Lead forms compounds in the \(+2\) and \(+4\) oxidation states. All lead(II) halides are known (and are known to be ionic). Only \(\mathrm{PbF}_{4}\) and \(\mathrm{PbCl}_{4}\) are known among the possible lead(IV) halides. Presumably lead(IV) oxidizes bromide and iodide ions, producing the lead(II) halide and the free halogen: Suppose \(25.00 \mathrm{~g}\) of a lead(IV) halide reacts to form \(16.12 \mathrm{~g}\) of a lead(II) halide and the free halogen. Identify the halogen.

Since we know that the mass of the reactant is conserved, and given the masses of lead(IV) halide and lead(II) halide, we can determine the mass of the halogen formed in the reaction: Mass of halogen = Mass of lead(IV) halide - Mass of lead(II) halide Mass of halogen = \(25.00\mathrm{~g} - 16.12\mathrm{~g} = 8.88\mathrm{~g}\)

We need to find the moles of lead and halogen in the reactant and products. We will use their molar masses for this calculation. Molar mass of lead (Pb): 207.2 g/mol Moles of Pb in lead(II) halide = \(\frac{16.12\text{g}}{207.2\text{g/mol}} = 0.0778\text{mol}\) Since moles of lead do not change throughout the reaction, there are 0.0778 mol of lead in the lead(IV) halide as well. Now, we can find the moles of halogen. Moles of halogen = \(\frac{8.88\mathrm{~g}}{x\mathrm{~g/mol}}\), where x is the molar mass of the halogen

Since only two possibilities for lead(IV) halides are given, we can narrow down to either F (fluorine) or Cl (chlorine). Molar mass of fluorine (F): 19.0 g/mol Molar mass of chlorine (Cl): 35.5 g/mol Calculate the number of moles for both halogens: Moles of F = \(\frac{8.88\mathrm{~g}}{19.0\mathrm{~g/mol}} = 0.467\mathrm{~mol}\) Moles of Cl = \(\frac{8.88\mathrm{~g}}{35.5\mathrm{~g/mol}} = 0.250\mathrm{~mol}\) Based on step 2, we have 0.0778 mol of lead. Therefore, the ratio of lead to halogen is: For PbF4: Mole ratio (Pb:F) = 0.0778:0.467 = 1:6 For PbCl4: Mole ratio (Pb:Cl) = 0.0778:0.250 = 1:3.21 The ratio of lead to halogen for PbCl4 is closer to the expected integer ratio (1:4) than for PbF4. Therefore, it's most likely that the unknown halogen is Chlorine (Cl). The reaction involves PbCl4 and the unknown halogen is Chlorine (Cl).