Question

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

• A. $0$
• B. $0.05$
• C. $0.5$
• D. $1.0$

Answer 1

The correct option is B $0.05$

Dipole moment $\left(\mu \right)=q\times d$

$\mu =0.38D$ (given) = $0.38\times {10}^{-18}$ esu cm

$q=\frac{\mu }{d}=\frac{0.38\times {10}^{-18}}{1.617\times {10}^{-8}}$

fractional charge = $\frac{particlecharge}{Totalcharge}$

$=\frac{0.38\times {10}^{-18}}{1.617\times {10}^{-8}\times 4.802\times {10}^{-10}}$

$=0.048$