Question

The chlorine end of the chlorine monoxide radical carries a charge of $+0.167\, e$. The bond length is $154.6\, pm$. Calculate the dipole moment of the radical in Debye units.

• A. 2.35 D
• B. 1.24 D
• C. 1.59 D
• D. 2.05 D

Answer 1

Answer: (B)

$\vec{\mu}= q \cdot d$
$q =1.167 e \frac{1.602 \times 10^{-19} c }{1 e ^{-}}$
$=2.675 \times 10^{-20} C$
$d =154.6 \,pm =154.6 \times 10^{-12} m$
So $\vec{\mu}=\left(2.675 \times 10^{-20} C \right)\left(154.6 \times 10^{-12}\right)$
$=4.136 \times 10^{-30} Cm$
$=\frac{4.136 \times 10^{-30}}{3.34 \times 10^{-30}}=1.24 \,D$