Problem 124 - Chemistry The Central Science In Si Units

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. \((\mathbf{e})\) 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?

Step-by-Step Solution

Answer

16.73 g of ClO2 can be produced.

Step 1: Lewis Structure for ClO2

To draw the Lewis structure for \(\mathrm{ClO}_2\), first calculate the total number of valence electrons: Chlorine (Cl) has 7 valence electrons, and each Oxygen (O) has 6. Therefore, the total is \(7 + 2 \times 6 = 19\) electrons. Distribute these electrons by forming a single Cl-O bond for each oxygen, accounting for 4 electrons. Place 11 more electrons around the two oxygens and chlorine to complete the octet, resulting in a structure with 1 unpaired electron on chlorine and oxygen.

Step 2: Understanding Reduction of ClO2

\(\mathrm{ClO}_2\) is easily reduced because it has an odd number of electrons, making it a free radical. Free radicals are generally unstable and highly reactive, eager to pair the unpaired electron, which makes \(\mathrm{ClO}_2\) able to readily accept electrons, hence it is easily reduced.

Step 3: Lewis Structure for ClO2-

For the \(\mathrm{ClO}_2^-\) ion, begin with the total number of valence electrons as in \(\mathrm{ClO}_2\), which is 19. Add one more electron for the negative charge, resulting in 20 electrons. Distribute the electrons similarly, but now you can place pairings or lone pairs more evenly, leading to each atom achieving an octet, resulting in all electrons being paired.

Step 4: Predict the O-Cl-O Bond Angle in ClO2-

The \(\mathrm{ClO}_2^-\) is a bent molecule following VSEPR theory. The central Cl atom forms two single bonds with oxygen and has two lone pairs, leading to a tetrahedral electron arrangement. The ideal tetrahedral bond angle is \(109.5^\circ\), but due to the lone pairs causing repulsion, the actual \(\mathrm{O-Cl-O}\) bond angle will be slightly less, likely around \(104.5^\circ\).

Step 5: Calculate Moles of NaClO2 and Cl2

Calculate moles of sodium chlorite: \( \text{Molar mass of } \mathrm{NaClO}_2 = 22.99 + 35.45 + 2 \times 16 = 90.44 \, \text{g/mol} \), so \(\frac{15.0 \, \text{g}}{90.44 \, \text{g/mol}} \approx 0.166 \, \text{moles} \). For \(\mathrm{Cl}_2\), use the ideal gas law: \( PV = nRT \) where \(P = 152.0 \, \text{kPa}, V = 2.00 \, \text{L}, R = 8.314 \, \text{L}\cdot\text{kPa/mol}\cdot\text{K}, T = 294 \, \text{K} (21^\circ \text{C})\). Rearrange to solve for \(n\): \( n = \frac{PV}{RT} = \frac{152 \times 2}{8.314 \times 294} \approx 0.124 \, \text{moles} \).

Step 6: Determine Limiting Reagent

Compare mole stoichiometry: According to the reaction \(\mathrm{Cl}_2 + 2 \mathrm{NaClO}_2 \to 2 \mathrm{ClO}_2 + 2 \mathrm{NaCl}\), 1 mole of \(\mathrm{Cl}_2\) reacts with 2 moles of \(\mathrm{NaClO}_2\). The ratio of \(0.124 \) moles \(\mathrm{Cl}_2 : 0.166\) moles \(\mathrm{NaClO}_2\) indicates that \(\mathrm{Cl}_2\) is the limiting reagent since there is not enough \(\mathrm{Cl}_2\) to react with all \(\mathrm{NaClO}_2\).

Step 7: Calculate Mass of ClO2 Produced

From the balanced equation, \(1 \text{ mole of } \mathrm{Cl}_2\) produces \(2\) moles of \(\mathrm{ClO}_2\). Since we have \(0.124\) moles of \(\mathrm{Cl}_2\), this will produce \(0.248\) moles \(\mathrm{ClO}_2\). Calculate the mass using the molar mass of \(\mathrm{ClO}_2 = 67.45 \, \text{g/mol}\). Therefore, the mass of \(\mathrm{ClO}_2\) produced is \(0.248 \times 67.45 \approx 16.73 \, \text{g}\).

