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Question
The dipole moment of H-Br molecule is 2.6×10^-30. It's bond length is 1.3A°. Then the percentage ionic character of H-Br ?
The dipole moment of H-Br molecule is 2.6×10^-30. It's bond length is 1.3A°. Then the percentage ionic character of H-Br ?
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Solution
The calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius.
μ = (electron charge) × (interatomic spacing) = (1.6⋅10^−19)C × (1.3⋅10^-10)m = 2.08×10^−29Cm
As you point out,
ionic character= [μobs]/[μ%] = [2.6×10^−30] / [2.08×10^−29] = 0.125
Or percentage ionic character =12.5 %
μ = (electron charge) × (interatomic spacing) = (1.6⋅10^−19)C × (1.3⋅10^-10)m = 2.08×10^−29Cm
As you point out,
ionic character= [μobs]/[μ%] = [2.6×10^−30] / [2.08×10^−29] = 0.125
Or percentage ionic character =12.5 %
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