Molecular AB has a bond length of 1.61,mathri | vedclass

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(B) The dipole moment $\mu$ is given by the formula $\mu = q 	imes d$,where $q$ is the charge and $d$ is the bond length.Given $\mu = 0.38\,D = 0.38

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(B) The dipole moment $\mu$ is given by the formula $\mu = q 	imes d$,where $q$ is the charge and $d$ is the bond length.Given $\mu = 0.38\,D = 0.38 	imes 10^{-18}\,esu\,cm$ and $d = 1.61\,\mathring{A} = 1.61 	imes 10^{-8}\,cm$.Calculating the actual charge $q$ in $esu$: $q = \frac{\mu}{d} = \frac{0.38 	imes 10^{-18}\,esu\,cm}{1.61 	imes 10^{-8}\,cm} \approx 2.36 	imes 10^{-11}\,esu$.The fractional charge is the ratio of the actual charge $q$ to the electronic charge $e_0 = 4.802 	imes 10^{-10}\,esu$.Fractional charge $= \frac{q}{e_0} = \frac{2.36 	imes 10^{-11}\,esu}{4.802 	imes 10^{-10}\,esu} \approx 0.049 \approx 0.05$.

List-$I$	List-$II$
$(1)$ ${\rm{HBr}}$	$(A)$ $0.23$
$(2)$ ${{\rm{H}}_2}{\rm{S}}$	$(B)$ $1.04$
$(3)$ ${\rm{N}}{{\rm{F}}_3}$	$(C)$ $1.07$
$(4)$ ${\rm{CC}}{{\rm{l}}_4}$	$(D)$ $0.79$
$(5)$ ${\rm{CHC}}{{\rm{l}}_3}$	$(E)$ $0.00$
	$(F)$ $0.95$

Molecular $AB$ has a bond length of $1.61\,\mathring{A}$ and a dipole moment of $0.38\,D$. The fractional charge on each atom (absolute magnitude) is $(e_0 = 4.802 \times 10^{-10}\,esu)$

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