# Problem: Use the dipole moments of HF and HCl (HCl µ = 1.08 D;{ m{D}}, HF µ = 1.82 D;{ m{D}}) together with the percent ionic character of each bond (HCl 20% ionic, HF 45% ionic) to estimate the bond length in each molecule.

###### FREE Expert Solution

Recall that dipole moment (μ) can be calculated using the following equation:

$\overline{){\mathbf{\mu }}{\mathbf{=}}{\mathbf{Q}}{\mathbf{·}}{\mathbf{r}}}$

where:
Q = charge of a bond with 100% ionic character = 1.6 x 10-19 C
r = distance between the atoms or bond length

On the other hand, the percent ionic character is given by:

where:
μobserved = observed or actual dipole moment
μ100% ionic char. = dipole moment of a bond with 100% ionic character

For HF:

Step 1: Calculate the dipole moment of a bond with 100% ionic character.

Given:

μ = 1.82 D
% ionic character = 45%.

μ100% ionic char. = 5.4 D

Step 2: Calculate the bond length of HF.

Given:

1 D = 3.34 x 10-30 C•m
μ = 5.4 D
Q = 1.6 x 10-19 C

r = 8.44x10-11 m

converto picometer (pm):     1 pm = 10-12 m

r = 84.4 pm

The bond length of the H-F bond is 84.4 pm.

For HCl: ###### Problem Details

Use the dipole moments of HF and HCl (HCl µ = 1.08 D, HF µ = 1.82 D) together with the percent ionic character of each bond (HCl 20% ionic, HF 45% ionic) to estimate the bond length in each molecule.