The dipole moment of a molecule AB is 0.50 D and the bond distance is

1.	The dipole moment of a molecule AB is 0.50 D and the bond distance is 1.41 Å Calculate the fractional charge \(\delta\) on A and B atom in AB molecule (electronic charge x 10-10 esu).
Answer» Dipole moment = q x r μ = 0.50D = 0.50 × 10⁻¹⁸ stat C cm r = 1.41 x 10^-8cm q = \(\frac{\mu}{r}\) = \(\frac{0.50\times10^{-18}\,stat \,C\,cm}{1.41\times10^{-8}cm}\) Now, Fraction of charge = \(\frac{Charge\,present}{Electronic\,charge}\) = \(\frac{0.35\times10^{-10}}{4.8\times10^{-10}}\) = 00.7 \(\therefore\) \(\delta_A\) = +0.07 and \(\delta_B\) = -0.07

The dipole moment of a molecule AB is 0.50 D and the bond distance is 1.41 Å Calculate the fractional charge \(\delta\) on A and B atom in AB molecule (electronic charge x 10-10 esu).

Answer»

Dipole moment = q x r

μ = 0.50D = 0.50 × 10⁻¹⁸ stat C cm

r = 1.41 x 10^-8cm

q = \(\frac{\mu}{r}\) = \(\frac{0.50\times10^{-18}\,stat \,C\,cm}{1.41\times10^{-8}cm}\)

Now, Fraction of charge = \(\frac{Charge\,present}{Electronic\,charge}\)

= \(\frac{0.35\times10^{-10}}{4.8\times10^{-10}}\) = 00.7

\(\therefore\) \(\delta_A\) = +0.07

and \(\delta_B\) = -0.07