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The dipole moment of KCI is 3.36 xx 10^(...

The dipole moment of `KCI` is `3.36 xx 10^(-29)Cm` The interatomic distance between `K^(o+)` and `CI^(Θ)` in this unit of `KCI` is `2.3 xx 10^(-10)m` Calculate the percentage ionic character of `KCI` .

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Verified by Experts

The correct Answer is:
`4.16xx10^(-29), 80.2%`
Dipole moment is calculated theoretically as `mu=q.d`
Here, `q=1.6xx10^(-19)C and d=2.6xx10^(-10)m, mu_("Theo")=1.6xx10^(-19)xx2.6xx10^(-10)=4.16xx10^(-29)cm`
% ionic character `=(mu_("obs"))/(mu_("Theo"))xx100=(3.336xx10^(-29))/(4.16xx10^(-29))xx100=80.2%`
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VMC MODULES ENGLISH-CHEMICAL BONDING-I & II-JEE Advanced (Archive)
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