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The dipole momnet of KCI is 3.336 xx 10^...

The dipole momnet of `KCI` is `3.336 xx 10^(-29) Cm` which indicates that it is a highly polar molecules. The inter atomic distance between `K^(o+)` and `CI^(Theta)` in this molecules is `2.6 xx10^(-10)` m Calculate the dipole moment of `KCI` molecule if there were opposite charges of one fundamental unit located at each nucleus Calculate the ionic character percentage of `KCI` .

Text Solution

Verified by Experts

Assuming `100%` ionic character in `KCI` calculated dipole momnet `=q xx d`
`mu_(cal) = 1.6 xx 10^(-19) xx 2.6 xx 10^(-10)Cm`
` = 4.1 xx 10^(-29)Cm`
Percentage ionic character ` =(mu_(obs))/(mu_(cal))xx100`
`=(3.336 xx 10^(-29) xx 100)/(4.1 xx 10^(-29))`
`=81.36%` .
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