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 1.50 atm at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

Step by step solution

01

a) Lewis structure for ClO2

First, count the total number of valence electrons. Chlorine has 7 valence electrons and each oxygen atom has 6 valence electrons. Thus, the total number of valence electrons is 1(7) + 2(6) = 19. 1. Place the least electronegative atom in the center (in this case, Cl). 2. Connect the atoms by single bonds. 3. Distribute the remaining valence electrons as lone pairs. 4. Verify that the central atom has an octet.

02

b) Reason for ClO2's easy reduction

ClO2 is easily reduced because chlorine is a highly electronegative element and can easily attract electrons. Additionally, the presence of oxygen, another highly electronegative element, further increases the overall electronegativity of the molecule.

03

c) Lewis structure for ClO2-

We follow the same steps as for ClO2, but now there is an additional electron due to the negative charge, resulting in a total of 20 valence electrons. 1. Place the least electronegative atom in the center (Cl). 2. Connect the atoms by single bonds. 3. Distribute the remaining valence electrons as lone pairs. 4. Verify that the central atom has an octet.

04

d) O-Cl-O bond angle in ClO2-

The central chlorine atom has three electron domains (two single bonds to oxygen atoms and one lone pair). This electron domain geometry is trigonal-planar, which predicts bond angles of approximately 120 degrees for the O-Cl-O bond.

05

e) Grams of ClO2 prepared

1. From the ideal gas law, calculate the moles of Cl2: PV = nRT -> n = PV /(RT) n(Cl2) = (1.50 atm)(2.00 L) / ((0.0821 L*atm/mol*K)(294 K)) = 0.1223 mol Cl2 2. Determine the limiting reactant from the given amounts of reactants: 1 mole of Cl2 reacts with 2 moles of NaClO2, so: 0.1223 mol Cl2 would need 0.2446 mol NaClO2 Given mass of NaClO2 = 15.0 g Molar mass of NaClO2 = 22.99 + 35.45 + 2(16.00) = 90.44 g/mol Moles of NaClO2 = 15.0g / 90.44 g/mol = 0.1659 mol NaClO2 is less than the required amount, so it is the limiting reactant. 3. Calculate the moles of ClO2 produced: 2 moles of NaClO2 form 2 moles of ClO2, so: 0.1659 mol NaClO2 will produce 0.1659 mol ClO2 4. Calculate the mass of ClO2 produced: Molar mass of ClO2 = 35.45 + 2(16.00) = 67.45 g/mol Mass of ClO2 = 0.1659 mol * 67.45 g/mol = 11.2 g Therefore, 11.2 grams of ClO2 can be prepared given the reaction conditions.

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