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Bond distance in $\mathrm{HF}$ is $9.17 \times 10^{-11} \mathrm{~m}$. Dipole moment of $\mathrm{HF}$ is $6.104 \times 10^{-30} \mathrm{Cm}$. The percentage ionic character in $\mathrm{HF}$ will be : (electron charge $=1.60 \times 10^{-19} \mathrm{C}$ )
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Verified Answer
The correct answer is:
$41.5 \%$
$41.5 \%$
Given $e=1.60 \times 10^{-19} \mathrm{C}$
$$
d=9.17 \times 10^{-11} \mathrm{~m}
$$
From $\mu=e \times d$
$$
\begin{aligned}
\mu & =1.60 \times 10^{-19} \times 9.17 \times 10^{-11} \\
& =14.672 \times 10^{-30}
\end{aligned}
$$
$\%$ ionic character
$=\frac{\text { Observed dipole moment }}{\text { Dipole moment for } 100 \%}$
$$
\begin{aligned}
& =\frac{6.104 \times 10^{-30}}{14.672 \times 10^{-30}} \times 100 \\
& =41.5 \%
\end{aligned}
$$
$$
d=9.17 \times 10^{-11} \mathrm{~m}
$$
From $\mu=e \times d$
$$
\begin{aligned}
\mu & =1.60 \times 10^{-19} \times 9.17 \times 10^{-11} \\
& =14.672 \times 10^{-30}
\end{aligned}
$$
$\%$ ionic character
$=\frac{\text { Observed dipole moment }}{\text { Dipole moment for } 100 \%}$
$$
\begin{aligned}
& =\frac{6.104 \times 10^{-30}}{14.672 \times 10^{-30}} \times 100 \\
& =41.5 \%
\end{aligned}
$$
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