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

Question

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 (A) 17 (B) 83 (C) 50 (D) Zero

Tardigrade Admin · Accepted Answer

Percent ionic character = (Obs. dipolemoment /Cal. dipolemoment) × 100 = (1.03 D/6.12 D) × 100 = 17 %

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

17

83

50

Zero

Percent ionic character
= $\frac{O b s . d i p o l e m o m e n t}{C a l . d i p o l e m o m e n t} \times 100$
= $\frac{1.03 D}{6.12 D} \times 100$ = 17 %

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

17

83

50

Zero

Percent ionic character = Cal.dipolemomentObs.dipolemoment​×100 = 6.12D1.03D​×100 = 17 %

Percent ionic character
= $\frac{O b s . d i p o l e m o m e n t}{C a l . d i p o l e m o m e n t} \times 100$
= $\frac{1.03 D}{6.12 D} \times 100$ = 17 %