Compound Z is chlorate with Cl(+5), formed under hot concentrated conditions

Steps:

• Assume 100 g Z: 45.1 g Cl (1.27 mol using atomic mass 35.5), 54.9 g O (3.43 mol using 16).
• Mole ratio Cl:O = 1:2.7, approximates Cl₂O₅ empirical formula.
• Oxidation state in Cl₂O₅: 5O contribute -10, so 2Cl contribute +10, each Cl +5.
• Cl(+5) compound forms via Cl₂ + hot concentrated NaOH → NaClO₃ (+5 in ClO₃⁻), matching ~45% Cl, ~55% O.

Why D is correct:

• Hot concentrated NaOH reacts with Cl₂ to yield NaClO₃ (Cl +5 per 3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O), aligning with composition and oxidation state.

Why the others are wrong:

• A: Cold dilute NaOH yields NaClO (Cl +1 via Cl₂ + 2NaOH → NaCl + NaClO + H₂O).
• B: Hot dilute NaOH favors NaClO (Cl +1), not efficient for +5 chlorate.
• C: Conditions correct but oxidation state mismatch; Z requires +5, not alternative.

Final answer: D