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Question

Bond length in HCI is 1.275 $Å$ and experimental value of dipole moment is 1.03 D. Percentage ionic character of HCI will be?

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Solution

$For HCl μ_{\exp} = 1 .03 D$
$\begin{matrix} Bond length = 1.275 Å \\ = 1.275 \times 10^{- 8} cm \end{matrix}$
$\begin{matrix} Charge of electron = 4.8 \times 10^{- 10} esu \\ 1 D = 10^{- 18} esucm \end{matrix}$
$\begin{matrix} % Ionic character of HCl = \frac{1.03 \times 10^{- 18} \times 100}{4.8 \times 10^{- 10} \times 1.275 \times 10^{- 8}} \\ = 16.83 % \end{matrix}$

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Bond length in HCI is 1.275 Å and experimental value of dipole moment is 1.03 D. Percentage ionic character of HCI will be?

For HCl μexp=1.03D Bond length =1.275Å=1.275×10−8cm Charge of electron =4.8×10−10esu 1D=10−18esucm% Ionic character of HCl=1.03×10−18×1004.8×10−10×1.275×10−8=16.83%

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Bond length in HCI is 1.275 $Å$ and experimental value of dipole moment is 1.03 D. Percentage ionic character of HCI will be?