In iron(III) oxalate Fe2(C2O4)3 the oxidation numbers of iron, carbon and oxygen are, in that order:

Question

BlackTom AI · Accepted Answer

Oxalate is the key: C2O4^2− has overall charge −2 and oxygen is normally −2 (except in peroxides). Use oxidation-number rules and charge balance.

1) Relevant knowledge points and concepts
- Sum of oxidation numbers in a neutral compound = 0.
- Sum of oxidation numbers in a polyatomic ion = the ion charge.
- Oxygen is usually −2 (unless in peroxide or O2− species).
- The formula name iron(III) indicates Fe is in the +3 oxidation state.

2) Step-by-step reasoning
- Write overall charge balance for Fe2(C2O4)3 (neutral compound):
  2(oxidation state of Fe) + 3(oxidation state of C2O4) = 0.
- Each oxalate ion C2O4 has charge −2, so 3 × (−2) = −6 total from oxalates.
- Therefore 2(Fe) + (−6) = 0 → 2(Fe) = +6 → Fe = +3.

3) Oxalate internal calculation (showing intermediate steps)
- For one C2O4^2−: let oxidation state of carbon be x (each carbon same).
  2x + 4(oxidation state of O) = −2.
- Oxygen = −2, so: 2x + 4(−2) = −2 → 2x − 8 = −2 → 2x = +6 → x = +3.
- Thus each carbon is +3 and each oxygen is −2.

4) Why the correct answer is correct
- Charge balance on the whole molecule gives Fe = +3.
- Charge balance within oxalate gives C = +3 and O = −2.
- These values satisfy all oxidation-number rules and the compound’s neutrality.

5) Why the other options are incorrect (brief)
- A (+II, +II, −I): Oxygen as −1 would imply peroxide-like oxygens, not applicable here; also Fe would not balance the −6 from oxalates.
- B (+III, −II, −II): This assigns carbon −2, contradictory to oxalate calculation and typical carbon chemistry in oxalate.
- C (+II, −III, −II): Fe +2 and C −3 do not satisfy overall neutrality and contradict the iron(III) naming.

6) Conclusion
Therefore the oxidation numbers are Fe +III, C +III, O −II. This matches the provided correct answer: D. + III, + III, - II.

In iron(III) oxalate Fe2(C2O4)3 the oxidation numbers of iron, carbon and oxygen are, in that order:单项选择题

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