Problem 119 - Chemistry: The Central Science

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 \(10.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(1.50 \mathrm{~atm}\) at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

Step-by-Step Solution

Answer

(a) The Lewis structure of \(\mathrm{ClO}_{2}\) has a central chlorine atom, with two single bonds to oxygen atoms, and three lone pairs on the chlorine atom. The chlorine atom has an unpaired electron, making it a radical. (b) \(\mathrm{ClO}_{2}\) is reduced readily due to its unstable electron configuration caused by the unpaired electron in the chlorine atom. (c) In the Lewis structure of \(\mathrm{ClO}_{2}^{-}\), the chlorine atom has a complete octet with four lone pairs, and each oxygen atom has a single bond and three lone pairs. (d) The \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in \(\mathrm{ClO}_{2}^{-}\) is expected to be less than \(109.5^\circ\) due to lone pair repulsion. (e) Approximately 7.48 grams of \(\mathrm{ClO}_{2}\) can be prepared from the given reaction conditions.

Step 1: (a) Lewis structure of \(\mathrm{ClO}_{2}\)

1. Count the total number of valence electrons in the molecule. Chlorine has 7 valence electrons, and Oxygen has 6 valence electrons. So, for \(\mathrm{ClO}_{2}\), we have a total of \(7 + (2\times6) = 19\) valence electrons. 2. Write the skeleton structure of the molecule. Place the central atom, chlorine, in the center with single bonds to each oxygen atom. 3. Distribute the remaining valence electrons on the atoms to fulfill the octet rule for each atom (except hydrogen). In this case, place 6 more electrons around each oxygen atom in the form of 3 lone pairs. 4. Check if the total number of electrons in the structure equals the total valence electrons calculated in step 1. Our structure currently has 14 electrons (2 on each bond, and 6 on each oxygen atom). We are missing 5 electrons. 5. Distribute the extra electrons on the central atom (chlorine) which now has 2 single bonds and 3 lone pairs. 6. The Lewis structure of \(\mathrm{ClO}_{2}\) is now complete. The chlorine atom has an extra electron unpaired, making it a radical.

Step 2: (b) Why is \(\mathrm{ClO}_{2}\) reduced readily?

The oxidizing power of \(\mathrm{ClO}_{2}\) results from its unstable electron configuration due to the unpaired electron in the chlorine atom. This makes the molecule highly reactive, and it can easily gain an electron, undergoing reduction in the process.

Step 3: (c) Lewis structure of \(\mathrm{ClO}_{2}^{-}\)

The chlorite ion, \(\mathrm{ClO}_{2}^{-}\), only differs from the neutral \(\mathrm{ClO}_{2}\) by gaining an extra external electron. So, take the Lewis structure of \(\mathrm{ClO}_{2}\) and add one electron to the chlorine atom. The central chlorine atom now has a complete octet with four lone pairs, while each oxygen atom retains the single bond and three lone pairs.

Step 4: (d) \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in \(\mathrm{ClO}_{2}^{-}\)

In the \(\mathrm{ClO}_{2}^{-}\) ion, chlorine is the central atom with two single bonds to the oxygen atoms and two lone pairs. This molecular geometry is described as bent or V-shaped. The presence of the lone pairs results in a repulsion that deviates the bond angle from the tetrahedral ideal. Therefore, the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle is expected to be less than \(109.5^\circ\) due to lone pair repulsion.

Step 5: (e) Calculating the grams of \(\mathrm{ClO}_{2}\) prepared

1. First, let's calculate the number of moles of \(\mathrm{NaClO}_{2}\): \(10.0 \,\mathrm{g} \times \frac{1 \, \mathrm{mol}}{90.44 \, \mathrm{g/mol}} \approx 0.111\, \mathrm{mol}\) 2. Now, convert the given pressure, volume, and temperature of chlorine gas into the number of moles using the ideal gas law (\(PV=nRT\)): \(n = \frac{PV}{RT} = \frac{(1.50 \, \mathrm{atm})(2.00 \, \mathrm{L})}{(0.0821 \, \mathrm{L \cdot atm/mol \cdot K})(294 \, \mathrm{K})} \approx 0.123\,\mathrm{mol}\) 3. Check the limiting reactant. From the chemical equation, 1 mole of \(\mathrm{Cl}_{2}\) reacts with 2 moles of \(\mathrm{NaClO}_{2}\). We have 0.123 moles of \(\mathrm{Cl}_{2}\) and 0.111 moles of \(\mathrm{NaClO}_{2}\). The limiting reactant is \(\mathrm{NaClO}_{2}\) since it will be completely consumed before the \(\mathrm{Cl}_{2}\). 4. Determine the moles of \(\mathrm{ClO}_{2}\) produced based on the limiting reactant: \(0.111 \, \mathrm{mol} \, \mathrm{NaClO}_{2} \times \frac{1 \, \mathrm{mol} \, \mathrm{ClO}_{2}}{1 \, \mathrm{mol}\, \mathrm{NaClO}_{2}} = 0.111 \, \mathrm{mol} \, \mathrm{ClO}_{2}\) 5. Convert the moles of \(\mathrm{ClO}_{2}\) to grams of \(\mathrm{ClO}_{2}\): \(0.111\, \mathrm{mol} \, \mathrm{ClO}_{2} \times \frac{67.45 \, \mathrm{g/mol}}{1 \, \mathrm{mol}} \approx 7.48 \, \mathrm{g}\) So, approximately 7.48 grams of \(\mathrm{ClO}_{2}\) can be prepared from this reaction.

Key Concepts

Lewis Structures

Understanding Lewis structures is essential in chemistry as they showcase how atoms in a molecule are bonded and help visualize the electronic structure. For chlorine dioxide

• The total number of valence electrons needs to be counted: Chlorine has 7, while each oxygen has 6, resulting in 19 electrons for ClO₂.
• Start by arranging the atoms with chlorine as the central atom bonded to oxygen atoms.
• Distribute the remaining electrons to satisfy the octet rule. Oxygen atoms should get the electrons needed, leaving any extras on chlorine.
• Don't forget, ClO₂ has an unpaired electron on chlorine, making it a radical.

For the ion ClO₂^-, an additional electron is added, filling an octet and resulting in a negative charge. Lewis structures are foundational, making it easier to predict molecule shapes and reactions.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between chemical species. Chlorine dioxide is a classic example:

• ClO₂ acts as an oxidizing agent. This means it readily accepts electrons from other substances, reducing itself.
• The presence of the unpaired electron makes ClO₂ unstable and highly reactive, making it a strong oxidant.
• In the case of chlorine dioxide gaining an electron, it transforms into the stable chlorite ion (ClO₂^-), illustrating its reduction.

Understanding redox reactions is vital in predicting the behavior of molecules in chemical processes. It explains why certain substances are easily reduced and highlights their reactivity.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry linking pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) to the number of moles (\(n\)) of a gas:\[PV = nRT\]In the context of this exercise, it helps us calculate how much chlorine gas can lead to the production of ClO₂:

• Use the law to find moles of Cl₂ using known values of pressure, volume, and temperature.
• Compare it with moles of NaClO₂ to identify the limiting reactant, which defines how much product is formed.
• Knowing the limiting reactant lets you determine the potential yield of ClO₂.

Mastering the Ideal Gas Law is key to solving various stoichiometric problems in chemistry, as it integrates several crucial properties of gases.