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: $${\mathrm{Cl}}_2(g)+2{\mathrm{NaClO}}_2(s)⟶2{\mathrm{ClO}}_2(g)+2\mathrm{NaCl}(s)$$ 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?

Short answer

The Lewis structure for ClO${}_2$ is O=Cl-O with an unpaired electron on the Cl atom. ClO${}_2$ is readily reduced due to the presence of this unpaired electron. For ClO${}_2^{-}$, the Lewis structure is O=Cl-O with a filled octet. The O-Cl-O bond angle in ClO${}_2^{-}$ is slightly less than 120°. By calculating the limiting reactant (Cl${}_2$) and moles of ClO${}_2$ produced, 0.8683 g of ClO${}_2$ can be prepared when 15.0 g of NaClO${}_2$ reacts with 2.00 L of Cl${}_2$ gas at 152.0 kPa and 21°C.

Step-by-step solution

1. Draw the Lewis structure for ClO${}_2$

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.

2. Why is ClO${}_2$ readily reduced?

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.

3. Draw the Lewis structure for ClO${}_2^{-}$

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.

4. Predict the O-Cl-O bond angle in ClO${}_2^{-}$

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°.

5. Calculate the mass of ClO${}_2$ produced

From the reaction: $${\mathrm{Cl}}_2(g)+2{\mathrm{NaClO}}_2(s)⟶2{\mathrm{ClO}}_2(g)+2\mathrm{NaCl}(s)$$ 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.

Key concepts

VSEPR Theory

Imagine trying to arrange your friends in a circle for a group photo. That’s similar to what VSEPR Theory does for molecules. The acronym stands for Valence Shell Electron Pair Repulsion. This theory helps us understand the arrangement of bonds and lone pairs around a central atom in a molecule.

In the case of $ext{ClO}_2^{-}$, chlorine is the central atom. It is surrounded by two oxygen atoms and a lone pair. VSEPR Theory tells us that these groups will arrange themselves to be as far apart as possible. This is to minimize the repulsion between electron pairs. In $ext{ClO}_2^{-}$, this arrangement results in a 'bent' shape.

• There are three total electron groups around chlorine: one lone pair and two bonds to oxygen.
• The electron pairs vibrate to decrease repulsion strength, leading to the molecular structure.

The repulsion from the lone pair is stronger than that between bonded pairs. Therefore, the ideal bond angle of 120° in a perfect trigonal planar shape becomes slightly less in the bent $ext{ClO}_2^{-}$ due to this extra repulsion.

Oxidizing Agent

An oxidizing agent is a chemical species that accepts electrons from another substance during a chemical reaction. When ${\text{ClO}}_2$ comes into play, it acts as a powerful oxidizing agent. This means that ${\text{ClO}}_2$ accepts electrons and gets reduced itself.

Here's why ${\text{ClO}}_2$ is so effective:

• It possesses an unpaired electron, making it highly reactive and eager to obtain that missing electron to stabilize itself.
• By gaining an electron, ${\text{ClO}}_2$ converts to the chlorite ion ${\text{ClO}}_2^{-}$, completing its electron shell and gaining a stable configuration.

This reactivity and stability-seeking facilitate its role in bleaching processes, where it oxidizes and breaks down the color within materials, leaving them paler.

Limiting Reactant

Imagine you’re baking cookies. You may have more than enough flour, but if you run out of sugar first, you can’t bake anymore. In chemistry, the limiting reactant is similar. It’s the ingredient you run out of first, limiting the amount of product you can form.

In the reaction between ${\text{Cl}}_2$ and ${\text{NaClO}}_2$ to form ${\text{ClO}}_2$, ${\text{Cl}}_2$ serves as the limiting reactant.

• The stoichiometry of the reaction is used to determine the limiting reactant by comparing mole ratios of the reactants supplied.
• Calculating the moles of ${\text{Cl}}_2$ shows that it provides fewer moles than needed when considering the balanced reaction equation.
• Because ${\text{Cl}}_2$ is consumed first, it dictates the maximum yield of ${\text{ClO}}_2$.

Recognizing the limiting reactant is crucial for figuring out how much product you can actually make, just like recognizing how much sugar you have tells you how many cookies you can bake.

Molecular Geometry

Molecular geometry refers to the three-dimensional shape formed by atoms in a molecule. It helps to predict properties such as boiling points, polarity, and reactivity.

For the ${\text{ClO}}_2^{-}$ ion, its molecular geometry is 'bent.' This shape comes from the arrangement of the three electron regions around the central chlorine atom: two bonding pairs (with oxygen) and one non-bonding lone pair. Here's what that means:

• Three electron regions would ideally form a trigonal planar shape, but the lone pair repulses more strongly than bonded pairs.
• This extra repulsion compresses the angle between the two $\text{O-Cl}$ bonds, making the geometry bent.

Understanding molecular geometry like this helps in predicting how molecules interact and react with others, influencing everything from the color of a substance to its solubility in water.