Question

The dipole moment ($\mu$) of BrF is 1.42 D and the bond length is 176 pm. The atomic charge distribution ($q$) in the molecule is _______ (rounded off to two decimal places). (Given: 1 D = $3.34 \times 10^{-30}$ C m; electronic charge $e = 1.60 \times 10^{-19}$ C).

Hint:

Always convert dipole moment into SI units and then divide by bond length. Expressing $q$ in terms of $e$ helps compare the fraction of electron charge involved in bond polarity.

Answer 1

Step 1: Formula. Dipole moment $\mu = q \times d$. Hence, \[ q = \frac{\mu}{d}. \] Step 2: Convert dipole moment. \[ \mu = 1.42 \times (3.34 \times 10^{-30}) = 4.74 \times 10^{-30}\, \text{C m}. \] Step 3: Convert bond length. \[ d = 176\ \text{pm} = 176 \times 10^{-12} = 1.76 \times 10^{-10}\, \text{m}. \] Step 4: Compute charge. \[ q = \frac{4.74 \times 10^{-30}}{1.76 \times 10^{-10}} \approx 2.69 \times 10^{-20}\, \text{C}. \] Step 5: Express in units of \(e\). \[ \frac{q}{e} = \frac{2.69 \times 10^{-20}}{1.60 \times 10^{-19}} \approx 0.168 \approx 0.17. \] Therefore, the atomic charge distribution is \(\mathbf{0.17\, e}\).