Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\), molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with $2.00 \mathrm{~L}\( of chlorine gas at a pressure of \)152.0 \mathrm{kPa}$ at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

To draw the Lewis structure for ClO\(_2\), start by calculating the total number of valence electrons in the molecule. Chlorine (Cl) has 7 valence electrons, and Oxygen (O) has 6 valence electrons. Hence, the total count of valence electrons for ClO\(_2\) is \(7 + 2(6) = 19\). Since Cl has the least electronegativity, it will be the central atom surrounded by O atoms. First, form single bonds between Cl and both O atoms. Now, 15 valence electrons remain to be distributed. Both the O atoms complete their octet, leaving one extra electron that cannot be paired. This unpaired electron will be placed on the central chlorine atom, giving it a formal charge of +1. The Lewis structure for ClO\(_2\) will be: O=Cl-O • Where the Cl atom has one unpaired electron and a +1 formal charge.

ClO\(_2\) can be readily reduced due to the presence of an unpaired electron on the central chlorine atom. This unpaired electron gives ClO\(_2\) a higher tendency to gain an electron to complete its octet and gain overall stability. Therefore, ClO\(_2\) acts as a strong oxidizing agent and itself gets easily reduced.

For ClO\(_2^{-}\), we again start by counting the total number of valence electrons. This time, one additional electron is added due to the negative charge: \(7 + 2(6) + 1 = 20\). In this structure, we will have 20 valence electrons to distribute. Just like the ClO\(_2\) case, we first form single bonds between Cl and the two O atoms. Distribute the remaining 16 valence electrons to complete the octet of oxygen atoms. In this case, since the total number of valence electrons is even, we don't have a spare electron, and all the electrons are combined to form stable octets. The Lewis structure for ClO\(_2^{-}\) is: O=Cl-O With all atoms surrounded by octets.

To predict the O-Cl-O bond angle in ClO\(_2^{-}\), we need to consider the molecular geometry of the ion. The Cl atom is surrounded by 3 electron regions: 2 bonding pairs (each with oxygen) and 1 non-bonding pair (lone pair). According to VSEPR Theory, the electron regions will attempt to stay as far apart from each other as possible to minimize repulsion. In this case, ClO\(_2^{-}\) molecule has a bent geometry, and the electron domains are arranged in a trigonal planar arrangement. The ideal bond angle for a trigonal planar arrangement is 120°. However, the lone pair-bonding pair repulsion is slightly stronger than bonding pair-bonding pair repulsion. As a result, the O-Cl-O bond angle will be less than 120°.

From the reaction: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2\mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ We are given: - Mass of NaClO\(_2\) = \(15.0\) g - Volume of Cl\(_2\) gas = \(2.00\) L - Pressure of Cl\(_2\) gas = \(152.0\) kPa - Temperature of Cl\(_2\) gas = \(21^\circ\) C \(=294.15\) K First, we need to find the limiting reactant. We will do this by calculating the moles of both reactants. 1. Moles of NaClO\(_2\) $$\text{moles of NaClO}_{2} =\frac{\text{mass}}{\text{molar mass}} = \frac{15.0 \text{ g}}{90.44 \text{ g/mol}} = 0.1658 \text{ moles}$$ 2. Moles of Cl\(_2\) For this, we will use the Ideal Gas Law, \(PV = nRT\). Rearranging the formula to solve for moles \(n\): $$n = \frac{PV}{RT} = \frac{152.0 \,\text{kPa} \times 2.00 \,\text{L}}{8.314 \,\text{J/mol K} \times 294.15 \,\text{K}}$$ Converting kPa to atm: $$1 \text{kPa} = 0.009869 \text{atm}$$ Therefore: $$n = \frac{152.0 \cdot 0.009869 \,\text{atm} \times 2.00 \,\text{L}}{0.0821 \,\text{L/atm mol K} \times 294.15 \,\text{K}} = 0.01288 \text{ moles}$$ Now, we can identify the limiting reactant with the given stoichiometry: $$\frac{\text{moles of NaClO}_{2}}{2} = 0.0829 \text{ moles}$$ $$\frac{\text{moles of Cl}_2}{1} = 0.01288 \text{ moles}$$ Since \(0.01288 < 0.0829\), Cl\(_2\) is the limiting reactant. The moles of ClO\(_{2}\) produced will be equal to the moles of limiting reactant, which is 0.01288 moles. Now, we can calculate the mass of ClO\(_{2}\) produced: $$\text{mass of ClO}_{2} = \text{moles} \times \text{molar mass} = 0.01288 \text{ moles} \times 67.45 \text{ g/mol} = 0.8683 \text{ g}$$ So, under these conditions, \(0.8683\) grams of ClO\(_{2}\) can be prepared.