calculate the percentage of ionic character of HCl molecule .Dipole moment of HCl is 1.03 D $ bond length is 1.275 A^0.electronic charge is 4.8 x 10^-10

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 $\times$ 100 =
 0.34 x 10^-29 C m / 2.04 × 10^–29 C m $\times$100 = 17%