Question

Calculate the percentage of ionic character of NaF(Sodium Fluoride). If EN of Na and F are 0.9 and 4 respectively.

Answer 1

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference ΔχΔχ.

Percent ionic character =(1−e−(Δχ/2)2)×100=(1−e−(Δχ/2)2)×100

But I'd like to correct the definition of percent ionic character in your question using dipole moment μμ (not Observed value of ionic character):

Percent ionic character = μobservedμcalculatedμobservedμcalculated ×100%×100%

Where μcalculatedμcalculated is calculated assuming a 100% ionic bond.

% of ionic character = 16x(∆EN) + 3.5x(∆EN)² ...(where ∆EN is electronegativity difference)

For e.g. in H-F,(∆EN = 2) % of ionic character = 16x2 + 3.5x(2)² = 32 + 14 = 46%
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I hope you help !!!!