Question

Calculate the fractional charge on bromine atom in HBr.
Given, dipole moment of HBr = 0.78 D
Bond distance of HBr=1.41 $\stackrel{\circ }{A},$ electronic charge
$e=4.8\times {10}^{-10}esu$

• A. 0.11
• B. -0.11
• C. -0.22
• D. 0.22

Answer 1

The correct option is B -0.11

$\mu =0.78D=0.78\times {10}^{-18}\text{esu cm}$
$d=1.41\stackrel{\circ }{A}=1.41\times {10}^{-8}\text{cm}$
$\mu =q\times d$
$0.78\times {10}^{-18}=q\times 1.41\times {10}^{-8}$
$q=\frac{0.78\times {10}^{-18}}{1.41\times {10}^{-8}}=5.53\times {10}^{-11}esu$
Fractional charge$=\frac{q}{e}=\frac{5.53\times {10}^{-11}}{4.48\times {10}^{-10}}=0.11$
Fractional charge on bromine atom should be negative. Hence, the correct answer is option (b).