Chlorine reacts with hot and concentrated `NaOH` and produces compounds (P) and (Q). Compound (P) gives white precipitate with silver nitrate solution. The average bond order between `Cl and O` atoms in (Q) is Report your answer up to two decimal places.

Answer

To solve the problem step by step, we will analyze the reactions and the compounds formed when chlorine reacts with hot and concentrated sodium hydroxide (NaOH).

### Step 1: Identify the Reaction
When chlorine (Cl₂) reacts with hot and concentrated NaOH, it undergoes a disproportionation reaction. The balanced equation for this reaction is:

\[ \text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO}_3 + \text{H}_2\text{O} \]

From this reaction, we can identify the products:
- Compound (P) is sodium chloride (NaCl).
- Compound (Q) is sodium chlorate (NaClO₃).

### Step 2: Analyze Compound (P)
We are told that compound (P) gives a white precipitate with silver nitrate (AgNO₃). When NaCl reacts with AgNO₃, it forms silver chloride (AgCl), which is a white precipitate:

\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} \downarrow + \text{NaNO}_3 \]

This confirms that compound (P) is indeed NaCl.

### Step 3: Analyze Compound (Q)
Now, we need to find the average bond order between chlorine (Cl) and oxygen (O) atoms in compound (Q), which is NaClO₃.

### Step 4: Structure of NaClO₃
The structure of sodium chlorate (NaClO₃) can be represented as follows:

- Sodium (Na) is a cation.
- Chlorine (Cl) is bonded to three oxygen (O) atoms.
- The structure can be drawn as:
- One Cl=O double bond
- Two Cl-O single bonds

### Step 5: Resonance Structures
Sodium chlorate has resonance structures. The resonance structures can be represented as follows:

1. One structure with Cl=O (double bond) and two Cl-O (single bonds).
2. The double bond can shift to one of the other oxygen atoms, creating two more resonance structures.

Thus, we can summarize the resonance structures:
1. Cl=O, Cl-O, Cl-O⁻
2. Cl-O, Cl=O, Cl-O⁻
3. Cl-O, Cl-O, Cl=O

### Step 6: Count Bonds
Now, we need to calculate the bond order. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}} \]

In each resonance structure:
- In the first structure, there is 1 double bond and 2 single bonds (total of 3 bonds).
- In the second structure, there is also 1 double bond and 2 single bonds (total of 3 bonds).
- In the third structure, again, there is 1 double bond and 2 single bonds (total of 3 bonds).

Total bonds across all structures = 3 (from each structure) × 3 (number of structures) = 9 bonds.

### Step 7: Calculate Average Bond Order
Now, we can calculate the average bond order:

\[ \text{Bond Order} = \frac{9 \text{ bonds}}{3 \text{ resonance structures}} = 3 \]

However, since we have 1 double bond and 2 single bonds in each structure, we need to consider the bond contributions:

- In each structure, the contribution to the bond order is:
- 2 bonds (from the double bond) + 1 bond (from each single bond) = 5 bonds total.

Thus, the average bond order is:

\[ \text{Bond Order} = \frac{5}{3} \approx 1.67 \]

### Final Answer
The average bond order between Cl and O atoms in NaClO₃ is **1.67**.

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