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Question

Calculate the percent ionic character of HF bond.
Given: χH=2.1, χF=4.0

A
43.04 %
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B
58.35 %
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C
34.28 %
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D
28.68 %
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Solution

The correct option is A 43.04 %
Using Hannay-Smith equation:
Percentage ionic character =
16(χAχB)+3.5(χAχB)2
Here, χA = E.N of fluorine = 4.0
χB = E.N of hydrogen = 2.1
Substituting values,
Percentage ionic character = 16(4.0 - 2.1) + 3.5(4.02.1)2
Percentage ionic character = 43.04 %

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