Question

If the dipole moment of AB molecule is given by 2.4 D and A-B the bond length is 1 A then % covalent character of the bond is:

Answer 1

$\mu$ is the dipole moment.
The formula to calculate the $\mu$ is
$\mu =e.d$ , where "e" is the charge of the electron and "d" is the bond legnth $1A^o$
${\mu }_{cal}=4.8\times {10}^{-10}e.s.u.\times {10}^{-8}cm=4.8\times {10}^{-18}esu.cm=4.8D.$
% of covalent character = $\frac{{\mu }_{obs}}{{\mu }_{cal}}\times 100$
$\therefore$ % ionic character $=\frac{2.4}{4.8}\times 100=50\%$
$\therefore$ % co-valent character = 50%