Question

If internuclear distance between $Cl$ atoms in $C{l}_2$ is $10$A & between $H$ atoms in $H_2$ is 2 A, then the internuclear distance between $H$ & $Cl$ (Electronegativity of $H=2.1$ & $Cl=3.0)$ is:

• A. $5.919$ A
• B. $4.911$ A
• C. $5.871$ A
• D. None of these

Answer 1

Answer: (A)

$5.919$ A

The radius of chlorine is one half the distance between two $Cl$ atoms.
$r_{Cl}=\frac{10A^0}{2}=5A^0$
The radius of hydrogen is one half the distance between two $H$ atoms.
$r_H=\frac{2A^0}{2}=1A^0$
The internuclear distance between $H$ and $Cl$ is
$r_{HCl}=r_H+r_{Cl}-0.09({\chi }_{Cl}-{\chi }_H)=1+5-0.09(3.0-2.1)=5.919A^0$
Thus, the internuclear distance between $H$ and $Cl$ is $5.919A^0$