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Calculate the percent ionic character if bond distance =1.42 angstom ,charge=4.8×10*10 e.s.u ,mu=1.08d find % i.c

Calculate the percent ionic character if bond distance =1.42 angstom ,charge=4.8×10*10 e.s.u ,mu=1.08d find % i.c

Grade:11

1 Answers

Vikas TU
14149 Points
one year ago
We know that
μ = e × d
For complete separation of unit charge the dipole moment is calculated as
μ=1.602×10^−19columb×1.275×10^−10m=2.04Cm
dipole moment of HCl = 1.03D = 0.34×10^−29Cm(1Cm=2.9979×10^29 D)
% of ionic character of HCl = Actual dipole moment/calculated dipole moment x 100
=0.34×10^−29Cm / 2.04×10^−29Cm×100 =  17%

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