MCQ
The dipole moment of $HBr$ is $1.6 \times {10^{ - 30}}\,cm$ and interatomic spacing is $1\,\mathop A\limits^o $. The $\%$ ionic character of $HBr$ is
- A$7$
- ✓$10$
- C$15$
- D$27$
✓
Answer
Correct option: B.
$10$
(b) Charge of ${e^ - } = 1.6 \times {10^{ - 19}}$
Dipole moment of $HBr = 1.6 \times {10^{ - 30}}$
Inter atomic spacing $ = 1\,{\rm{{\mathop A\limits^o }}} = 1 \times {10^{ - 10}}\,m$
$\%$ of ionic character in
$HBr = \frac{{{\rm{dipole}}\,{\rm{moment}}\,{\rm{of}}\,HBr \times 100}}{{{\rm{inter}}\,{\rm{spacing}}\,{\rm{distance}}\, \times q}}$
$ = \frac{{1.6 \times {{10}^{ - 30}}}}{{1.6 \times {{10}^{ - 19}} \times {{10}^{ - 10}}}} \times 100$
$ = {10^{ - 30}} \times {10^{29}} \times 100 = {10^{ - 1}} \times 100$ $ = 0.1 \times 100 = 10\% $
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