The substance chlorine monoxide, \(\mathrm{ClO}(g)\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D},\) and the \(\mathrm{Cl}-\mathrm{O}\) bond length is \(160 \mathrm{pm} .(\mathbf{a})\) Determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in units of the electronic charge, \(e .(\mathbf{b})\) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion ClO \(^{-}\) exists. What is the formal charge on the Cl for the best Lewis structure for \(\mathrm{ClO}^{-} ?\)

Given: Dipole moment (P) = 1.24 D Bond length (d) = 160 pm = \(1.60 \cdot 10^{-10} m\) Electric charge (e) = \(1.60 \cdot 10^{-19} C\) Dipole moment (D) is related to the charge and bond length by the formula: D = q * d Where q is the magnitude of the charges on Cl and O atoms, and d is the bond length. We can calculate q as follows: q = \(\frac{D}{d}\) q = \(\frac{1.24D}{1.60 \cdot 10^{-10} m}\) q = \(\frac{1.24 \times 3.34 \times 10^{-30} \mathrm{C}\/{m}}{1.60 \times 10^{-10} \mathrm{m}}\) q = \(2.57 \times 10^{-20}\mathrm{C}\) Now, we have the charge magnitude and need to express it in units of electric charge (e). q = \(2.57 \times 10^{-20} \mathrm{C}\) Magnitude of charges in units of e: q = \(\frac{2.57 \times 10^{-20}\mathrm{C}}{1.60 \times 10^{-19}\mathrm{C}}\) q = 0.16e The magnitude of the charges on Cl and O atoms is 0.16e.

From the periodic table, the electronegativity values for Cl and O are: Electronegativity of Cl = 3.16 Electronegativity of O = 3.44 Since the electronegativity of oxygen (O) is higher, it will have a partial negative charge in the ClO molecule.

Using formal charges as a guide, we can draw the Lewis structure for ClO as follows: O - Cl ⋅ Oxygen has a single bond to Cl, and Cl has 3 lone electron pairs, as Cl has 7 valence electrons, and one is shared with O to create the bond. Formal charge (FC) = valence electrons - (lone electrons + 1/2(bonded electrons)) Formal charge for Cl = 7 - (6 + 1/2(2)) = 0 Formal charge for O = 6 - (4 + 1/2(2)) = 0 With both formal charges equal to 0, this is the dominant Lewis structure for ClO.

To determine the formal charge on Cl in the ClO⁻ anion, we first draw the Lewis structure: O = Cl ⋅ The extra electron is placed on Cl, making it negatively charged. The bond is now a double bond, and there are 2 lone electron pairs on Cl and O. Formal charge for Cl in ClO⁻ = 7 - (4 + 1/2(4)) = -1 The formal charge on Cl in the best Lewis structure for ClO⁻ is -1.