Question

The bond length of HCl molecule is $1.275 \, \mathring{A}$ and its dipole moment is 1.03 D. The ionic character of the molecule (in percent) (charge of electron $4.8 \times 10^{-10}$ esu) is

• A. 100
• B. 67.3
• C. 33.6
• D. 16.83

Answer 1

Answer: (D)

% of ionic character $= \frac{\mu_{experimental}}{\mu_{theoretical}}\times100$
$μ_{ theoretical} = e \times d = 4.8 \times 10^{-10} \,esu \times 1.275 \times 10^{-8} \,cm$
$= 6.12 \times 10^{-18 }\,esu\, cm = 6.12 \,D$
$μ_{ experimental } = 1.03\, D$
$\therefore \quad\%$ of ionic character $= \frac{1.03}{6.12}\times 100 = 16.83$