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 ________. (Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Hot, concentrated NaOH means chlorate, so share five bonds' worth of order over three Cl–O links.

1. Identify the products. With hot, concentrated alkali, chlorine disproportionates all the way to chlorate. NaCl gives a white precipitate of AgCl with silver nitrate, so (X) is NaCl and (Y) is NaClO₃.

   3Cl₂ + 6NaOH →(hot, conc.) 5NaCl + NaClO₃ + 3H₂O

2. Structure of chlorate. In ClO₃⁻, chlorine forms one Cl-O⁻ single bond and two Cl=O double bonds. Resonance spreads these equally over all three Cl–O bonds.
3. Average bond order. Total bond order is 1 + 2 + 2 = 5, shared over 3 bonds, so the average is 5/3 ≈ 1.67.
4. So. The average Cl–O bond order in (Y) rounds to 2.

The common slip: 1. Taking (Y) as NaOCl, the product with cold dilute NaOH, gives a single Cl–O bond and an answer of 1.

The idea to keep. Cold, dilute alkali gives hypochlorite. Hot, concentrated alkali gives chlorate. For a resonance ion, average bond order = total bonds between the two atom types ÷ number of such links.

Answer from the NTA final answer key. Solution by PrepWiser.