A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with silver nitrate gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y'. The compound 'X' is treated with hot concentrated NaOH, a compound Z is formed. The average bond order between Cl and O atom in (Z) is ________.

Explanation

To solve the problem step by step, we will analyze the information provided and deduce the necessary compounds and their properties.

### Step 1: Identify the gas 'X'
The problem states that a gas 'X' is passed through water to form a saturated solution. The solution, when treated with silver nitrate, gives a white precipitate. This indicates that 'X' must be a gas that forms hydrochloric acid (HCl) in solution, as HCl reacts with silver nitrate (AgNO3) to form silver chloride (AgCl), which is a white precipitate.

**Conclusion**: Gas 'X' is chlorine (Cl2).

### Step 2: Reaction of 'X' with water
When chlorine gas (Cl2) is dissolved in water, it forms a mixture of hydrochloric acid (HCl) and hypochlorous acid (HOCl):
\[ Cl_2 + H_2O \rightarrow HCl + HOCl \]

### Step 3: Reaction with magnesium ribbon
The saturated solution of HCl will react with magnesium (Mg) ribbon, producing magnesium chloride (MgCl2) and hydrogen gas (H2):
\[ Mg + 2HCl \rightarrow MgCl_2 + H_2 \]

Here, 'Y' is identified as hydrogen gas (H2).

### Step 4: Treatment of 'X' with hot concentrated NaOH
Next, we treat chlorine gas (Cl2) with hot concentrated sodium hydroxide (NaOH). This reaction produces sodium chloride (NaCl), sodium hypochlorite (NaClO), and sodium chlorate (NaClO3):
\[ Cl_2 + 2NaOH \rightarrow NaCl + NaClO3 + H_2O \]

The compound 'Z' formed in this reaction is sodium chlorate (NaClO3).

### Step 5: Determine the average bond order between Cl and O in 'Z'
To find the average bond order between the chlorine (Cl) and oxygen (O) atoms in sodium chlorate (NaClO3), we need to analyze its structure. Sodium chlorate has the following Lewis structure:
- Cl is bonded to three oxygen atoms: one with a double bond and two with single bonds.
- The resonance forms of ClO3^- show that there are multiple ways to arrange the double bond.

The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}} \]

In NaClO3:
- Total bonds between Cl and O: 3 (one double bond and two single bonds).
- Number of resonance structures: 3.

Calculating the bond order:
\[ \text{Bond Order} = \frac{3}{3} = 1 \]

However, since one of the bonds is a double bond, we can also consider the effective bond order to be:
\[ \text{Effective Bond Order} = \frac{4}{3} \approx 1.33 \]

### Final Answer
The average bond order between Cl and O atoms in compound 'Z' (NaClO3) is approximately **1.33**.

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