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) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q)\) ? (e) 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 named dichlorine heptoxide. The balanced equation for its formation is \(\mathrm{Cl}_{2(g)} + \frac{7}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Cl}_{2} \mathrm{O}_{7(l)}\). Its boiling point of \(81^{\circ} \mathrm{C}\) is surprising due to its highly oxidized nature, and it is more reactive towards \(\mathrm{H}^{+}(a q)\) ions. The oxidation state of \(\mathrm{Cl}\) in this compound is +7, with an electron configuration of \([\mathrm{Ne}]3\mathrm{s}^{0}3\mathrm{p}^{0}\).

Step 1: 1. Name of the product

Based on Table 2.6, the name of the product \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is dichlorine heptoxide.

Step 2: 2. Balanced equation

To form \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) from the elements, we need one molecule of chlorine (\(\mathrm{Cl}_{2}\)) and 3.5 molecules of oxygen (\(\mathrm{O}_{2}\)) to combine and produce one molecule of dichlorine heptoxide. The balanced equation can be written as: \[ \mathrm{Cl}_{2(g)} + \frac{7}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Cl}_{2} \mathrm{O}_{7(l)} \]

Step 3: 3. Boiling point

The boiling point of \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is \(81^{\circ} \mathrm{C}\). This may be surprising because highly oxidized molecules like \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) have strong oxygen-chlorine bonds and may be expected to have higher boiling points. However, due to the electron deficiency of the central atom, the molecule is highly reactive and unstable and can be broken down into simpler molecules in a lower boiling point range.

Step 4: 4. Reactivity towards \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q)\)

Since \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a highly oxidized molecule, it is expected to be more reactive towards \(\mathrm{H}^{+}(a q)\), which will help it achieve a more stable oxidation state. The molecule will likely act as an oxidizing agent, reacting with the \(\mathrm{H}^{+}(a q)\) ions.

Step 5: 5. Oxidation state and electron configuration of \(\mathrm{Cl}\)

Considering the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to have the \(-2\) oxidation state, the two chlorine atoms will need to share a total oxidation state of +14, as there are 7 oxygen atoms. Thus, each chlorine atom has an oxidation state of +7. The electron configuration of \(\mathrm{Cl}\) in this oxidation state is obtained by removing 7 electrons from its ground state configuration. The ground state configuration of \(\mathrm{Cl}\) is \([\mathrm{Ne}]3\mathrm{s}^{2}3\mathrm{p}^{5}\). When 7 electrons are removed, the configuration becomes \([\mathrm{Ne}]3\mathrm{s}^{0}3\mathrm{p}^{0}\), effectively leaving the chlorine atoms in an excited state with the same electron configuration as neon.

Key Concepts

Chemical Reactions

When we discuss chemical reactions, it's all about transforming substances. Here, chlorine and oxygen react to form dichlorine heptoxide, a compound with the formula \(\text{Cl}_2\text{O}_7\). This transformation is an example of a synthesis reaction, where simpler substances combine to form a more complex product.
In a balanced chemical equation, each side of the equation must have the same number of each type of atom. This reflects the principle of conservation of mass. For dichlorine heptoxide, the balanced equation is:

• \(\text{Cl}_{2(g)} + \frac{7}{2} \text{O}_{2(g)} \rightarrow \text{Cl}_2\text{O}_{7(l)}\)

Here, you find 2 chlorine atoms and 7 oxygen atoms on both sides, maintaining balance. Balancing chemical equations requires careful counting of atoms and sometimes using fractional coefficients, which you see in the use of \(\frac{7}{2} \text{O}_2\).

Oxidation States

Oxidation states are a way to keep track of electrons during chemical changes. In dichlorine heptoxide, the oxygen is typically assigned an oxidation state of \(-2\). Given there are 7 oxygen atoms, they contribute a total oxidation state of \(-14\). For the compound to be neutral, the two chlorine atoms must jointly possess a \(+14\) oxidation state.
This means each chlorine atom holds an oxidation state of \(+7\). This is significant because it tells us that chlorine is in a very high and oxidized state, making \(\text{Cl}_2\text{O}_7\) a powerful oxidizing agent. The compound will likely be more reactive with positively charged species like \(\text{H}^+\). This is because such reactions allow chlorine to "accept" electrons, moving closer to a lower, more stable oxidation state.

Electron Configuration

Electron configurations describe the arrangement of electrons in an atom or molecule. They are crucial for understanding chemical reactivity. A neutral chlorine atom has the electron configuration \([\text{Ne}]3s^23p^5\), using a noble gas shorthand.
However, when chlorine is in the \(+7\) oxidation state, as in \(\text{Cl}_2\text{O}_7\), 7 electrons are removed, resulting in a configuration of \([\text{Ne}]\). This effectively empties its valence shell, making the chlorine atoms akin to neon in terms of electron configuration. This gives an understanding of why chlorine in this compound is highly reactive, as it energetically "desires" to gain electrons to fill these idealized empty shells back to a more stable state.