A diatomic molecule has a dipole moment of 1. | vedclass

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Question

(A) The fractional charge is calculated as the ratio of experimental dipole moment to the theoretical dipole moment (assuming $100\%$ ionic character)

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) The fractional charge is calculated as the ratio of experimental dipole moment to the theoretical dipole moment (assuming $100\%$ ionic character).
$	ext{Fractional charge} = \frac{\mu_{	ext{exp.}}}{\mu_{	ext{cal.}}}$
Given $\mu_{	ext{exp.}} = 1.2 \ D = 1.2 	imes 10^{-18} \ esu \ cm$.
The bond distance $d = 1 \ \mathring{A} = 10^{-8} \ cm$.
The theoretical dipole moment $\mu_{	ext{cal.}}$ for a full electronic charge $(e = 4.8 	imes 10^{-10} \ esu)$ is:
$\mu_{	ext{cal.}} = q 	imes d = (4.8 	imes 10^{-10} \ esu) 	imes (10^{-8} \ cm) = 4.8 	imes 10^{-18} \ esu \ cm$.
$	ext{Fractional charge} = \frac{1.2 	imes 10^{-18}}{4.8 	imes 10^{-18}} = 0.25$.
Expressing this as $	imes 10^{-2}$:
$0.25 = 25 	imes 10^{-2}$.

$A$ diatomic molecule has a dipole moment of $1.2 \ D$. If the bond distance is $1 \ \mathring{A}$,then fractional charge on each atom is.........$\times 10^{-2}$ (Given $1 \ D = 10^{-18} \ esu \ cm$ )

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