Question

The observed dipole moment of HCl is 1.03D. If the inter atomic distance is $1.27\stackrel{\circ }{A}$, calculate its % ionic character :

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$\mu =\delta \times d[1D=3.33\times {10}^{-30}cm]$
$\delta =\mu /d=\frac{1.03\times 3.33\times {10}^{-30}}{1.27\times {10}^{-10}}$
$=2.7\times {10}^{-20}C$
$\therefore$ $\%$ ionic character =$\frac{\delta }{{\delta }_{max}}\times 100$
$=\frac{2.7\times {10}^{-27}}{16\times {10}^{-19}}\times 100$
$=16.875\%\approx 17\%$