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 equation for the reaction between the bromate ion and xenon difluoride is: \[\mathrm{BrO}_{3}^- + \mathrm{XeF}_{2} \rightarrow \mathrm{BrO}_{4}^- + \mathrm{Xe} + 2\,\mathrm{HF}\]. The oxidation state of Br in the bromate ion (\(\mathrm{BrO}_{3}^-\)) is +5 and in the perbromate ion (\(\mathrm{BrO}_{4}^-\)) is +7.

Step-by-step solution

(a) Write the balanced equation

To write the balanced equation, we first need to identify the reactants and products. Reactants: Bromate ion (\(\mathrm{BrO}_{3}^-\)) and xenon difluoride (\(\mathrm{XeF}_{2}\)) Products: Perbromate ion (\(\mathrm{BrO}_{4}^-\)), xenon (\(\mathrm{Xe}\)), and hydrofluoric acid (\(\mathrm{HF}\)) Now, we can write the equation and balance it: \[\mathrm{BrO}_{3}^- + \mathrm{XeF}_{2} \rightarrow \mathrm{BrO}_{4}^- + \mathrm{Xe} + 2\,\mathrm{HF}\] Checking the conservation of mass, we have: - Total atoms of bromine: 1 on both sides - Total atoms of oxygen: 3 on the reactant side and 4 on the product side - Total atoms of xenon: 1 on both sides - Total atoms of fluorine: 2 on the reactant side and 2 on the product side The balanced equation is: \[\mathrm{BrO}_{3}^- + \mathrm{XeF}_{2} \rightarrow \mathrm{BrO}_{4}^- + \mathrm{Xe} + 2\,\mathrm{HF}\]

(b) Determine oxidation states of Br

To find the oxidation states of bromine in the Br-containing species, we need to consider the electron distribution in each molecule/ion. First, let's find the oxidation state of Br in the bromate ion (\(\mathrm{BrO}_{3}^-\)): The oxidation state of oxygen is generally -2. With three oxygen atoms having a total charge of (-2 x 3 = -6), and the ion charge of the bromate ion being -1, the oxidation state of Br in the bromate ion should be: 1. \(x - 6 = -1\) (where \(x\) is the oxidation state of Br) 2. \(x = -1 + 6\) 3. \(x = +5\) Now, let's find the oxidation state of Br in the perbromate ion (\(\mathrm{BrO}_{4}^-\)): Again, the oxidation state of oxygen is -2. With four oxygen atoms, having a total charge of (-2 x 4 = -8), and the ion charge of the perbromate ion being -1, the oxidation state of Br in the perbromate ion should be: 1. \(x - 8 = -1\) (where \(x\) is the oxidation state of Br) 2. \(x = -1 + 8\) 3. \(x = +7\) The oxidation state of Br in bromate ion (\(\mathrm{BrO}_{3}^-\)) is +5, and in perbromate ion (\(\mathrm{BrO}_{4}^-\)) is +7.

Key concepts

Understanding Bromate Ion

In chemistry, understanding the bromate ion is crucial as it plays a significant role in various reactions. The bromate ion is represented by the chemical formula \( \mathrm{BrO}_3^- \). This denotes a bromine ion combined with three oxygen atoms, carrying a net charge of -1. The presence of this negative charge is key to understanding its behavior in chemical reactions, particularly in oxidation-reduction processes.

The oxidation state of an atom in a chemical compound helps us understand how electrons are distributed among the atoms. For the bromine atom in the bromate ion, its oxidation state is +5. This can be determined by knowing that oxygen typically has an oxidation state of -2. With three oxygens, the total comes to -6. To balance the overall charge to -1, bromine must have an oxidation state of +5.

Bromate ions are often utilized as oxidizing agents due to their ability to effectively transfer oxygen atoms during reactions. This behavior is exemplified in the oxidation reactions, like the one that produces the perbromate ion, where the bromate ion undergoes further changes in its electron distribution.

Delving Into Perbromate Ion

The discovery and synthesis of the perbromate ion \( \mathrm{BrO}_4^- \) marked an exciting advancement in chemistry, combining bromine with four oxygen atoms to achieve a unique ion formation. The perbromate ion carries a charge of -1, similar to its bromate counterpart, but hosts a different oxidation state for bromine, which is +7. This is due to the additional oxygen atom present in perbromate, compared to the bromate ion.

To calculate the oxidation state of bromine in the perbromate ion, we follow the same principles: with oxygen's oxidation state standing at -2, four oxygens give a total of -8. For the ion's charge to be -1, bromine needs to possess an oxidation state of +7.

This high oxidation state underscores perbromate's strong oxidizing capabilities. Such characteristics are harnessed in specialized chemical reactions, where the perbromate ion can facilitate oxidation by accepting electrons from other substances, contributing to a myriad of potential applications in chemical synthesis and analysis.

Mastering Chemical Equation Balancing

Chemical equation balancing is a fundamental skill in chemistry, ensuring that electrons and atoms are conserved in reactions. In any chemical equation, it's essential to have the same number of each type of atom on both the reactant and product sides, abiding by the law of conservation of mass.

To balance a chemical equation, such as the one converting bromate ions to perbromate ions, you start by identifying the reactants and products. Then, adjust coefficients—the numbers placed in front of compounds—to ensure each element is evenly represented on both sides of the equation. This ensures that neither mass nor charge is unbalanced.

For the reaction discussed here, the equation is balanced as follows:

• Identify reactants: \( \mathrm{BrO}_{3}^- \) and \( \mathrm{XeF}_2 \)
• Identify products: \( \mathrm{BrO}_{4}^- \), \( \mathrm{Xe} \), and \( \mathrm{HF} \)
• Balance each element gradually, starting from xenon, which is already balanced, to oxygen, fluorine, and finally hydrogen

Ultimately, balancing chemical equations requires practice and patience, but once mastered, it unveils the fascinating stoichiometric relationships inherent in chemical reactions. This balanced snapshot of a chemical process ensures clarity in understanding how different compounds interact during these transformative interactions.