Problem 63 - Chemistry The Central Science

Question

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) . (a) What is the name of this product (see Table 2.6) ? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?(\mathbf{d})\) If the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is considered to have the \(-2\) oxidation state, what is the oxidation state of the Cl? What is the electron configuration of Cl in this oxidation state?

Step-by-Step Solution

Answer

The product of the reaction is named "Dichlorine heptoxide". The balanced equation for the formation of dichlorine heptoxide is \(\mathrm{Cl}_2 (g) + \cfrac{7}{2} \mathrm{O}_2 (g) \rightarrow \mathrm{Cl}_2 \mathrm{O}_7(l)\). \(\mathrm{Cl}_2 \mathrm{O}_7\) would be more reactive toward \(\mathrm{OH}^{-}\) ions. The oxidation state of Cl in \(\mathrm{Cl}_2 \mathrm{O}_7\) is +7, with the electron configuration of \(1s^2 2s^2 2p^6 3s^1\).

Step 1: (a) Determining the name of the product

The product of the reaction between chlorine and oxygen is given as \(\mathrm{Cl}_2 \mathrm{O}_7\). Based on the periodic table, this compound is named as "Dichlorine heptoxide".

Step 2: (b) Writing a balanced equation for the formation of \(\mathrm{Cl}_2 \mathrm{O}_7(l)\)

To write a balanced equation for the formation of dichlorine heptoxide, we need to combine the elements \(\mathrm{Cl}_2\) and \(\mathrm{O}_2\) in the proper ratio to form \(\mathrm{Cl}_2 \mathrm{O}_7\). The balanced equation is: \[ \mathrm{Cl}_2 (g) + \cfrac{7}{2} \mathrm{O}_2 (g) \rightarrow \mathrm{Cl}_2 \mathrm{O}_7(l) \]

Step 3: (c) Reactivity of \(\mathrm{Cl}_2 \mathrm{O}_7\) toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q)\)

Dichlorine heptoxide (\(\mathrm{Cl}_2 \mathrm{O}_7\)) is a strong oxidizing agent with a high positive oxidation state of chlorine. Since it seeks to gain electrons, it would react more readily with the species offering electrons. In this case, \(\mathrm{H}^{+}\) ions would not react with the compound as they do not readily offer electrons. On the other hand, \(\mathrm{OH}^{-}\) ions have an extra electron and can therefore offer electrons to the compound. So, \(\mathrm{Cl}_2 \mathrm{O}_7\) would be more reactive toward \(\mathrm{OH}^{-}\) ions.

Step 4: (d) Oxidation state of Cl and electron configuration

In \(\mathrm{Cl}_2 \mathrm{O}_7\), oxygen has an oxidation state of -2. To balance the charges, the sum of the oxidation states for the entire compound must be zero. Since there are 2 Cl atoms and 7 O atoms, we can set up the equation: \[ 2x + 7(-2) = 0 \] Solving for x, we find the oxidation state of chlorine: \[ x = \cfrac{14}{2} \] \[ x = 7 \] The oxidation state of Cl in \(\mathrm{Cl}_2 \mathrm{O}_7\) is +7. The electron configuration of Cl in its ground state is \(1s^2 2s^2 2p^6 3s^2 3p^5\). In the +7 oxidation state, Cl loses all the electrons from the 3p orbital and one electron from the 3s orbital, giving the electron configuration: \[ 1s^2 2s^2 2p^6 3s^1 \]

Key Concepts

Chemical Reaction Balance

Understanding the balance of chemical reactions is a fundamental skill in chemistry, one that allows us to predict the outcomes of chemical interactions and ensure that matter is neither created nor destroyed in the process. When chlorine reacts with oxygen to form dichlorine heptoxide, we follow the law of conservation of mass. To balance the equation for the reaction, we must have the same number of each type of atom on both sides of the equation.

For the formation of \(\mathrm{Cl}_2 \mathrm{O}_7(l)\), the balanced equation is found to be:
\[ \mathrm{Cl}_2 (g) + \cfrac{7}{2} \mathrm{O}_2 (g) \rightarrow \mathrm{Cl}_2 \mathrm{O}_7(l) \]
This equation indicates that one molecule of chlorine gas reacts with three and a half molecules of oxygen gas to produce one molecule of dichlorine heptoxide. Since oxygen gas naturally exists as diatomic molecules (\(O_2\)), we use the fraction \(\cfrac{7}{2}\) to balance the seven oxygen atoms needed on the product side.

Oxidation States

Determining Oxidation States

The concept of oxidation states, also referred to as oxidation numbers, allows us to understand the transfer of electrons in chemical reactions and is essential when dealing with compounds like dichlorine heptoxide. Each atom in a molecule has an oxidation state that denotes its potential to gain or lose electrons.

For the molecule \(\mathrm{Cl}_2 \mathrm{O}_7\), we can calculate the oxidation state of chlorine by considering that oxygen has a usual oxidation state of -2. The compound is neutral, so the overall charge is zero. Through the algebraic process shown in the exercise solution, we discover that chlorine has an oxidation state of +7 in dichlorine heptoxide.

Chemical Implications

This high positive oxidation state implies that chlorine is in a highly oxidized form, indicating a significant electron deficiency. Such a state makes dichlorine heptoxide a powerful oxidizing agent, reacting readily with species that can donate electrons—in this case, favoring a reaction with hydroxide ions (\(\mathrm{OH}^{-}\)) over hydronium ions (\(\mathrm{H}^{+}\)).

Electron Configuration

The electron configuration is a description of the arrangement of an atom's electrons in its orbitals, which dictates how an atom behaves chemically. For chlorine (Cl), found in the Periodic Table under Group 17, the ground state electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^5\). This configuration shows a full 1s, 2s, and 2p subshell, and five of the six spots filled in the 3p subshell.

When chlorine's oxidation state is +7, as it is in dichlorine heptoxide, it has lost electrons compared to its ground state configuration. The removal of electrons follows a specific order, known as the Aufbau principle, which generally sees electrons lost from the outermost shells first.

In the +7 oxidation state, chlorine has lost all the electrons in the 3p subshell (5 electrons) and one from the 3s subshell. The resulting electron configuration is \(1s^2 2s^2 2p^6 3s^1\), reflecting a much more electron-deficient chlorine that has a strong affinity for gaining electrons, hence its reactivity as an oxidizing agent.