Calculate percentage ionic character of HCl given that observed dipole moment is 1.03D and bond length is 1.275 angstrom.

Dear student
 
Formula used :
μ = e × d
For complete separation of unit charge (100% ionic character) the dipole moment calculated as:
μ = 1.602 × 10–19 coulomb × 1.275 × 10–10 m
= 2.04 C m
Actual dipole moment of HCl = 1.03 D = 0.34 x 10–29 C m (1 C m = 2.9979×1029 D.)
% ionic character of HCl = Actual dipole moment/calculated dipole moment × 100 =
0.34 x 10-29 C m / 2.04 × 10–29 C m ×100 = 17%
 
Regards
Arun (askIITians forum expert)

Formula used :
μ = e × d
For complete separation of unit charge (100% ionic character) the dipole moment calculated as:
μ = 1.602 × 10–19 coulomb × 1.275 × 10–10 m
= 2.04 C m
Actual dipole moment of HCl = 1.03 D = 0.34 x 10–29 C m (1 C m = 2.9979×1029 D.)
% ionic character of HCl = Actual dipole moment/calculated dipole moment × 100 =
0.34 x 10-29 C m / 2.04 × 10–29 C m ×100 = 17%