H – F bond distance in HF molecule is 0.9Ao . Dipole moment of HF is 6×10−30Cm . The percentage ionic character of H−F will be : (charge of electron = 1.6×10−19C )

% ionic character = 6×10-301.6×10-19×9×10-11×100

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H – F bond distance in HF molecule is $0.9 A^{o}$ . Dipole moment of HF is $6 \times 10^{- 30} C m$ . The percentage ionic character of $H - F$ will be : (charge of electron = $1.6 \times 10^{- 19} C$ )

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% ionic character = $\frac{6 \times 10^{- 30}}{1.6 \times 10^{- 19} \times 9 \times 10^{- 11}} \times 100$

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