MCQ
If $HCl$ molecule is completely polarized, so expected value of dipole moment is $6.12\,D \,(deby),$ but experimental value of dipole moment is $1.03\,D.$ Calculate the percentage ionic character
- ✓$17$
- B$83$
- C$50$
- D$0$
✓
Answer
Correct option: A.
$17$
(a) $\%$ of ionic character =$\frac{{{\rm{Experimental\, value \,of \,dipole \,moment}}}}{{{\rm{Expected \,value \,of \,dipole \,moment}}}}$
$ = \frac{{1.03}}{{6.12}} \times 100 = 16.83\% \approx 17\% $
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