Bond length of hcl is 2.07 and it is 15% ionic. Find dipole moment
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μ=e×dμ=e×d
For complete separation of unit charge the dipole moment is calculated as
μ=1.602×10−19columb×1.275×10−10m=2.04Cmμ=1.602×10−19columb×1.275×10−10m=2.04Cm
Actual dipole moment of HCl = 1.03D = 0.34×10−29Cm(1Cm=2.9979×1029D)0.34×10−29Cm(1Cm=2.9979×1029D)
%% of ionic character of HCl = Actual dipole moment/calculated dipole moment x 100
=0.34×10−29Cm2.04×10−29Cm×100=17%
For complete separation of unit charge the dipole moment is calculated as
μ=1.602×10−19columb×1.275×10−10m=2.04Cmμ=1.602×10−19columb×1.275×10−10m=2.04Cm
Actual dipole moment of HCl = 1.03D = 0.34×10−29Cm(1Cm=2.9979×1029D)0.34×10−29Cm(1Cm=2.9979×1029D)
%% of ionic character of HCl = Actual dipole moment/calculated dipole moment x 100
=0.34×10−29Cm2.04×10−29Cm×100=17%
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