Problem 22 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Chlorine reacts with hot and concentrated \(\mathrm{NaOH}\) and produces compounds \((X)\) and \((Y)\). Compound \((X)\) gives white precipitate with silver nitrate solution. The average bond order between \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in \((Y)\) is

Step-by-Step Solution

Answer

The average bond order between Cl and O in (Y) is \(\frac{4}{3}\).

Step 1: Write the Reaction Equation

When chlorine reacts with hot and concentrated sodium hydroxide (\(\text{NaOH} \)), the reaction produces sodium chloride (\(\text{NaCl}\)) as compound \((X)\) and sodium chlorate (\(\text{NaClO}_3\)) as compound \((Y)\): \[ 3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O} \].

Step 2: Identify Compound (X)

Compound \((X)\) is identified as sodium chloride (\(\text{NaCl}\)) because it gives a white precipitate with silver nitrate due to the formation of silver chloride (\(\text{AgCl}\)).

Step 3: Analyze Compound (Y)

Compound \((Y)\) is sodium chlorate (\(\text{NaClO}_3\)). We need to determine the average bond order of the \(\text{Cl}\)-\(\text{O}\) bonds in chlorate ion \((\text{ClO}_3^-)\).

Step 4: Calculate Bond Order in Chlorate Ion

The chlorate ion \(\text{ClO}_3^-\) has three equivalent \(\text{Cl-O}\) bonds. The molecule can be described as a resonance hybrid of several structures, where \(\text{Cl}\) shares a double bond with one \(\text{O}\) atom at a time due to delocalized electrons. Bond order is calculated as the total number of bonds divided by the number of resonance structures, which is \[ \text{Bond Order} = \frac{4}{3} \] for each \(\text{Cl-O}\) bond.

Key Concepts

Sodium Chloride Formation

When chlorine gas reacts with hot and concentrated sodium hydroxide, it results in the formation of sodium chloride (\( \text{NaCl} \)) among other compounds. This reaction is not only important in laboratory synthesis but also plays a role in industrial processes. Here's a simple way to understand the formation of sodium chloride in this reaction:

• In the chemical equation,\[ 3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O} \], sodium chloride emerges as product \((X)\).
• Sodium chloride's significance in the solution is evident as it can be easily identified by adding silver nitrate, which reacts with sodium chloride to form a white precipitate of silver chloride \( (\text{AgCl}) \).

Remember, the presence of the white precipitate is a definitive test for chloride ions, and hence confirms the formation of sodium chloride.

Resonance Structures

In chemistry, resonance structures are different ways of drawing the same molecule that help illustrate how electrons are distributed. For sodium chlorate (\( \text{NaClO}_3 \)) as compound \((Y)\) in our reaction, the concept of resonance is crucial:

• In the chlorate ion (\( \text{ClO}_3^- \)), the distribution of electrons can be represented by different forms.
• All these forms suggest that the central chlorine atom shares a double bond alternately with each of the oxygen atoms, thus illustrating the concept of delocalized electrons.

To put it simply, resonance structures don't change the actual chemical structure but help us understand possible interactions. This idea is fundamental in visualizing how electrons can spread over different parts of a molecule, providing stability.

Bond Order Calculation

A bond order quantifies the number of bonds between two atoms. For the chlorate ion (\( \text{ClO}_3^-\)), calculating the bond order provides insights into the molecular structure:

• The bond order is calculated using the formula:\[\text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}}\]
• In the chlorate ion, there are four bonds shared among three chlorine-oxygen connections, resulting in a bond order of \( \frac{4}{3} \).This suggests partial sharing of double bonds due to resonance.
• A bond order greater than 1 but less than 2 indicates partial double bond characteristics, contributing to the molecule's stability.

Understanding bond order helps predict reactivity, stability, and bond length, making it a valuable tool in chemical analysis.