MCQ
Molecular $AB$ has a bond length of $1.61\,\mathop A\limits^o $ and a dipole moment of $0.38\, D$. The fractional charge on each atom (absolute magnitude) is $(e_0\, = 4.802\times10^{-10}\, esu)$
- A$0.5$
- ✓$0.05$
- C$0$
- D$1.0$
✓
Answer
Correct option: B.
$0.05$
b
$1\,e = 1.602 \times {10^{ - 19}}\,C$
$1\,e = 1.602 \times {10^{ - 19}}\,C$
$1\,esu = 3.33 \times {10^{ - 10}}\,C$
$\frac{{1\,e}}{{1\,esu}} = \frac{{1.602 \times {{10}^{ - 19}}\,C}}{{3.33 \times {{10}^{ - 10}}\,C}}$
$1\,e = 4.802 \times {10^{ - 10}}\,esu\,$
Dipole moment $= q \times $ distance
$ \Rightarrow \,1\,D \approx {10^{ - 18}}\,esu\,cm$
$0.38 \times {10^{ - 18}}\,esu\,cm = q \times (1.61 \times {10^{ - 8}}\,cm)$
$q = 2.36 \times {10^{ - 11}}\,esu$
$q = \frac{{2.36 \times {{10}^{ - 11}}\,esu}}{{4.802 \times {{10}^{ - 10}}\,esu}}$
$q = 0.049$
$q \approx 0.05$ fractional charge
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