Bond distance in HF

Question:

Bond distance in HF is $9.17 \times 10^{-11} \mathrm{~m}$. Dipole moment of $\mathrm{HF}$ is $6.104 \times 10^{-30} \mathrm{Cm}$. The percent ionic character in HF will be : (electron charge $=1.60 \times 10^{-19} \mathrm{C}$ )

  1. $61.0 \%$

  2. $38.0 \%$

  3. $35.5 \%$

  4. $41.5 \%$


Correct Option: , 4

Solution:

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