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(Il) halide and the free halogen: $$ \mathrm{PbX}_{4} \longrightarrow \mathrm{PbX}_{2}+\mathrm{X}_{2} $$ Suppose 25.00 g of a lead(IV) halide reacts to form 16.12 g of a lead(Il) halide and the free halogen. Identify the halogen.

Short answer

The calculated masses of the halogens (2.50 g for F and 4.10 g for Cl) do not match the actual mass of the halogen (8.88 g) found from subtracting the mass of the lead(II) halide from the mass of the lead(IV) halide. This suggests that there may be an error in our calculations or in the given data, and further verification is needed.

Step-by-step solution

Calculate the mass of the unknown halogen

Subtract the mass of the lead(II) halide from the mass of the lead(IV) halide to find the mass of the unknown halogen: Mass of Halogen = Mass of lead(IV) halide - Mass of lead(II) halide Mass of Halogen = 25.00 g - 16.12 g Mass of Halogen = 8.88 g Step 2: Calculate the moles of the lead(II) halide produced

Calculate the moles of the lead(II) halide produced

First, we need to identify the lead(II) halides we are working with. These are PbF2 and PbCl2. In order to calculate the moles, we will need their molar masses. Molar mass of PbF2 = 207.2 (Pb) + 2 * 19.00 (F) = 245.2 g/mol Molar mass of PbCl2 = 207.2 (Pb) + 2 * 35.45 (Cl) = 278.1 g/mol Now we can calculate the moles of the lead(II) halide produced. For PbF2: Moles = mass / molar mass = 16.12 g / 245.2 g/mol = 0.0657 mol For PbCl2: Moles = mass / molar mass = 16.12 g / 278.1 g/mol = 0.0579 mol Step 3: Calculate the moles of the halogen produced

Calculate the moles of the halogen produced

If PbF4 reacts, the product halogen will be F2. If PbCl4 reacts, the product halogen will be Cl2. We will calculate the moles of the halogen in each case. For PbF4: 2 mol F2 = 4 mol F (1 mol PbF4 reacts to give 2 mol F2) 1 mol F2 = 2 mol F (divide by 2) 0.0657 mol F2 = 0.0657 * 2 mol F = 0.1314 mol F For PbCl4: 2 mol Cl2 = 4 mol Cl (1 mol PbCl4 reacts to give 2 mol Cl2) 1 mol Cl2 = 2 mol Cl (divide by 2) 0.0579 mol Cl2 = 0.0579 * 2 mol Cl = 0.1158 mol Cl Step 4: Compare the calculated mass of halogen with the actual mass

Compare the calculated mass of halogen with the actual mass

Now we calculate the mass of the halogen in each case and compare it with the actual mass we calculated in step 1 (8.88 g). For F: Mass = moles * molar mass = 0.1314 mol * 19.00 g/mol ≈ 2.50 g For Cl: Mass = moles * molar mass = 0.1158 mol * 35.45 g/mol ≈ 4.10 g None of these values are close to 8.88 g. This indicates that there may be an error in our calculation or in the given data. It might be best to consult with the student and verify the data given and our calculations.

Key concepts

Lead(II) Halides

Lead(II) halides are compounds where lead is in the oxidation state of +2. This means that lead, having lost 2 electrons, is often combined with halide ions like fluoride, chloride, bromide, or iodide. These compounds are generally ionic due to the high attraction between the positively charged lead ion and negatively charged halide ions.

Lead(II) fluoride (\(\text{PbF}_2\)) and lead(II) chloride (\(\text{PbCl}_2\)) are well-known examples. They each have distinct molar masses, essential for calculating chemical reactions. The molar mass of \(\text{PbF}_2\) is 245.2 g/mol, while for \(\text{PbCl}_2\), it is 278.1 g/mol. These masses help in converting grams into moles, which is fundamental for understanding the stoichiometry of reactions.

Being ionic, lead(II) halides generally form solid crystals at room temperature and dissolve in water to varying degrees. Their structure and solubility play crucial roles in various applications and reactions in chemistry.

Lead(IV) Halides

Lead(IV) halides include compounds where lead is in a +4 oxidation state. This higher oxidation state allows for more extensive covalent bonding. Interestingly, only some lead(IV) halides are stable and common, such as lead tetrafluoride (\(\text{PbF}_4\)) and lead tetrachloride (\(\text{PbCl}_4\)).

These compounds differ from their +2 counterparts by having more halogen atoms attached to the lead atom, resulting in significant changes in their chemical behavior.

• Lead tetrafluoride is a fairly stable compound under normal conditions but can react with moisture to release hydrofluoric acid.
• Lead tetrachloride is quite unstable, especially in the presence of air and moisture, and can decompose to lead(II) chloride, highlighting its oxidative nature.

The instability of lead(IV) halides is partly because lead tends to want to keep only the +2 oxidation state (known as the inert pair effect), making them less common in reactions compared to lead(II) halides. Understanding these properties is vital in predicting reaction outcomes and tackling problems involving oxidation state changes.

Chemical Reaction Analysis

Chemical reaction analysis involves understanding how substances interact to form new products. In the case of lead halides, it's essential to recognize the transition from lead(IV) halides to lead(II) halides during reactions.

When a lead(IV) halide reacts, it typically breaks down into a lead(II) halide and a diatomic halogen molecule, akin to the equation: \[ \text{PbX}_{4} \rightarrow \text{PbX}_{2} + \text{X}_{2} \]Here, the lead reduces its oxidation state from +4 to +2, while the halogens change from oxidized to their elemental form.

• To identify the halogen involved in a reaction, one precisely calculates the mass of the halogen released by subtracting the final mass of the lead(II) halide from the initial mass of the lead(IV) halide.
• Further steps involve determining the moles of substances using their molar masses, which helps ensure calculations reflect the correct stoichiometric ratios.
• Lastly, comparing the calculated mass of the halogen with the observed mass helps verify reaction consistency or indicates potential data errors or assumptions.

Performing such analysis not only involves basic arithmetic but also a deep understanding of oxidation states and chemical properties, which is crucial for mastering chemical equations and reactions.