Question

Bond distance in HF is 9.17 × 10⁻¹¹ m. Dipole moment of HF is 6.104 × 10⁻³⁰ Cm. The percentage of ionic character in HF will be ______.

(electron charge = 1.60 × 10⁻¹⁹ C)

Options

• 61%

• 38%

• 35.5%

• 41.5%

Answer 1

Bond distance in HF is 9.17 × 10⁻¹¹ m. Dipole moment of HF is 6.104 × 10⁻³⁰ Cm. The percentage of ionic character in HF will be 41.5%.

Explanation:

Given: electron charge (e) = 1.60 × 10⁻¹⁹ C

d = 9.17 × 10⁻¹¹ m

From µ = e × d

µ = 1.60 × 10⁻¹⁹ × 9.17 × 10⁻¹¹

= 14.672 × 10⁻³⁰

% of ionic character = `"Observed dipole moment"/"Dipole moment for 100% ionic bond"`

= `(6.104 xx 10^-30)/(14.672 xx 10^-30) xx 100`

= 41.5%