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: $$ \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{NaClO}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ClO}_{2}(\mathrm{~g})+2 \mathrm{NaCl}(\mathrm{s}) $$ 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?

Short answer

The Lewis structure for ClO$_{2}$ is : \[ : \overset{..}{O} - \underbrace{\underline{\overset{..}{C}\mathrm{l}}}_{3\, \mathrm{lp}} - \overset{..}{O} : \] ClO$_{2}$ is reduced readily due to its high instability, with 3 lone pair electrons on the central Chlorine atom. The Lewis structure for ClO$_{2}^{-}$ is : \[ : \overset{..}{O} - \underbrace{\underline{\overset{..}{C}\mathrm{l}}}_{2\, \mathrm{lp}} - \overset{..}{O} :^{-} \] The O-Cl-O bond angle in ClO$_{2}^{-}$ is slightly less than \(109.5 ^{\circ}\) due to the repulsion caused by lone pair electrons. Finally, using the given reaction conditions, the reaction will yield 7.456 grams of ClO$_{2}$.

Step-by-step solution

Identify the total number of valence electrons

For Chlorine, there are 7 valence electrons, and for each Oxygen atom, there are 6 valence electrons. So, the total number of valence electrons is 7 + 6 + 6 = 19.

Arrange the atoms

As Cl is the central atom, place it in the center and connect the Oxygens on each side. Cl - O - O.

Distribute electrons and create bonds

First, create single bonds between the central Cl and each O atom. Subtract 4 electrons for the two single bonds, leaving 15 electrons. Distribute these remaining electrons as lone pairs on each Oxygen atom and the remaining electrons on the central Chlorine atom.

Finalize the Lewis structure

The resulting Lewis structure would be: \[ : \overset{..}{O} - \underbrace{\underline{\overset{..}{C}\mathrm{l}}}_{x\, \mathrm{lp}} - \overset{..}{O} : \] Since the Cl atom has 7 valence electrons, x is 3 and the final structure is: \[ : \overset{..}{O} - \underbrace{\underline{\overset{..}{C}\mathrm{l}}}_{3\, \mathrm{lp}} - \overset{..}{O} : \] (b) Ease of reduction of ClO2

Reason for the ease of reduction

In ClO2, there are 3 lone pair electrons on the central Chlorine atom, creating a highly unstable and reactive structure. This high instability makes it very easy for ClO2 to be reduced by gaining one electron. (c) Lewis structure for ClO2-

Determine the valence electrons

Since ClO2- gains an extra electron, there will now be a total of 20 valence electrons.

Distribute electrons and create bonds

Similar to ClO2, create single bonds between the Cl and O atoms. This utilizes 4 electrons, leaving 16 electrons which are distributed as lone pairs on the Oxygen atoms and the remaining electrons on the central Chlorine atom.

Finalize the Lewis structure

The resulting Lewis structure for ClO2- would be: \[ : \overset{..}{O} - \underbrace{\underline{\overset{..}{C}\mathrm{l}}}_{2\, \mathrm{lp}} - \overset{..}{O} :^{-} \] (d) O-Cl-O bond angle in ClO2-

Identify the electron domain geometry

Since there are 4 electron domains (two single bonds and two lone pair electrons) around the central Cl atom, the electron domain geometry would be tetrahedral.

Identify the molecular geometry and bond angle

With 2 lone pair electrons and 2 bonding electrons around the Cl atom, the molecular geometry would be a bent (V-shaped) structure. The O-Cl-O bond angle in ClO2- will be slightly less than \(109.5 ^{\circ}\) due to the repulsion caused by lone pair electrons. (e) Calculate the grams of ClO2 produced

Identify the limiting reactant

First, we need to find the number of moles of Cl2 and NaClO2, and determine the limiting reactant. For NaClO2, we have 10 g/\(\mathrm{molar \ mass}\) and for Cl2, we use the ideal gas law PV=nRT: \(\frac{1.50 \mathrm{~atm} \times 2.00 \mathrm{~L}}{0.0821 \mathrm{~L.atm/mol.K} \times (273+21) \mathrm{K}}\).

