Question

If the dipole moment of $AB$ molecule is given by $1.2\, D$ and $A-B$ the bond length is $1\,Å$ then $\%$ ionic character of the bond is
[Given: 1 debye $=10^{-18}$ esu. $cm$ ]

• A. 75
• B. 50
• C. 60
• D. 25

Answer 1

Answer: (D)

$\mu_{\text {Cal }}=4.8 \times 10^{-10}$ e.s. $u \times 10^{-8} cm$
$=4.8 \times 10^{-18}$ esu cm
$=4.8 D$
Hence $\%$ ionic character $=\frac{\mu_{\text {experimental }}}{\mu_{\text {caculated }}} \times 100 =25\%$