ is given by the sum of the covalent radii of atoms $A$ and $B$,i.e.,$d_{AB} = r_A + r_B$. For $H_2$,the internuclear distance is $74 \, pm$,so the covalent radius of $H$ is $r_H = \frac{74}{2} = 37 \, pm$. For $Cl_2$,the internuclear distance is $198 \, pm$,so the covalent radius of $Cl$ is $r_{Cl} = \frac{198}{2} = 99 \, pm$. Therefore,the bond length of $HCl$ is $d_{HCl} = r_H + r_{Cl} = 37 + 99 = 136 \, pm$.

The internuclear distances in $H_2$ and $Cl_2$ molecules are $74 \, pm$ and $198 \, pm$ respectively. What will be the bond length of $HCl$ in $pm$?

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