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  4. 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)

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Q. Bond distance in $HF$ is $9.17 \times 10^{-11}\, m$. Dipole moment of $HF$ is $6.10^4 \times 10^{-30}\, Cm$ . The percentage ionic character in $HF$ will be : (electron charge $= 1.60 \times 10^{-19}\, C$)

JEE MainJEE Main 2013Chemical Bonding and Molecular Structure

Solution:

Givene $e = 1.60 \times 10^{-19}\,C$
$d=9.17\times10^{-11}\,m$
From $\mu=e\times d$
$\mu=1.60\times10^{-19}\times9.17\times10^{-11}$
$=14.672\times10^{-30}$
$\%$ ionic character
$=\frac{Observed\, dipole\, moment}{Dipole\, moment\, for\, 100\%\,ionic\, bond}$
$=\frac{6.104\times10^{-30}}{14.672\times10^{-30}}\times100$
$=41.5\%$