Question

Bond distance in HF is 9.17 x 10-11m. Dipole
moment of HF is 6.104 x 10-30 Cm. The
percentage of ionic character in HF will be
(electron charge = 1.60 x 10-19 C)​

Answer 1

The  percentage of ionic character in HF is 41.6%

Given:

Bond distance = 9.17 × 10⁻¹¹ m

Dipole moment = 6.104 × 10⁻³⁰ Cm

Charge of electron = 1.60 × 10⁻¹⁹ C

Explanation:

The ionic character in HF is given by the formula:

% HF = (Observed dipole moment)/(Calculated dipole moment) × 100

The observed dipole moment is given in the question.

The calculated dipole moment is given as:

Dipole moment = Charge of electron × Bond distance

On substituting the values, we get,

Dipole moment = 1.60 × 10⁻¹⁹ × 9.17 × 10⁻¹¹

Dipole moment = 14.67 × 10⁻³⁰ Cm

Now, the percentage of ionic character is:

% HF = (6.104 × 10⁻³⁰)/(14.67 × 10⁻³⁰) × 100

% HF = 0.416 × 100

∴ % HF = 41.6%