Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The dipole moment ($\mu$) is given by $\mu =q\times r$, where $q$ is the charge and $r$ is the bond length.

Given $\mu =1.2$ D and $r=1\times 10^{-8}$ cm.

Convert D to C cm: $1$ D $=3.33564\times 10^{-30}$ C m $=3.33564\times 10^{-28}$ C cm.

So, $\mu =1.2\times 3.33564\times 10^{-28}$ C cm $=4.002768\times 10^{-28}$ C cm.

The charge $q=\frac{\mu }{r}=\frac{4.002768\times 10^{-28}Ccm}{1\times 10^{-8}cm}=4.002768\times 10^{-20}$ C.

The charge of an electron is $e=1.602\times 10^{-19}$ C.

Fraction of charge $=\frac{q}{e}=\frac{4.002768\times 10^{-20}C}{1.602\times 10^{-19}C}\approx 0.25$.

The final answer is $\boxed{0.25}$.