The dipole moment of HBr is 2.60×10^-30 Cm and the interatomic distance is 141 pm.What is the percentage ionic character of HBr?
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absolute dipole moment = electron charge * interatommic distance.
μ100% = (1.6*10^-19)C*(1.41*10^-10)m
μ100% = 2.26*10^-29 Cm
and,
% ionic character = μ(obs) / μ100%
=>2.6*10^-30 / 2.26*10^-29
=> 0.12
=> 12%
hope this will help you..
μ100% = (1.6*10^-19)C*(1.41*10^-10)m
μ100% = 2.26*10^-29 Cm
and,
% ionic character = μ(obs) / μ100%
=>2.6*10^-30 / 2.26*10^-29
=> 0.12
=> 12%
hope this will help you..
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