Question

If the observed dipole moment of $HCl$ is $1.03\text{D}$ and the distance between the atoms is $1.27\times {10}^{-8}\text{cm}$, what will be the percentage ionic character of $HCl$?

• A. 83%
• B. 25%
• C. 17%
• D. 90%

Answer 1

The correct option is C 17%

$\%ioniccharacter=\frac{Observeddipolemoment}{theoriticaldipolemoment}\times 100$
If `HCl` is `100%` ionic
$HCl\to H^{+}+C{l}^{-}$
$q=1.6\times {10}^{-19}C$ charge on a electron
$d=1.27\times {10}^{-8}cm$
$d=1.27\times {10}^{-8}\times {10}^{-2}$ meter
$1D=3.335\times {10}^{-30}$ coulomb-meter
Theoritical dipole moment
$=\frac{q\times d}{3.335\times {10}^{-30}}=6.09D$
$\%ioniccharacter=\frac{1.03}{6.09}\times 100=17\%$ approximately