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Question

The observed dipole moment of HBr is 2.60×1030 and the interatomic spacing is 1.41˚A. What is the percentage ionic character of HBr?

A
12%
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B
15%
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C
5%
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D
10%
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Solution

The correct option is A 12%
% ionic character in a molecule =observeddipolemomentexpecteddipolemoment×100
Common unit of interatomic spacing=A° (Angstrom)
Hδ+BrδH+Br
Expected dipole moment=Q×d
Interatomic spacing=1.41A°
=1.41×108 cm
=141×1010 cm
Expected dipole moment=1.6×1019×141×1010
=255.6×1029 cm
Observed dipole moment=2.6×1030 cm
% Ionic character=2.6×1030225.6×1029×100
=2.6225.6×1000
=26225.6=11.55 % 12 %.

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