Question

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.

Short answer

The unknown halogen is Chlorine (Cl), as the reaction involves PbCl4 and the mole ratio of lead to chlorine is closer to the expected integer ratio (1:4) compared to the alternative halogen, Fluorine.

Step-by-step solution

Determine the mass of 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 = \(25.00\mathrm{~g} - 16.12\mathrm{~g} = 8.88\mathrm{~g}\)

Calculate moles of lead and halogen

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

Identify 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).

Key concepts

Oxidation States

In chemistry, an oxidation state (often called an oxidation number) is used to describe the partitioning of electrons among atoms in a molecule. Specifically, it indicates the degree of oxidation of an atom in a chemical compound. For lead in the context of lead halides, we observe two primary oxidation states: +2 and +4.
When lead is in the +2 oxidation state, it tends to form ionic halides such as lead(II) chloride (PbCl₂). However, in its +4 oxidation state, lead can form covalent halides, like PbF₄ or PbCl₄.
Understanding oxidation states is critical because it helps predict the type of chemical reactions lead compounds undergo. In redox reactions, elements change their oxidation states by gaining or losing electrons. For example, when a lead(IV) halide reacts with bromide or iodide ions, it oxidizes them, simultaneously reducing itself to the +2 state, i.e., forming a lead(II) halide.

Molar Mass

The molar mass of a compound is its molecular weight expressed in grams per mole. It is a crucial concept in stoichiometry as it enables the conversion between mass and moles of a substance. In this exercise, lead (Pb) has a molar mass of 207.2 g/mol.
With this information, one can calculate how many moles of lead exist within a given mass of lead halide. For example, the moles of lead in a substance are calculated using the formula:

• Moles = \( \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \)

When working with lead(IV) halides, we use this relation to determine the mole quantity. It's helpful for evaluating how many moles of other reactants or products might form during a chemical reaction.

Chemical Reaction

A chemical reaction involves the transformation of one or more substances into different substances via the breaking and forming of chemical bonds. In the exercise with lead(IV) halides, the reaction we're examining is an oxidation-reduction (redox) process.
Here, lead(IV) halide reacts and reduces to lead(II) halide while the halide ions are oxidized. This is an example of a reaction where substances are undergoing changes in their oxidation states synchronously, demonstrating principles of electron transfer. The basic form of this reaction can be described as:

• \( \text{PbX}_4 \rightarrow \text{PbX}_2 + X_2 \)

"X" represents a generic halogen. In this situation, determining exactly which halogen X represents becomes an essential part of solving the exercise.

Mole Ratio

Mole ratio is an expression of the relationship between the amounts, usually in moles, of any two compounds involved in a chemical reaction. This ratio is essential for balanced chemical equations and stoichiometry. It shows us how much of one reactant is needed to complete the reaction with another.
In the case of lead(IV) halide reacting to form lead(II) halide, we analyze the mole ratio between lead and the different potential halogens to find which specifically is reacting. By comparing the calculated mole ratios to the expected stoichiometric ratios, one can infer which compound is most likely involved.
For our reaction: we looked at the theoretical Pb to halogen ratios in compounds PbF₄ and PbCl₄. By calculating the actual moles of halogens and matching the observed mole ratio, we concluded chlorine with its closest expected ratio (1:4) fits the reaction scenario best.