The iodine monobromide molecule, IBr, has a bond length of \(249 \mathrm{pm}\) and a dipole moment of \(1.21 \mathrm{D} .(\mathbf{a})\) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, \(e\).

Determine the electronegativity of both Iodine (I) and Bromine (Br) atoms. From the periodic table, we have the following approximate electronegativity values: Iodine (I): 2.66 Bromine (Br): 2.96 Since Bromine is more electronegative than Iodine, the electrons in the I-Br bond will be closer to Bromine, making it slightly negatively charged.

As we established in the previous step, Bromine is more electronegative than Iodine, and thus has a partial negative charge. This means that the dipole moment vector points towards the Bromine atom. Answer to part (a): Bromine atom is expected to have a negative charge.

The given dipole moment is in Debye (D), and we need to convert it to SI units (Coulomb meter, C m) before proceeding to the next step. The conversion factor is 1 D = \(3.336 \times 10^{-30}\) C m. Dipole moment in SI units: \(1.21 \, \mathrm{D} \times \frac{3.336 \times 10^{-30} C m}{1 \, \mathrm{D}} \approx 4.03 \times 10^{-30}\) C m

To calculate the effective charges on Iodine and Bromine atoms in IBr, we will use the following equation for the dipole moment: \( \mathrm{Dipole\, moment} = \mathrm{Effective\, charge\, difference} \times \mathrm{Bond\, length} \) Let the effective charge on Iodine be \(q_I\) and on Bromine be \(q_{Br}\). Since the total charge of the molecule is neutral, we have: \(q_I + q_{Br} = 0 \) \( q_{Br} = -q_I \) Now, we plug the values for the dipole moment and bond length into the equation: \(4.03 \times 10^{-30}\, \mathrm{C\, m} = (q_{Br} - q_I) \times 249 \times 10^{-12}\, \mathrm{m} \) Rearrange and solve for \(q_I\): \(q_I = \frac{4.03 \times 10^{-30}\, \mathrm{C\, m}}{249 \times 10^{-12}\, \mathrm{m}} = 1.62 \times 10^{-18}\) C With \(q_{Br} = -q_I\), we have: \(q_{Br} = -1.62 \times 10^{-18}\) C Now, divide the effective charges by the electronic charge \(e (1.6 \times 10^{-19}\, \mathrm{C})\) to get the effective charges in units of e: \(q_I (e) = \frac{1.62 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 10.1\, e \) \(q_{Br} (e) = \frac{-1.62 \times 10^{-18}}{1.6 \times 10^{-19}} \approx -10.1\, e \) Answer to part (b): The effective charges on the I and Br atoms in IBr are approximately 10.1e and -10.1e, respectively.