Problem 63 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\mathrm{Cl}\) ? What is the electron configuration of \(\mathrm{Cl}\) in this oxidation state?

Step-by-Step Solution

Answer

The product \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is called Dichlorine heptoxide. The balanced equation for its formation is \(\mathrm{Cl}_2(g) + \frac{7}{2}\mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)\). It would be more reactive towards \(\mathrm{H}^{+}(aq)\) ions. The oxidation state of Cl in the compound is +7, and its electron configuration in this state is \([\mathrm{Ne}]\).

Step 1: (a) Name of the compound

From Table 2.6, the name of the compound \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is Dichlorine heptoxide.

Step 2: (b) Balanced equation for the formation of Cl2O7

To write a balanced equation for the formation of \(\mathrm{Cl}_{2}\mathrm{O}_{7}(l)\) from the elements, we'll start with: Chlorine (gas): \(\mathrm{Cl}_2\) Oxygen (gas): \(\mathrm{O}_2\) The formation of the compound will then be: $$\mathrm{Cl}_2(g) + \mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)$$ Balancing the equation, we get: $$\mathrm{Cl}_2(g) + \frac{7}{2}\mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)$$

Step 3: (c) Reactivity towards H+ or OH- ions

Given that \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) is a compound with a high oxidation state of chlorine (which we will determine in part (d)), it can be expected that it will be a powerful oxidizing agent. Therefore, it will have a higher reactivity towards \(\mathrm{H}^{+}(aq)\) rather than \(\mathrm{OH}^{-}(aq)\) ions, as the H+ ions can easily reduce the high oxidation state of Cl in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\).

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

As given, the oxygen in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) has an oxidation state of -2. To find the oxidation state of Cl, let's use the formula: $$2x + 7(-2) = 0$$ Where x represents the oxidation state of Cl. Solving for x, we get: $$x = +7$$ So, the oxidation state of Cl is +7 in this compound. The electron configuration of a neutral chlorine atom is \([\mathrm{Ne}]3s^23p^5\). To achieve an oxidation state of +7, chlorine has to lose all 5 valence electrons in the 3p orbital and 2 valence electrons in the 3s orbital. Thus, the electron configuration of chlorine in the +7 oxidation state will be the same as that of a neon atom: $$[\mathrm{Ne}]$$

Key Concepts

Oxidation States

In a chemical reaction, oxidation states help us understand how electrons are transferred between atoms. The oxidation state is essentially the charge that an atom "feels" because of its electron configuration. In the case of \(\mathrm{Cl}_{2}\mathrm{O}_{7}\), each oxygen atom is assigned an oxidation state of \(-2\). For the chlorine atoms, to find their oxidation states, we use the equation \[2x + 7(-2) = 0\]where \(x\) is the oxidation state of chlorine. Solving for \(x\), we find it is \(+7\). This means chlorine is in a high oxidation state, making \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) a potent oxidizing agent. High oxidation states like \(+7\) indicate that an atom has lost several electrons, which can affect its reactivity significantly. A compound with a higher oxidation state is often more reactive towards reducing agents like \(\mathrm{H}^{+}\) ions because these ions can provide the electrons needed to lower the oxidation state.

Electron Configuration

The electron configuration of an element describes the distribution of electrons in its atomic orbitals. For chlorine in its neutral state, the electron configuration is \[[\mathrm{Ne}]3s^23p^5\],indicating that chlorine has 17 electrons, fitting into its 3 energy levels with the outermost level having 7 electrons. However, when chlorine reaches an oxidation state of \(+7\) in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\), it has lost these 7 electrons. So, its electron configuration resembles that of neon, which is \([\mathrm{Ne}]\). This is because chlorine loses its entire valence shell in the process of reaching this high oxidation state. Understanding electron configurations is crucial for predicting how atoms engage in reactions, as it helps indicate what atoms will tend to do during a chemical reaction—like gaining or losing electrons to form bonds or breaking them.

Chemical Equations

Chemical equations are shorthand notations to describe chemical reactions. They show the reactants starting a reaction and the products formed by it. In this exercise, we balanced a chemical equation for the formation of \(\mathrm{Cl}_{2}\mathrm{O}_{7}\)from its elements as \[\mathrm{Cl}_2(g) + \frac{7}{2}\mathrm{O}_2(g) \rightarrow \mathrm{Cl}_{2}\mathrm{O}_{7}(l)\].This equation illustrates that one mole of chlorine gas reacts with \(\frac{7}{2}\) moles of oxygen gas to produce one mole of dichlorine heptoxide. Balancing equations is important because it reflects the conservation of mass: no atoms are lost or gained during a chemical reaction. Each side of the equation must have the same number of atoms for each element involved. Balanced chemical equations are foundational for understanding how substances interact and transform in chemical reactions.