Dipole moment of HCl = 1.03 D,HI = 0.38 D. | vedclass

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Question

(C) The dipole moment is given by the formula $\mu = \delta 	imes d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond lengt

Vedclass · Accepted Answer

$3.3: 1$

Vedclass · Answer

(C) The dipole moment is given by the formula $\mu = \delta 	imes d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.Therefore,the fraction of charge is $\delta = \frac{\mu}{d}$.For the ratio of charges in $HCl$ and $HI$:$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{\mu_{HCl}}{d_{HCl}} 	imes \frac{d_{HI}}{\mu_{HI}}$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 $3.3 : 1$.

Dipole moment of $HCl = 1.03 \ D$,$HI = 0.38 \ D$. 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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