The iodine monobromide molecule, IBr, has a bond length of \(2.49 \mathrm{~A}\) and a dipole moment of \(1.21 \mathrm{D}\). (a) Which atom of the molecule is expected to have a negative charge? Explain. (b) Calculate the effective charges on the \(\mathrm{I}\) and \(\mathrm{Br}\) atoms in IBr, in units of the electronic charge, \(e\).

From the periodic table, we know the electronegativities of Iodine and Bromine are 2.5 and 2.8 respectively. Now, calculate the difference in electronegativities: Electronegativity difference = \(|2.8 - 2.5|\) = \(0.3\) Since the electronegativity difference is positive, it means the Bromine (Br) atom is more electronegative than Iodine (I), and will have a partial negative charge.

The Bromine (Br) atom is expected to have a negative charge because it is more electronegative than the Iodine (I) atom.

We are given the bond length as \(2.49 \mathrm{~A}\) and the dipole moment as \(1.21 \mathrm{D}\). In order to calculate the effective charges, we need to convert these units into SI units. Convert the bond length into meters: \(2.49 \mathrm{~A} * (1 \times 10^{-10} \mathrm{m/A}) = 2.49 \times 10^{-10} \mathrm{m}\) Convert the dipole moment into Coulomb meters: \(1.21 \mathrm{D} * (3.336 \times 10^{-30} \mathrm{C\cdot m/D}) = 4.03616 \times 10^{-30} \mathrm{C\cdot m}\) Now, we can relate the dipole moment to the effective charge and bond length using the following formula: Dipole moment (µ) = Effective charge (Q) × Bond length (r) So, we can calculate the effective charge (Q) as follows: Effective charge (Q) = Dipole moment (µ) / Bond length (r) = \(4.03616 \times 10^{-30} \mathrm{C\cdot m} / 2.49 \times 10^{-10} \mathrm{m} = 1.62094 \times 10^{-20} \mathrm{C}\).

Since the electronic charge (e) is equal to \(1.602 \times 10^{-19} \mathrm{C}\), we can express the effective charge in terms of the electronic charge by dividing the effective charge by the electronic charge: Effective charge (Q) in units of electronic charge (e) = \(1.62094 \times 10^{-20} \mathrm{C} / 1.602 \times 10^{-19} \mathrm{C} = 0.1011\)

In the IBr molecule, the effective charges on the Iodine (I) and Bromine (Br) atoms are approximately +0.1011e and -0.1011e, respectively.