1. Reaction of chlorine with hot, concentrated NaOH

Chlorine reacts with hot and concentrated sodium hydroxide as:

$3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$

So the products are:

• $(X) = NaCl$
• $(Y) = NaClO_3$

2. Identification of compound (X)

It is given that $(X)$ gives a white precipitate with silver nitrate solution.

Since:

$NaCl + AgNO_3 \rightarrow AgCl\downarrow + NaNO_3$

and $AgCl$ is a white precipitate, this confirms:

$X = NaCl$

Hence,

$Y = NaClO_3$

3. Structure of chlorate ion in $(Y)$

In sodium chlorate, the anion is $ClO_3^-$.

We need the average bond order between $Cl$ and $O$ atoms in $ClO_3^-$.

The chlorate ion has resonance structures. The major canonical forms contain:

• two $Cl=O$ double bonds
• one $Cl-O^-$ single bond

These single and double bonds are delocalized over the three oxygen atoms.

4. Average bond order calculation

Total bond order over the three $Cl-O$ bonds:

$2 + 2 + 1 = 5$

Average bond order:

$\text{Average bond order} = \frac{5}{3} = 1.666\ldots$

So,

$\boxed{1.67}$

For an integer-type/numerical format, the value is approximately:

$\boxed{1.66 \text{ to } 1.67}$

5. Comparison with stored correct answer

Stored correct answer: $1.66$ to $1.67$

Our derived answer matches this range.