Key Concepts

Chlorine Dioxide

Chlorine dioxide is a chemical compound represented by the formula \( \mathrm{ClO}_2 \). It is a yellowish-green gas and is highly valued as a bleaching agent in various industrial applications, notably in the paper and water treatment industries. Chlorine dioxide functions as a potent oxidizing agent, which means it can accept electrons from other substances, facilitating the oxidation reaction. This happens because the \( \mathrm{ClO}_2 \) molecule has an odd number of valence electrons, leading to one unpaired electron. As a free radical, \( \mathrm{ClO}_2 \) is highly reactive. This unique characteristic makes it susceptible to reduction during chemical reactions, often converting itself to the chlorite ion, \( \mathrm{ClO}_2^-\), as it gains additional electrons.
Understanding the Lewis structure for \( \mathrm{ClO}_2 \) is key to recognizing its stability issues. Chlorine has 7 valence electrons while oxygen has 6, bringing the total count for \( \mathrm{ClO}_2 \) to 19 electrons. These electrons form bonds and lone pairs in a way that leaves one unpaired electron, rendering \( \mathrm{ClO}_2 \) reactive. This reactivity is harnessed in oxidation reactions, where \( \mathrm{ClO}_2 \) undergoes reduction by accepting electrons and transforming into a more stable form.

Oxidation-Reduction Reactions

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two substances. In any redox reaction, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. This electron transfer results in changes to the oxidation states of the substances involved.
In the context of chlorine dioxide, \( \mathrm{ClO}_2 \) acts generally as an oxidizing agent. During the bleaching process, \( \mathrm{ClO}_2 \) itself is reduced while oxidizing the substances it interacts with. The reduction in this scenario involves the gain of electrons, giving rise to the chlorite ion, \( \mathrm{ClO}_2^-\).
Every redox reaction is composed of two half-reactions: one for oxidation and one for reduction. In the case of \( \mathrm{ClO}_2 \), the reduction half-reaction can be written as:
\[ \mathrm{ClO}_2 + e^- \rightarrow \mathrm{ClO}_2^- \]
This setup enables an easy understanding of how electron transfer results in the molecular changes, characteristic of all oxidation-reduction reactions. The overall redox balance is usually maintained by equalizing the gain and loss of electrons across the reaction components.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) Theory helps predict the 3D shape of molecules. According to VSEPR theory, the shape of a molecule is determined by the repulsions between all of the electron pairs in the valence shell of the central atom. This includes both bonding pairs (those electrons shared in a bond) and lone pairs (those not in bonding).
In the chlorite ion, \( \mathrm{ClO}_2^-\), chlorine forms two single bonds with oxygen atoms, and there are two lone pairs of electrons on the central chlorine atom. These lone pairs repel each other and the bonding pairs, pushing the \( \mathrm{O-Cl-O} \) bonds closer together and forming a bent shape. Typically, a molecule with four groups of electrons (including lone pairs) around a central atom will have a geometry corresponding to tetrahedral electron arrangement. However, due to the increased repulsion from the lone pairs, the bond angles are compressed.
The general tetrahedral angle is \(109.5^\circ\), but the presence of two lone pairs will reduce this angle in the \( \mathrm{ClO}_2^-\) ion to roughly \(104.5^\circ\). This distortion exemplifies how VSEPR theory helps us understand and predict the molecular geometry adjustments based on electron pair interactions.

Limiting Reagent

The concept of a limiting reagent is central to understanding how chemical reactions proceed. In a chemical reaction, the limiting reagent is the substance that is entirely consumed first, thus limiting the amount of product that can be formed.
For the reaction of chlorine \((\mathrm{Cl}_2)\) and sodium chlorite \((\mathrm{NaClO}_2)\) to form chlorine dioxide \((\mathrm{ClO}_2)\), determining the limiting reagent involves comparing the mole ratios of the reactants based on the balanced equation:

• \( \mathrm{Cl}_2(g) + 2 \mathrm{NaClO}_2(s) \rightarrow 2 \mathrm{ClO}_2(g) + 2 \mathrm{NaCl}(s) \)

In this setup, stoichiometry dictates that 1 mole of chlorine reacts with 2 moles of sodium chlorite.
To find the limiting reagent, calculate the moles of each reactant:

• First, calculate the moles of \( \mathrm{NaClO}_2 \) from its mass.
• Use the ideal gas law \(( PV = nRT )\) to find the moles of \( \mathrm{Cl}_2 \).

If the available moles of \( \mathrm{Cl}_2 \) are less than what is needed for the corresponding moles of \( \mathrm{NaClO}_2 \), then \( \mathrm{Cl}_2 \) is the limiting reagent. This is crucial for calculating the maximum amount of product, \( \mathrm{ClO}_2 \), that can be generated during the reaction, based solely on the limiting reagent's initial amount.