# HBr molecule has internuclear distance of 1.27 ×10^–10 m. The electronic charge is 4.8 ×10^–10 esu. Observed dipole moment is 1.03 D. find % ionic character of the bond.

Q: HBr molecule has internuclear distance of 1.27 × 10–10 m. The electronic charge is 4.8 × 10–10 esu. Observed dipole moment is 1.03 D. find % ionic character of the bond.

Solution: % ionic character $\large = \frac{\mu_{observed}}{\mu_{Theo}} \times 100$

$\large = \frac{1.03 \times 100}{(\frac{2.035 \times 10^{-29}}{3.33 \times 10^{-30}})}$

= 16.8