If dipole moment of AB molecule is 1.6 × 10−29 Cm and distance of A − B bond is 1.6 Å, then percentage ionic character of the compound will be
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Answer:
62.5% Answer
Explanation:
The diapole moment of 100% ionic character will be:
μ_ionic = q×d = 1.6×10^−19 C × 1.6× 10^-10 m
=2.56×10^-29 C m
The percentage ionic character is =
% Ionic Character = 62.5%