If HCl molecule is completely polarized, so expected value of dipole moment is 6.12D (debye), but experimental value of dipole moment is 1.03D. Calculate the percentage ionic character

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Solution

The correct option is A 17
% Ionic character $= \frac{1.03}{6.12} \times 100 = 17$ % approx.

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If molecule is completely polarized, so expected value of dipole moment is 6.12D (debye), but experimental value of dipole moment is 1.03D. Calculate the percentage ionic character

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If HCl molecule is completely polarized, so expected value of dipole moment is 6.12D (debye), but experimental value of dipole moment is 1.03D. Calculate the percentage ionic character

The correct option is A 17% Ionic character =1.036.12×100=17% approx.

The correct option is A 17
% Ionic character $= \frac{1.03}{6.12} \times 100 = 17$ % approx.