Problem 116 - Chemistry and Chemical Reactivity

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 \(\mathrm{ClO}_{2}\). 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, \(\mathrm{O}_{3}\) 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 \(1050 \mathrm{mm} \mathrm{Hg}\) in a 1.45-L flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?

Step-by-Step Solution

Answer

ClO2− has 20 valence electrons and a bent shape; O3 has a larger bond angle. Produce 11.64 g of ClO2.

Step 1: Calculate Valence Electrons in ClO2

The molecule ClO2 consists of one Cl atom and two O atoms. Chlorine has 7 valence electrons, and each oxygen atom has 6 valence electrons. So, the total number of valence electrons is:\[ (1 \times 7) + (2 \times 6) = 19. \]

Step 2: Draw Electron Dot Structure for ClO2-

To draw the structure of the chlorite ion \( \text{ClO}_{2}^{-} \), note that it has one additional electron due to the negative charge. Total valence electrons are:\[ 19 + 1 = 20. \]Cl is the central atom. Form single bonds with each O and distribute remaining electrons to satisfy octet. Two single bonds and lone pairs are possible on Cl, with one oxygen having a double bond to account for all electrons and charge.

Step 3: Determine Hybridization and Shape of ClO2-

The central Cl atom is surrounded by three regions of electron density (two bonded oxygens and one lone pair). Thus, the hybridization is \( \text{sp}^{2} \). The shape of \( \text{ClO}_{2}^{-} \) ion is bent due to repulsion between lone pairs.

Step 4: Compare Bond Angles of O3 and ClO2-

In \( \text{O}_3 \), the bond angle is around \(120^\circ\) typical of sp^2 hybridization. The \( \text{ClO}_{2}^{-} \) with lone pair repulsion has a smaller bond angle than \(120^\circ\), typically around \( 110^\circ\) or less, making \( \text{O}_3 \) have the larger bond angle.

Step 5: Calculate Mass of Produced ClO2 Using Stoichiometry

First, find moles of \( \text{NaClO}_{2} \):Molar mass is approximately \( 90.44 \text{ g/mol} \). Therefore,\[ \text{moles of } \text{NaClO}_{2} = \frac{15.6}{90.44} \approx 0.1725 \text{ mol}. \]From the balanced equation:\[ 2 \text{ NaClO}_{2} \rightarrow 2 \text{ ClO}_{2}, \]this implies 1:1 mole ratio. So, moles of \( \text{ClO}_{2} \) produced = \( 0.1725 \text{ mol} \).Then, find mass of \( \text{ClO}_{2} \):Molar mass of \( \text{ClO}_{2} \approx 67.45 \text{ g/mol} \).\[ \text{mass of } \text{ClO}_{2} = 0.1725 \times 67.45 \approx 11.64 \text{ g}. \]

Key Concepts

Valence Electrons

In the molecule Chlorine Dioxide (\(\text{ClO}_2\)), it's essential to understand how many valence electrons are present. Knowing the number of valence electrons helps us figure out how atoms bond and the structure they form. In \(\text{ClO}_2\), chlorine (Cl) has 7 valence electrons, while each oxygen (O) atom has 6 valence electrons.
To find the total number, we add them up:

• 1 Chlorine atom: \(1 \times 7 = 7\)
• 2 Oxygen atoms: \(2 \times 6 = 12\)

Adding these gives 19 valence electrons. These electrons are crucial in determining the molecule's structure and bonding.

Electron Dot Structure

An electron dot structure, or Lewis structure, is a visual way to represent how valence electrons are distributed in a molecule. For the chlorite ion (\(\text{ClO}_2^-\)), there are 20 valence electrons (due to an extra electron from the negative charge).
In forming the \(\text{ClO}_2^-\) structure:

• Chlorine (Cl) is the central atom.
• Each oxygen (O) is connected by a single bond.
• Lone pairs of electrons fill remaining spaces to satisfy the octet rule.

The structure typically includes two single bonds and possibly one oxygen with a double bond, allowing for an arrangement that accounts for all electrons and the negative charge.

Hybridization

Hybridization explains how atomic orbitals mix to form new hybrid orbitals suitable for pair bonding. For the \(\text{ClO}_2^-\) ion, chlorine is at the center and has three electron regions around it (from two oxygens and one lone pair).
Thus, the hybridization is \(\text{sp}^2\). This configuration helps stabilize the molecule and predicts the bent shape of the \(\text{ClO}_2^-\) ion, as electrons will spread out to minimize repulsion.

Bond Angles

The bond angle is the geometric angle between two adjacent bonds. In ozone (\(\text{O}_3\)), the bond angle is around \(120^\circ\) because of \(\text{sp}^2\) hybridization and lack of lone pair repulsion.
For \(\text{ClO}_2^-\), the lone pair on chlorine causes electron repulsion, reducing the bond angle from \(120^\circ\) to approximately \(110^\circ\) or less. Consequently, \(\text{O}_3\) has a larger bond angle than \(\text{ClO}_2^-\), showcasing how lone pairs influence molecular geometry.

Stoichiometry

Stoichiometry involves calculating reactants and products in chemical reactions. In producing \(\text{ClO}_2\) from \(\text{NaClO}_2\) and chlorine, it's vital to understand this balance to determine the mass of \(\text{ClO}_2\) produced.
Starting with 15.6 g of \(\text{NaClO}_2\):

• Molar mass = \(90.44 \text{ g/mol}\).
• Moles \(= \frac{15.6}{90.44} = 0.1725 \text{ mol}\).

One mole of \(\text{NaClO}_2\) produces one mole of \(\text{ClO}_2\). Hence, you get \(0.1725\) moles of \(\text{ClO}_2\).
To calculate mass, use molar mass \(= 67.45 \text{ g/mol}\): \[\text{Mass of } \text{ClO}_2 = 0.1725 \times 67.45 \approx 11.64 \text{ g}.\] This demonstrates the relevance of stoichiometry in predicting product amounts based on reactant quantities.