Question

The molecule $AB$ has a bond length of $ 1.61\mathring{A} $ & a dipole moment of $0.38 D$. The fractional charge on each atom is: (charge on $\left.1 e ^{-}=-4.8 \times 10^{-10} esu \right)$

• A. 0.5
• B. 0.05
• C. 0
• D. 1

Answer 1

Answer: (B)

$1D =10^{-18}$ esu cm

$m = q \times d$

$0.380= q \times 1.61\mathring{A}$

$0.38 \times 10^{-18} esu = q 1.61 \times 10^{-8} cm$

$q =\frac{0.38 \times 10^{-18}}{1.61 \times 10^{-8}}=0.23 \times 10^{-10} esu$

Fractional charge on each atom

$=\frac{0.23 \times 10^{-10} esu }{4.8 \times 10^{-10} esu }$

$=0.05$