Question:

Molecule AB has a bond length of 1.617 \(\mathrm{\AA}\) and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is : (\(e_0 = 4.802\times10^{-10}\) esu)

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Use \(\mu = q\times d\) with \(1\ \text{D} = 10^{-18}\) esu cm, and divide the charge q by \(e_0\).
Updated On: Nov 11, 2026
  • 0
  • 0.05
  • 0.5
  • 1.0
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The Correct Option is B

Solution and Explanation

\(\mu = 0.38\ \text{D} = 0.38\times10^{-18}\) esu cm; \(d = 1.617\ \text{Å} = 1.617\times10^{-8}\) cm.
Charge \(q = \frac{\mu}{d} = \frac{0.38\times10^{-18}}{1.617\times10^{-8}} = 2.35\times10^{-11}\) esu.
Fractional charge \(= \frac{q}{e_0} = \frac{2.35\times10^{-11}}{4.802\times10^{-10}} = 0.049 \approx 0.05\).
Answer: (2) 0.05
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