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: $$ \mathrm{Pb} \mathrm{X}_{4} \longrightarrow \mathrm{PbX}_{2}+\mathrm{X}_{2} $$ 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.

Understanding the oxidation states of lead is crucial when studying lead halides chemistry. Lead can exhibit multiple oxidation states, with +2 (lead(II)) and +4 (lead(IV)) being the most common. The oxidation state indicates the degree of oxidation of an atom in a compound, essentially showing how many electrons an atom has lost or gained.
In lead(II) halides, which are known to be ionic, the lead atom has lost two electrons, resulting in a +2 oxidation state. In contrast, lead(IV) halides show a lead atom that has lost four electrons. However, not all lead(IV) halides are stable. Lead(IV) is a strong oxidizing agent, and as such, it can react with halides like bromide (Br-) and iodide (I-) to reduce itself to lead(II), simultaneously oxidizing the halides to their elemental form, as shown in the reaction from the exercise:
\[\begin{equation}\mathrm{PbX}_{4} \longrightarrow \mathrm{PbX}_{2} + \mathrm{X}_{2}\end{equation}\]

Stoichiometry calculations are a fundamental aspect of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. These calculations are based on the principle of conservation of mass and the mole concept. Using the balanced chemical equation, we can deduce the stoichiometry of the reaction, which helps us understand the proportions in which compounds react and form products.

Steps in Stoichiometry

• First, identify the balanced chemical equation for the reaction.
• Convert all given information (like mass) into moles, using the molar mass of the substances involved.
• Use the mole ratio from the balanced equation to find out the amount of products or reactants.
• Convert moles back into desired units (mass, volume, particles) if necessary.

In the case of the lead(IV) halides exercise, using stoichiometry we find equal moles of lead(IV) halide, lead(II) halide, and the diatomic halogen molecule. By equating the moles from the given masses using their molar masses and solving for the molar mass of the diatomic halogen, we can identify the specific halogen involved in the reaction.

The molar mass of a substance is the mass of one mole of that substance and is typically expressed in grams per mole (g/mol). It is a physical property that is essential for converting between the mass of a substance and the amount of substance in moles. To calculate the molar mass of a compound, such as lead halides, you sum up the molar mass of each constituent atom.
For instance, lead has a molar mass of 207.2 g/mol. If we consider a halide like chloride (Cl), with a molar mass of 35.5 g/mol, the molar mass of lead(II) chloride, PbCl2, would be calculated as follows:
\[\begin{equation}\text{Molar mass of PbCl}_2 = \text{Molar mass of Pb} + 2(\text{Molar mass of Cl}) \end{equation}\]\[\begin{equation}= 207.2 \; \text{g/mol} + 2(35.5 \; \text{g/mol}) = 278.2 \; \text{g/mol}\end{equation}\]Understanding molar mass allows students not only to perform stoichiometry calculations accurately but also to predict and understand the physical properties of a substance, like its density or how it behaves under different conditions.