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Question

Calculate the percent ionic character of HF bond.
Given: χH=2.1, χF=4.0
  1. 43.04 % 
  2. 58.35 %
  3. 34.28 %
  4. 28.68 %


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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