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

Question

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

Aicharya · Accepted Answer

Given $e=1.60 	imes 10^{-19} \mathrm{C}$ $ d=9.17 	imes 10^{-11} \mathrm{~m} $ From $\mu=e 	imes d$ $ \begin{aligned} \mu & =1.60 	imes 10^{-19} 	imes 9.17 	imes 10^{-11} \ & =14.672 	imes 10^{-30} \end{aligned} $ $\%$ ionic character $=\frac{	ext { Observed dipole moment }}{	ext { Dipole moment for } 100 \%}$ $ \begin{aligned} & =\frac{6.104 	imes 10^{-30}}{14.672 	imes 10^{-30}} 	imes 100 \ & =41.5 \% \end{aligned} $

Bond distance in $\mathrm{HF}$ is $9.17 \times 10^{-11} \mathrm{~m}$. Dipole moment of $\mathrm{HF}$ is $6…

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