Dipole moment of HCl = 1.03 D and HI = 0.38 | vedclass

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(C) From the definition of dipole moment,$\mu = \delta 	imes d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.The

Vedclass · Accepted Answer

$3.3: 1$

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(C) From the definition of dipole moment,$\mu = \delta 	imes d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.Therefore,$\delta = \frac{\mu}{d}$.The ratio of the fraction of electric charge for $HCl$ and $HI$ is given by:$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{\mu_{HCl}}{d_{HCl}} 	imes \frac{d_{HI}}{\mu_{HI}}$Substituting the given values:$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{1.03 	imes 1.6}{1.3 	imes 0.38} = \frac{1.648}{0.494} \approx 3.33: 1$.Thus,the ratio is approximately $3.3: 1$.

Dipole moment of $HCl = 1.03 \ D$ and $HI = 0.38 \ D$. Bond length of $HCl = 1.3 \ \mathring{A}$ and $HI = 1.6 \ \mathring{A}$. The ratio of the fraction of electric charge,$\delta$,existing on each atom in $HCl$ and $HI$ is:

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