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Chemistry Question on Bond Parameters

The bond length of HClHCl bond is 2.29×10−10 m2.29 \times 10^{-10}\, m. The percentage ionic character of HClHCl, if measured dipole moment is 6.226×10−30 C m6.226 \times 10^{-30}\, C \,m is

A

8%

B

20%

C

17%

D

50%

Answer

17%

Explanation

Solution

μcal=q×d=1.6×10−19×2.29×10−10{\mu_{cal}} = q \times d = 1.6 \times 10^{-19 } \times 2.29 \times 10^{-10}.
=3.664×10−29 C m= {3.664 \times 10^{-29}\, C\, m}
% ionic character = μexpμcal×100\frac{\mu_{exp}}{\mu_{cal}} \times 100
=6.226×10−303.664×10−29×100=17%= \frac{6.226 \times 10^{-30}}{3.664 \times 10^{-29}} \times 100 = 17 \%