The dipole moment of HCl is 1.03 D. If the H | vedclass

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(D) Dipole moment is the product of the magnitude of charge and the separation between charges. $P = q 	imes d$ For a fully ionic bond,one electron i

Vedclass · Accepted Answer

$17$

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(D) Dipole moment is the product of the magnitude of charge and the separation between charges. $P = q 	imes d$ For a fully ionic bond,one electron is transferred from hydrogen to chlorine. Magnitude of charge,$q = 1.6 	imes 10^{-19} \ C$. Bond length,$d = 1.26 \ \mathring{A} = 1.26 	imes 10^{-10} \ m$. Theoretical dipole moment,$P_{theoretical} = 1.6 	imes 10^{-19} \ C 	imes 1.26 	imes 10^{-10} \ m = 2.016 	imes 10^{-29} \ Cm$. Since $1 \ D = 3.335 	imes 10^{-30} \ Cm$,$P_{theoretical} = \frac{2.016 	imes 10^{-29}}{3.335 	imes 10^{-30}} \ D \approx 6.045 \ D$. Percentage ionic character = $\frac{	ext{Experimental dipole moment}}{	ext{Theoretical dipole moment}} 	imes 100$ $= \frac{1.03 \ D}{6.045 \ D} 	imes 100 \approx 17.04 \ \%$. Thus,the percentage of ionic character is approximately $17 \ \%$.

The dipole moment of $HCl$ is $1.03 \ D$. If the $H-Cl$ bond distance is $1.26 \ \mathring{A}$,what is the percentage of ionic character in the $H-Cl$ bond? ............. $\%$

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