Question

The gaseous potassium chloride molecule has a measured dipole moment of $10.0\text{D}$, which indicates that it is a very polar molecule. The separation between the nuclei in this molecule is $2.67\times {10}^{–8}\text{cm}$. The percentage covalent character in KCl molecule is:

• A. $78\text{\%}$
• B. $22\text{\%}$
• C. $80\text{\%}$
• D. $25\text{\%}$

Answer 1

The correct option is B $22\text{\%}$

Given, ${\mu }_{observed}=10\text{D}$
For compound $\left(KCl\right)$ to be $100\text{\%}$ ionic, charge on each atom $=4.8\times {10}^{-10}\text{esu}$
$\therefore$ Theoretical dipole moment, ${\mu }_{theoretical}=(4.8\times {10}^{-10}\text{esu})(2.67\times {10}^{-8}\text{cm})$
${\mu }_{theoretical}=12.8\times {10}^{-18}\text{esu.cm}=12.8\text{D}$
$\therefore \text{\% ionic character}=\frac{{\mu }_{observed}}{{\mu }_{theoretical}}\times 100=\frac{10}{12.8}\times 100\text{\%}$ $=78.125\text{\%}$
$\text{Hence, \% covalent character}=100-78.125\text{\%}=21.875\text{\%}$ ~ 22 %