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 = $4.8\times {10}^{-10}$ e.s.u

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

Answer 1

The correct option is B $-0.11$

Let the fractional charge on H = $+\delta$ so charge on Br = $-\delta$
Given, dipole moment $\mu$ of HBr = 0.78 D = $0.78\times {10}^{-18}esu.cm,d=1.41\stackrel{\circ }{A}$ = $1.41\times {10}^{-8}$ cm
We know, dipole moment $\mu$ = $q\times d$
So, q = $\frac{0.78\times {10}^{-18}esu.cm}{1.41\times {10}^{-8}cm}$ = 0.55 $\times {10}^{-10}$ esu
We can also say,
Fractional charge $\delta$ = $\frac{q}{e}$ = $\frac{0.55\times {10}^{-10}}{4.8\times {10}^{-10}}$ = 0.11
Therefore, ${\delta }_{H^{+}}$ = +0.11 and ${\delta }_{B{r}^{-}}$ = $-0.11$