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Chemical Bonding and Molecular Structure
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30/08/2021 10:05 am
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Bond distance in HF is 9.17 × 10–11 m. Dipole moment of HF is 6.104 × 10-30 Cm. The percentage ionic character in HF will be: (electron charge = 1.60 × 10-19 C)
(a) 61.0%
(b) 38.0%
(c) 35.5%
(d) 41.5%
1 Answer
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30/08/2021 10:10 am
Answer (d) 41.5%
Explanation:
Given e = 1.60 × 10–19 C
d = 9.17 × 10–11 m
From μ = e × d
μ = 1.60 × 10–19 × 9.17 × 10–11
= 14.672 × 10–30
% ionic character
= \(\frac{Observed\;dipole\;moment}{Calculated\;Dipole\;moment}\) x 100
= \(\frac{6.104 \times 10^{-30}}{14.672 \times 10^{-30}}\) x 100
= 41.5 %
This post was modified 3 years ago by admin