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?

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: \[\mathrm{BrO}_{3}^- + \mathrm{XeF}_{2} \rightarrow \mathrm{BrO}_{4}^- + \mathrm{Xe} + 2\,\mathrm{HF}\]

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.