Question

A Chlorine gas \(\left(\mathrm{Cl}_{2}\right)\) is used as a disinfectant in municipal water supplies, although chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) and ozone are becoming more widely used. \(\mathrm{ClO}_{2}\) is a better choice than \(\mathrm{Cl}_{2}\) in this application because it leads to fewer chlorinated by-products, which are themselves pollutants. (a) How many valence electrons are in \(\mathrm{ClO}_{2} ?\) (b) The chlorite ion, \(\mathrm{ClO}_{2}^{-}\), is obtained by reducing ClO \(_{2} .\) Draw a possible electron dot structure for \(\mathrm{ClO}_{2}^{-} .\) (Cl is the central atom.) (c) What is the hybridization of the central Cl atom in \(\mathrm{ClO}_{2}^{-} ?\) What is the shape of the ion? (d) Which species has the larger bond angle, \(\mathbf{O}_{3}\) or \(\mathrm{ClO}_{2}^{-} ?\) Explain briefly. (e) Chlorine dioxide, \(\mathrm{ClO}_{2},\) a yellow-green gas, can be made by the reaction of chlorine with sodium chlorite: \(2 \mathrm{NaClO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+2 \mathrm{ClO}_{2}(\mathrm{g})\) Assume you react \(15.6 \mathrm{g}\) of \(\mathrm{NaClO}_{2}\) with chlorine gas, which has a pressure of \(1050 \mathrm{mm}\) Hg in a \(1.45-\mathrm{L}\). flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?

Short answer

(a) 19 electrons. (b) Draw the structure as Cl central with resonating O. (c) Hybridization is \(sp^2\), shape is bent. (d) \(\text{O}_3\) has larger angle. (e) 14.13 g \(\text{ClO}_2\) formed.

Step-by-step solution

Determine Valence Electrons in ClO2

To determine the number of valence electrons in \( \text{ClO}_2 \), sum the valence electrons of each element. Chlorine (Cl) has 7 valence electrons, and each oxygen (O) atom has 6. Thus, the total is \( 7 + 2 \times 6 = 19 \) valence electrons.

Drawing Lewis Structure for ClO2^-

For \( \text{ClO}_2^- \), add an extra electron to the 19 counted for \( \text{ClO}_2 \), giving 20 valence electrons. Start by placing chlorine in the center, with a single bond to each oxygen. Distribute the remaining electrons to satisfy the octet rule. One resonance structure: \(Cl:O^- \, || :\O^ :: \) with a formal negative charge on one oxygen.

Determine Hybridization of Cl in ClO2^-

Count the number of regions of electron density around the central Cl in \( \text{ClO}_2^- \), including lone pairs and bonded atoms. There are two bonded oxygens and one lone pair, totaling 3 regions. This corresponds to \( sp^2 \) hybridization.

Identify Shape of ClO2^- Ion

With 3 regions of electron density (2 bonds, 1 lone pair) and \( sp^2 \) hybridization, \( \text{ClO}_2^- \) forms a bent or V-shape, similar to water.

Compare Bond Angles in O3 and ClO2^-

\( \text{O}_3 \) is typically bent, with a bond angle slightly less than 120° due to resonance. \( \text{ClO}_2^- \) is expected to have a smaller bond angle due to electron pair repulsion being greater than atom repulsion, further reducing its angle from close to 120°.

Calculate Mass of ClO2 Formed

First, calculate moles of \( \text{NaClO}_2 \)` used:\[\text{Molar Mass of NaClO}_2 \approx 74.44 \, \text{g/mol} \]. Moles = \(\frac{15.6 \, \text{g}}{74.44 \, \text{g/mol}} \approx 0.2095 \, \text{mol}\). According to the reaction, chlorine gas-to-\(\text{ClO}_2 \) is 1:1. Next, calculate chlorine gas moles:Use PV=nRT to find moles of chlorine: convert pressure to atm (\(1050 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} \approx 1.382 \, \text{atm} \)). T = 295 K (22°C + 273), V = 1.45 L. Use R = 0.0821 L atm/mol K.\[n = \frac{(1.382 \, \text{atm}) (1.45 \, \text{L})}{(0.0821 \, \text{L atm/mol K}) (295 \, \text{K})} \approx 0.085 \, \text{mol of Cl}_2\]. Since \(\text{NaClO}_2\) is the limiting reactant, 0.2095 mol \(\text{ClO}_2\) can be formed. Finally, calculate mass of \(\text{ClO}_2\): \[\text{Molar Mass of ClO}_2 \approx 67.45 \, \text{g/mol}\]\[\text{Mass} = \text{Molar Mass} \times \text{Moles Formed} = 67.45 \, \text{g/mol} \times 0.2095 \, \text{mol} \approx 14.13 \, \text{g}\].

