1. Tardigrade
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  3. Chemistry
  4. If the molecule of HCl were totally polar, the expected value of dipole moment is 6.12 D (debye) but the experimental value of dipole moment was 1.03 D. Calculate the percentage ionic character

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Q. If the molecule of HCl were totally polar, the expected value of dipole moment is 6.12 D (debye) but the experimental value of dipole moment was 1.03 D. Calculate the percentage ionic character

Chemical Bonding and Molecular Structure

Solution:

Percent ionic character
= $\frac{Obs. dipolemoment }{Cal. dipolemoment} \times 100$
= $\frac{1.03 \, D}{6.12 \, D} \times 100$ = 17 %