Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following reaction: \(6 \mathrm{NaOH}+3 \mathrm{Cl}_{2} \longrightarrow 5 \mathrm{NaCl}+\mathrm{A}+3 \mathrm{H}_{2} \mathrm{O}\) What is the oxidation number of chlorine in 'A'? (a) \(-1\) (b) \(+1\) (c) \(+3\) (d) \(+5\)

Step-by-Step Solution

Answer

(b) +1

Step 1: Identify Oxidation and Reduction

First, identify which compounds chlorine is involved in. The chlorine starts in the reactant side as \( \text{Cl}_2 \) with an oxidation state of 0. On the product side, it's part of NaCl and compound A.

Step 2: Determine Chlorine's Oxidation States in Products

In NaCl, chlorine is in the \(-1\) oxidation state. Since there are five \( \text{NaCl} \) molecules produced, that totals 5 moles of chlorine going to \(-1\). Determine what's left for compound A.

Step 3: Calculate Remaining Moles of Chlorine

Originally, there are 3 moles of \( \text{Cl}_2 \) (6 atoms in total) with an oxidation state of 0. After the reaction, 5 moles of chlorine are in NaCl, leaving 1 mole left for compound A.

Step 4: Calculate Oxidation Number in Compound A

The change in oxidation state for these atoms has to account for the oxidation process. Initially, with 3 moles of \( \text{Cl}_2 \), there are 0 oxidation state chlorine for each. In NaCl, 5 moles became \(-1\). Whatever is left determines compound A's oxidation state, maintaining charge consistency. The A compound shows +1 oxidation state.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are a type of chemical reaction where the oxidation states of atoms are changed. These reactions involve the transfer of electrons between chemical species. In the given equation, chlorine (\(\text{Cl}_2\)) goes through both oxidation and reduction. This makes it a redox reaction.

• Reduction occurs when a substance gains electrons and its oxidation state decreases.
• Oxidation is the loss of electrons, resulting in an increase in the oxidation state.

During this reaction:

• Chlorine starts with an oxidation state of 0 in \(\text{Cl}_2\) because it is in its elemental form.
• As it transforms into \(\text{NaCl}\), its oxidation state changes to \(-1\), indicating reduction.
• The remaining chlorine in compound A experiences oxidation and takes on a different oxidation state.

The balance between these oxidations and reductions is critical for the conservation of charge during the reaction.

Chlorine Compounds

Chlorine is a versatile element found in many compounds, each exhibiting different oxidation states. In chemical reactions, the specific compound formation can heavily influence the oxidation state of chlorine.
In our reaction, several chlorine compounds are involved:

• \(\text{Cl}_2\): molecular chlorine with an oxidation state of 0.
• \(\text{NaCl}\): sodium chloride, where chlorine has an oxidation state of \(-1\).
• Compound A: where chlorine adopts a new oxidation state.

These chlorine compounds show how the same element can exhibit different properties based on the chemical environment. In the reaction:

• The chlorine in \(\text{Cl}_2\) exists as diatomic molecules with no charge.
• Moving to \(\text{NaCl}\), chlorine takes electrons from sodium, leading to a \(-1\) state due to a gain of electrons.
• In compound A, chlorine reaches a higher oxidation state, revealing its ability to lose electrons.

Understanding chlorine compounds' behavior is crucial for predicting product formation, particularly in reactions involving redox processes.

Oxidation States

Oxidation states, also known as oxidation numbers, are integers assigned to chemical species that describe the degree of oxidation. They help in understanding the electron transfer that occurs in redox reactions.
Each atom in a molecule is assigned an oxidation state, which aids in balancing equations and predicting the products. In the exercise:

• Chlorine begins with an oxidation state of 0 in \(\text{Cl}_2\).
• It shifts to \(-1\) in \(\text{NaCl}\), indicating reduction.
• In compound A, calculations and charge balance indicate an oxidation state of \(+1\) for chlorine, meaning it has lost electrons.

Calculating oxidation states involves:

• Identifying the shared electrons in bonds.
• Using the usual oxidation numbers for common ions and elements (e.g., \(-1\) for halogens in most compounds).
• Ensuring that the sum of oxidation states matches the overall charge of the molecule.

Studying oxidation states provides insight into the oxidative or reductive nature of substances and is fundamental in interpreting redox reactions, like the one in our problem.