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

We have chlorine reacting with hot and concentrated sodium hydroxide. The balanced equation actually taught in the N.C.E.R.T. text is

$$3\,\text{Cl}_2 + 6\,\text{NaOH}\;(\text{hot, conc.}) \;\longrightarrow\; 5\,\text{NaCl} + \text{NaClO}_3 + 3\,\text{H}_2\text{O}$$

From this single reaction we can immediately identify the two products that contain chlorine in different oxidation states:

$$\text{NaCl}\;(\text{X}) \quad\text{and}\quad \text{NaClO}_3\;(\text{Y}).$$

Compound (X) is $$\text{NaCl}$$. When treated with silver-nitrate solution $$\text{AgNO}_3$$, it forms a curdy white precipitate of $$\text{AgCl}$$, exactly as described in the statement of the question, so our identification of (X) is consistent.

Therefore compound (Y) must be $$\text{NaClO}_3$$. In aqueous solution this salt furnishes the chlorate ion $$\text{ClO}_3^-$$, and it is in this ion that we have to discuss the average bond order of the Cl-O bonds.

First, recall the definition that will be used. For a species that shows resonance, the conventional (Pauling) bond order is defined as

$$\text{Bond order} \;=\; \frac{\text{Total bond multiplicity in all canonical forms}}{\text{Number of equivalent bonds}\;\times\;\text{Number of canonical forms}}.$$

Put colloquially, we count the value (1 for a single, 2 for a double, etc.) of each Cl-O bond in every possible resonance structure, add them all up, and then divide by the number of resonance structures times the three equivalent Cl-O positions. Every Cl-O bond in the real molecule has the same “average” order obtained in this way.

Now let us draw the lowest-energy (formal-charge-minimised) resonance structures of $$\text{ClO}_3^-$$. In each of them chlorine is hypervalent (uses a 3d orbital) and expands its octet. The pattern is

Cl has two Cl=O double bonds and one Cl-O− single bond.

Because any one of the three oxygen atoms can be the one that bears the negative charge, we obtain three canonical forms, all of equal weight:

  Structure 1: O=Cl=O with one O−
  Structure 2: O=Cl-O−=O
  Structure 3: O−-Cl=O=O

Let us now tabulate the bond multiplicities.

• In any one canonical form there are exactly two Cl=O double bonds. Each double bond carries a multiplicity of 2.
• In the same canonical form there is exactly one Cl-O single bond. A single bond carries a multiplicity of 1.

Therefore, in a single canonical form the total Cl-O bond multiplicity is

$$2 + 2 + 1 \;=\; 5.$$

There are exactly $$3$$ such canonical forms. Hence the grand total of Cl-O bond multiplicity summed over all canonical forms is

$$5 \times 3 \;=\; 15.$$

The number of equivalent Cl-O bond positions is $$3$$ (because there are three oxygens attached to the same chlorine). The Pauling definition now gives

$$\text{Average bond order} \;=\; \frac{15}{3 \times 3} \;=\; \frac{15}{9} \;=\; \frac{5}{3} \;=\; 1.67.$$\

Thus every Cl-O bond in the chlorate ion possesses an average order of $$\displaystyle 1.67$$, exactly as required.

So, the answer is $$1.67$$.