Question

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?

Short answer

The short answer to the question is as follows: a) The Lewis structure for ClO2 is [Cl=O-O]-, after distributing the valence electrons and ensuring the central atom has an octet. b) ClO2 is readily reduced because of the high electronegativity of chlorine and oxygen, which easily attract electrons. c) The Lewis structure for ClO2- is [Cl=O-O*]-, with an additional electron due to the negative charge. d) The O-Cl-O bond angle in ClO2- is approximately 120 degrees because of the trigonal-planar electron domain geometry. e) Given the reaction conditions, 11.2 grams of ClO2 can be prepared.

Step-by-step solution

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.

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.

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.

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.

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.

Key concepts

Lewis Structure

The Lewis structure is a handy way of representing molecules by showing the arrangement of atoms and the electrons involved in bonding. When dealing with the molecule of chlorine dioxide \(\mathrm{ClO}_{2}\), it's important to first calculate the total valence electrons. Chlorine provides 7, while each oxygen contributes 6.Let's break it down:

\[\quad 1 \times 7 + 2 \times 6 = 19 \, \text{valence electrons}. \]

To draw the Lewis structure for \(\mathrm{ClO}_{2}\):

• Place the least electronegative atom (Cl) in the center.
• Connect each oxygen atom to chlorine with a single line (bond). This uses up 4 of the 19 valence electrons.
• Distribute the remaining 15 electrons around the oxygen atoms first to satisfy their octet requirement, then place any leftovers around chlorine.
• Check your structure. If Cl does not have an octet, it might share some electrons with oxygen to form double bonds.

This representation helps us understand the molecule's geometry and possible reactions.

Reduction Reactions

Reduction reactions are processes where a molecule, atom, or ion gains electrons. In the case of \(\mathrm{ClO}_{2}\), these reactions are significant. Why? The chlorine dioxide molecule will gain an electron to become a more stable chlorite ion, \(\mathrm{ClO}_{2}^{-}\). This is possible because of the electronegative nature of chlorine and oxygen.In essence, during the oxidation process, the material with which \(\mathrm{ClO}_{2}\) reacts loses electrons, while \(\mathrm{ClO}_{2}\) itself gains them, thus:

• \(\mathrm{ClO}_{2} + e^{-} \rightarrow \mathrm{ClO}_{2}^{-}\)

The ability of \(\mathrm{ClO}_{2}\) to attract and hold onto an extra electron makes it a strong oxidizing agent, facilitating this reduction.

Limiting Reactant Calculation

The limiting reactant is the substance that determines the maximum amount of product that can be formed in a chemical reaction. In our example, we consider the reaction of chlorine (\(\mathrm{Cl}_{2}\)) and sodium chlorite (\(\mathrm{NaClO}_{2}\)) to form chlorine dioxide (\(\mathrm{ClO}_{2}\)).Let's walk through how to determine which reactant limits the reaction:

• Imagine you have \(15.0\,\text{g of}\,\mathrm{NaClO}_{2}\). First, convert this mass into moles:\[m (\mathrm{NaClO}_{2}) = \frac{15.0\,\text{g}}{90.44\,\text{g/mol}} = 0.1659\,\text{moles}\,\mathrm{NaClO}_{2}.\]
• Using the ideal gas law, calculate the moles of \(\mathrm{Cl}_{2}\):\[ \left( 1.50\,\text{atm} \right) \cdot \left( 2.00\,\text{L} \right) = n \cdot \left( 0.0821\,\text{L} \cdot \text{atm/mol} \cdot \text{K} \right) \cdot \left( 294\,\text{K} \right) \] which gives \(n = 0.1223\,\text{moles of}\,\mathrm{Cl}_{2}.\)
• From the balanced chemical reaction, you need 1 mole of \(\mathrm{Cl}_{2}\) for every 2 moles of \(\mathrm{NaClO}_{2}\). Therefore:\[0.1223\,\text{moles of}\,\mathrm{Cl}_{2}\) would require \(0.2446\,\text{moles of}\,\mathrm{NaClO}_{2}\]
• With only \(0.1659\,\text{moles of}\,\mathrm{NaClO}_{2}\) available, \(\mathrm{NaClO}_{2}\) is the limiting reactant.

This tells us that the amount of \(\mathrm{ClO}_{2}\) we can produce depends on the \(\mathrm{NaClO}_{2}\) we have, making it the reactant that limits the creation of our product. Remember, always identify the limiting reactant in a reaction to know how much product you can expect to yield.