Question

The experimental value of dipole moment of $HCl$ is $1.05D$. The length of the $H-Cl$ bond is $1.250A^{\circ }$. The percentage of ionic character in $HCl$ is: (Given, $\frac{2}{3.34}=0.6$)

Answer 1

Given:
The bond length of $H-Cl=1.200\stackrel{\circ }{A}$
Experimental dipole moment of $HCl=1.05D$

$\therefore$ Dipole moment assuming 100 % ionic bond $=e\times bondlength=1.6\times {10}^{-19}\times 1.250\times {10}^{-10}$
$=2.000\times {10}^{-29}\text{Cm}$

Unit conversion Cm to D
$3.34\times {10}^{-30}Cm=1D$
$\therefore \frac{2.000\times {10}^{-29}}{3.34\times {10}^{-30}}\text{D}\approx 6\text{D}$

We know,
$\text{\% ionic character}=\frac{\text{Observed dipole moment}}{\text{Calculated dipole moment (assuming 100 \% ionic bond)}}\times 100$
$=\frac{1.05}{6}\times 100$
$=17.5\%$