Question

Although the \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{IO}_{4}^{-}\) ions have been known for a long time, \(\mathrm{BrO}_{4}^{-}\) was not synthesized until \(1965 .\) The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction?

Short answer

The balanced chemical equation for the given reaction is \(\mathrm{BrO_3^- + 2\ XeF_2 \rightarrow BrO_4^- + 2\ Xe + 2\ HF}\). The oxidation states of Br in the Br-containing species in this reaction are +5 in the bromate ion \(\mathrm{(BrO_3^-)}\) and +7 in the perbromate ion \(\mathrm{(BrO_4^-)}\).

Step-by-step solution

Write the unbalanced equation for the given reaction

The reaction between bromate ion \(\mathrm{(BrO_3^-)}\) and xenon difluoride \(\mathrm{(XeF_2)}\) produces perbromate ion \(\mathrm{(BrO_4^-)}\), xenon \(\mathrm{(Xe)}\), and hydrofluoric acid \(\mathrm{(HF)}\). So, the unbalanced equation for the given reaction is: \[\mathrm{BrO_3^- + XeF_2 \rightarrow BrO_4^- + Xe + HF}\]

Balance the chemical equation

To balance the chemical equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation: \(\mathrm{BrO_3^- + 2\ XeF_2 \rightarrow BrO_4^- + 2\ Xe + 2\ HF}\) Now, we have a balanced equation for the given reaction: \[\mathrm{BrO_3^- + 2\ XeF_2 \rightarrow BrO_4^- + 2\ Xe + 2\ HF}\]

Determining the oxidation states of Br in Br-containing species

We need to find the oxidation states of Br in the bromate ion \(\mathrm{(BrO_3^-)}\) and the perbromate ion \(\mathrm{(BrO_4^-)}\). The oxidation state of oxygen is generally -2. For the bromate ion: Let the oxidation state of Br = x \[\mathrm{x + 3\ (-2) = -1}\] \[\mathrm{x = -1 + 6}\] \[\mathrm{x = +5}\] So, the oxidation state of Br in \(\mathrm{BrO_3^-}\) is +5. For the perbromate ion: Let the oxidation state of Br = y \[\mathrm{y + 4\ (-2) = -1}\] \[\mathrm{y = -1 + 8}\] \[\mathrm{y = +7}\] Hence, the oxidation state of Br in \(\mathrm{BrO_4^-}\) is +7. To sum up, we have the balanced chemical equation for the given reaction: \[\mathrm{BrO_3^- + 2\ XeF_2 \rightarrow BrO_4^- + 2\ Xe + 2\ HF}\] The oxidation states of Br in the Br-containing species in this reaction are +5 in the bromate ion \(\mathrm{(BrO_3^-)}\) and +7 in the perbromate ion \(\mathrm{(BrO_4^-)}\).

Key concepts

Bromate Ion

The bromate ion is represented by the chemical formula \( \mathrm{BrO_3^-} \). It is an oxyanion of bromine and plays a significant role in chemical reactions involving oxidation. Understanding the oxidation state is crucial to grasp how bromate ions participate in chemical processes.
The oxidation state of an atom in a molecule indicates the degree of oxidation or loss of electrons. For bromate ions, bromine is in the oxidation state of +5.

• Each oxygen atom contributes an oxidation state of -2.
• The total combined oxidation state for the three oxygen atoms is -6.
• This results in a net charge of -1 for the bromate ion, as calculated by setting an equation: \( x + 3(-2) = -1 \), where \( x \) is the oxidation state of bromine.
• Solving this gives \( x = +5 \).

Bromate ions find common applications in water treatment and analytical chemistry, primarily due to their oxidizing properties.

Xenon Difluoride

Xenon difluoride, represented by the chemical formula \( \mathrm{XeF_2} \), vastly differs from the typical properties of xenon, a noble gas. It's a stable, colorless crystalline solid and one of the few compounds where xenon acts as a central atom forming stable bonds.
This compound is highly reactive as a fluorinating agent due to the unique ability of xenon to form chemical bonds. In the context of the oxidation process, xenon difluoride reacts with the bromate ion to form the perbromate ion, indicating its strong oxidizing power. The reaction also illustrates a rare instance of the noble gas xenon being involved in redox chemistry.
The compound plays significant roles in:

• Fluorination reactions, often used in organic chemistry to introduce fluorine atoms into other molecules.
• Laboratory syntheses where non-reactive elements require activation.

Despite its high reactivity, xenon difluoride’s applications are typically limited to controlled environments, as its potent fluorinating ability can lead to hazardous reactions if not handled properly.

Perbromate Ion

The perbromate ion, which is denoted by \( \mathrm{BrO_4^-} \), is an even more oxygen-rich anion compared to the bromate ion. It features bromine in an oxidation state of +7, showcasing bromine at one of its highest oxidation potential states.
For this ion:

• The oxidation state of bromine (\( y \)) is calculated based on the four oxygen atoms each contributing -2.
• The equation is \( y + 4(-2) = -1 \).
• Solving gives \( y = +7 \), indicating a complete loss of electrons by bromine in this compound form.

The synthesis of perbromate ion using xenon difluoride and bromate ion represents a classic redox reaction where bromine is oxidized further. This extreme oxidation state is not commonly observed due to the general instability of perbromates, hence their synthesis and existence aren't as widespread as other bromine oxyanions.
Perbromate ions are primarily produced in research or experimental settings, often used to study the chemistry of high oxidation state compounds.