Calculate the percent ionic character of $H - F$ bond.
Given: $χ_{H} = 2.1, χ_{F} = 4.0$

43.04 %

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58.35 %

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34.28 %

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28.68 %

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Solution

The correct option is A 43.04 %
Using Hannay-Smith equation:
Percentage ionic character =
16( $χ_{A} - χ_{B}) + 3.5 (χ_{A} - χ_{B})^{2}$
Here, $χ_{A}$ = E.N of fluorine = 4.0
$χ_{B}$ = E.N of hydrogen = 2.1
Substituting values,
Percentage ionic character = 16(4.0 - 2.1) + 3.5( $4.0 - 2.1)^{2}$
Percentage ionic character = 43.04 %

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