OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 2

✅ Correct Answer

B: NaClO₄

💡 Key Knowledge

• Oxidation States: The Roman numeral (VII) in chlorate(VII) indicates that chlorine has an oxidation state of +7 in this polyatomic ion.
• Ionic Charges: Sodium is in Group 1, so it always forms a +1 ion (Na⁺). The question states the chlorate(VII) ion has a 1− charge. Combining Na⁺ and ClO₄⁻ in a 1:1 ratio yields a neutral compound.
• Deriving the Formula: Let the unknown formula of the chlorate(VII) ion be ClOₓ⁻. Using oxidation states: oxidation state of Cl + x × (oxidation state of O) = total ion charge. Therefore, (+7) + x(−2) = −1, which gives −2x = −8, so x = 4. This gives the formula ClO₄⁻ .

🧠 Exam Technique

Do not confuse the Roman numeral oxidation state with the number of oxygen atoms! A common mistake is assuming chlorate(VII) contains 7 oxygen atoms (option C). Always calculate the formula using the rules of oxidation numbers rather than blindly matching digits.

❌ Common Errors

• Option C (NaClO₇): Assuming the Roman numeral (VII) directly represents 7 oxygen atoms instead of an oxidation state of +7.
• Option A (NaClO₃): Confusing chlorate(VII) with chlorate(V) (where chlorine has an oxidation state of +5).
• Option D (NaClO₈): Incorrectly balancing or misinterpreting the stoichiometry of chlorine and oxygen.