Reaction conditions determine the product and Cl oxidation state in Z Steps:

• Calculate moles from mass %: Cl = 45.1/35.5 ≈ 1.27 mol, O = 54.9/16 ≈ 3.43 mol.
• Find ratio Cl:O ≈ 1:2.7, closest to 1:3 for ClO₃ empirical formula.
• Determine oxidation state of Cl in ClO₃: +5 (Cl +5, 3O at -2 balances to -1 charge in chlorate ion).
• Match preparation: hot concentrated NaOH + Cl₂ yields NaClO₃ (chlorate, Cl +5); other conditions yield hypochlorite (Cl +1). Why D is correct:
• Hot concentrated NaOH with Cl₂ disproportionates to chlorate (3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O), where Cl oxidation state is +5. Why the others are wrong:
• A: Cold dilute NaOH forms hypochlorite (Cl₂ + 2NaOH → NaClO + NaCl + H₂O), Cl +1.
• B: Hot dilute NaOH primarily forms hypochlorite, not chlorate +5.
• C: Cold concentrated NaOH forms hypochlorite, Cl +1. Final answer: D