Calculate the moles of each reactant

Number of moles of NaClO2 = \(\frac{10.0}{90.44} = 0.1106\) moles (Where the molar mass of NaClO2 is 90.44 g/mol) Number of moles of Cl2 = \(\frac{1.50 \mathrm{~atm} \times 2.00 \mathrm{~L}}{0.0821 \mathrm{~L.atm/mol.K} \times 294 \mathrm{K}} = 0.1227\) moles

Calculate the mole ratio

Mole ratio of NaClO2 to Cl2 = \(\frac{0.1106}{0.1227}=0.901\) Compared to the stoichiometric ratio of 2:1, NaClO2 is the limiting reactant.

Calculate the moles of ClO2 formed

Using the stoichiometric ratio of 2 moles of ClO2 formed per 2 moles of NaClO2, we have: 0.1106 moles NaClO2 will produce 0.1106 moles of ClO2.

Calculate the grams of ClO2 formed

Using the molar mass of ClO2 (67.45 g/mol), the final mass of ClO2 produced = 0.1106 moles × 67.45 g/mol = 7.456 g. So, the reaction will yield 7.456 grams of ClO2.

Key concepts

Chlorine Dioxide (ClO2)

Chlorine dioxide, or ClO2, is a chemical compound that plays an important role as a bleaching agent in the paper and textile industry. The understanding of its Lewis structure is essential as it helps predict the compound's reactivity and properties. In the Lewis structure of ClO2, the central chlorine atom is surrounded by two oxygen atoms with single bonds, and the chlorine has three lone pairs of electrons. This accounts for a total of 19 valence electrons: 7 from chlorine and 6 from each oxygen atom.

The presence of unpaired electrons in this structure contributes to the compound's ability to participate in oxidation-reduction reactions, as these lone pairs can readily accept an additional electron during the reduction process. This is what makes ClO2 a powerful oxidizing agent.

Oxidation-Reduction Reactions

Understanding oxidation-reduction reactions is crucial in chemistry, especially when dealing with substances like chlorine dioxide that act as oxidizing agents. Oxidation-reduction (redox) reactions involve the transfer of electrons between species. The substance that gains electrons undergoes reduction, while the substance that loses electrons is oxidized.

In the case of ClO2, due to its electron configuration and high reactivity, it's often involved in redox reactions within industrial processes. The fact that it can readily accept an electron makes it susceptible to reduction, turning into chlorite (ClO2-), while oxidizing another substance in the process. This characteristic is particularly advantageous in bleaching operations where electron transfer is essential for breaking down color-causing compounds.

Molecular Geometry

Molecular geometry is the three-dimensional arrangement of atoms within a molecule, which can be predicted using the VSEPR (Valence Shell Electron Pair Repulsion) theory. For ClO2 and its ion ClO2-, the molecular geometry is determined by the total number of electron pairs around the central chlorine atom.

Chlorine dioxide has a bent geometry due to the repulsion between the lone pairs and the bonding pairs of electrons. For the chlorite ion (ClO2-), with an additional electron, a bent shape is also present, with the O-Cl-O bond angle being slightly less than the tetrahedral angle of 109.5 degrees. This is because the lone pairs occupy more space and thus, push the bonding pairs closer together, reducing the bond angle.

Stoichiometry Calculations

Stoichiometry is the study of the quantitative relationships, or ratios, within chemical reactions. When calculating how much product can be obtained from given reactants, as in the synthesis of ClO2 from chlorine gas and sodium chlorite, stoichiometry allows for precise predictions.

By applying stoichiometric principles and the ideal gas law, we can determine the limiting reactant and calculate the exact amount of ClO2 produced from a reaction. It involves using the molar masses of the reactants and the product, as well as the balanced chemical equation, to find that 10.0 grams of sodium chlorite and 2.00 liters of chlorine gas at 1.50 atm and 21°C will yield 7.456 grams of ClO2. Understanding how to carry out these stoichiometry calculations is fundamental to not only chemistry but to various industrial and lab processes where precise chemical formulation is required.