Question

Calculate the % ionic character in $HCl$ molecule . Given bond length of $HCl$ is $1.275$ Å and ${\mu }_{HCl}=1.03debye$

Answer 1

Dipole moment of $HCl\left({\mu }_{HCl}\right)$ is given by ,

${\mu }_{HCl}=\delta \times d$

$\because$ $\mu =1.03D=1.03\times {10}^{-18}esucm$

and $d=1.275\AA =1.275\times {10}^{-8}cm$

$\therefore$ $1.03\times {10}^{-18}=\delta \times 1.275\times {10}^{-8}$

$\therefore$ $\delta =0.808\times {10}^{-10}esu$

$\because$ $4.803\times {10}^{-10}esu$ charge , % ionic nature of $HCl=100$

$\therefore$ $0.808\times {10}^{-10}esu$ charge , % ionic nature of $HCl$

$=\frac{100\times 0.808\times {10}^{-10}}{4.803\times {10}^{-10}}=16.82$ %