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding. These electrons determine how atoms interact and bond with each other. Every element has a specific number of valence electrons that can be figured out from its position on the periodic table.
For chlorine, found in group 17, there are 7 valence electrons. Oxygen, in group 16, has 6 valence electrons. When considering \( \text{ClO}_2 \), the total valence electrons will be the sum of chlorine's and the two oxygens' valence electrons:

• Chlorine (Cl) = 1 x 7 = 7
• Oxygen (O) = 2 x 6 = 12
• Total = 7 + 12 = 19

Valence electrons provide the basis for understanding how molecules are structured and predict the nature of bonding between atoms.

Lewis Structures

Lewis structures, commonly known as electron dot structures, are diagrams that represent the valence electrons around atoms within a molecule. They are invaluable tools for visualizing the electronic structure and understanding how molecules bond.
In such diagrams, elements are represented by their chemical symbol with dots around them symbolizing valence electrons. Bonds between atoms are represented as lines.
For the chlorite ion \( \text{ClO}_2^- \), remember it has 20 valence electrons, as it gains an additional electron due to the negative charge. The Lewis structure can be drawn as:

• Place Cl in the center and connect it with single bonds to each O.
• Distribute remaining electrons to fulfill the octet requirement for each atom.
• Place the extra electron as a lone pair or account it with resonance structures.

Lewis structures help predict the shapes and physical properties of molecules.

Hybridization

Hybridization in chemistry refers to the concept of mixing atomic orbitals to form new hybrid orbitals. These are associated with specific geometric adaptations of binding atoms.
For the \( \text{ClO}_2^- \) ion, this involves the central Cl atom. We consider the number of electron-density regions around Cl. This includes both bonded atoms and lone pairs.
In our case:

• Two oxygen atoms form sigma bonds with chlorine.
• One lone pair on chlorine contributes to electron density.

Therefore, there are 3 regions of electron density leading to \( sp^2 \) hybridization. This provides us insight into how these molecular orbitals form the bonding architecture seen in compounds.

Molecular Geometry

Understanding molecular geometry is essential for grasping how molecular shapes result from the spatial arrangement of atoms bonded together in a molecule. It enables predictions of molecular behavior and interactions.
For \( \text{ClO}_2^- \), with \( sp^2 \) hybridization and one lone pair, one can infer that the molecular geometry is bent or V-shaped. This results from the repulsion between the lone pair and bonds, causing a compression of bond angles.
The actual bond angle in \( \text{ClO}_2^- \) will be less than 120° due to the lone pair repulsion, affecting how the atoms orient in space. Understanding such geometry is vital in predicting complexity and chemical reactivity of molecules.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships, or ratios, between reactants and products in chemical reactions. It is based on the conservation of mass and the stoichiometric coefficients from balanced equations.
Taking the reaction for the creation of \( \text{ClO}_2 \):
\[2 \text{NaClO}_2 + \text{Cl}_2 \rightarrow 2 \text{NaCl} + 2 \text{ClO}_2\]
we employ stoichiometry to find the mass of \( \text{ClO}_2 \) produced. Start with moles of reactants, in this case, sodium chlorite and chlorine gas. Use the ideal gas law, \( PV=nRT \), to determine the moles of chlorine. Then use molar masses to convert grams to moles and back as needed.
By comparing the mole ratio from the reaction, determine the limiting reactant to calculate maximum product yield. Such calculations illustrate the practical use of stoichiometry in predicting the quantities of substances consumed and produced in reactions.