How to Determine Oxidation States

Oxidation states are bookkeeping. They’re the hypothetical charges you’d see if every bond were ionic, and they have one job: to make redox electron-counting tractable. The chromium atom in dichromate isn’t really sitting on a +6 charge — but assigning it +6 lets you balance the half-reaction with Cr going to +3 and immediately know that three electrons moved per chromium. Without oxidation numbers, you’re trying to balance redox by inspection. With them, it becomes algebra. The rules are a strict priority list — when two rules conflict, the higher-priority one wins.

The priority list

Apply these in order. When two rules disagree, the earlier rule wins.

1. Free elements: 0. Fe metal, O₂, S₈, P₄, Cl₂ — every atom is at oxidation state 0.
2. Monatomic ions: their charge. Na⁺ → +1; Cl⁻ → −1; Fe³⁺ → +3; S²⁻ → −2.
3. Fluorine: always −1 in compounds. F has the highest electronegativity, period.
4. Oxygen: usually −2. Exceptions: peroxides (O–O bonds, e.g., H₂O₂, Na₂O₂) → −1; superoxides (KO₂) → −½; OF₂ → +2 (rule 3 beats rule 4).
5. Hydrogen: usually +1. Exception: metal hydrides (NaH, CaH₂, LiAlH₄) → −1.
6. Group 1 metals: +1 in compounds.
7. Group 2 metals: +2 in compounds.
8. Sum rule. Oxidation states across a neutral compound sum to 0; across a polyatomic ion they sum to the ion’s charge.

Apply known rules first, then use the sum rule to solve for whatever is unknown.

The recipe

1. Tag every atom you can from rules 1–7.
2. Write the sum-rule equation: total = 0 (neutral) or = charge (ion).
3. Solve for the unknown.
4. Verify by re-summing.

Worked example: H₂SO₄

• H: +1 (rule 5), 2 atoms
• O: −2 (rule 4), 4 atoms
• S: ?

Sum = 0: 2(+1) + S + 4(−2) = 0 → S = +6

Verify: 2 + 6 − 8 = 0. ✓

Worked example: dichromate ion Cr₂O₇²⁻

• O: −2, 7 atoms
• Cr: ?, 2 atoms
• Total charge: −2

2(Cr) + 7(−2) = −2 → 2(Cr) = 12 → Cr = +6

Both Cr atoms sit at +6. The reduction in chromate-to-Cr³⁺ chemistry is therefore a 3-electron change per chromium.

Worked example: KMnO₄

• K: +1 (rule 6)
• O: −2
• Mn: ?

+1 + Mn + 4(−2) = 0 → Mn = +7

This is Mn at its maximum oxidation state — which is why permanganate is such an aggressive oxidizer.

Worked example: a peroxide

Na₂O₂ (sodium peroxide).

• Na: +1 (rule 6, higher priority than oxygen’s “usually −2”)
• O: ?

2(+1) + 2(O) = 0 → O = −1

This is one of the rare cases where rule 4’s exception kicks in. The O–O bond in the peroxide ion (O₂²⁻) means each oxygen averages −1, not −2. You spot peroxides by looking for the O₂²⁻ unit or by recognizing common ones (H₂O₂, Na₂O₂, BaO₂).

Worked example: identifying redox

Fe + CuSO₄ → FeSO₄ + Cu

• Fe(s): 0 (free element)
• Cu in CuSO₄: SO₄²⁻ contributes −2, so Cu = +2
• Fe in FeSO₄: same logic, Fe = +2
• Cu(s): 0

Fe: 0 → +2, lost 2 electrons → oxidized (reducing agent) Cu: +2 → 0, gained 2 electrons → reduced (oxidizing agent)

Two electrons transferred per Fe/Cu pair. Now you can balance the half-reactions properly.

A few less-obvious cases

• OF₂. Rule 3 beats rule 4. F = −1 (always), so 2(−1) + O = 0 → O = +2.
• CaH₂. Rule 7 beats rule 5. Ca = +2, so 2 + 2H = 0 → H = −1.
• NH₄⁺. H = +1, sum = +1: N + 4(+1) = +1 → N = −3.
• Thiosulfate S₂O₃²⁻. O = −2, sum = −2: 2S + 3(−2) = −2 → S average = +2. (The two sulfurs are actually inequivalent — central S is +5, terminal is −1 — but the “average +2” works for redox bookkeeping.)

Traps people fall into

• Forgetting peroxides. H₂O₂ has O at −1, not −2. Same for Na₂O₂, BaO₂. If you assign O = −2 here, every other atom in the compound comes out wrong.
• Forgetting metal hydrides. NaH, LiAlH₄, CaH₂ — H is −1. If you assign H = +1 to NaH, you’d conclude Na = −1, which is nonsense for a Group 1 metal.
• Setting an ion’s sum to zero. SO₄²⁻‘s oxidation states sum to −2, not 0. The minus-2 charge has to live somewhere on the bookkeeping.
• Treating average oxidation states as physically real. In Fe₃O₄ the iron averages +8/3, but the actual structure is one Fe²⁺ and two Fe³⁺. The average is a redox-counting tool, not a description of the bonding.

Practice

Try these against the Oxidation State Calculator:

1. Assign all oxidation states in HNO₃.
2. Mn in MnO₂?
3. Assign all oxidation states in Na₂Cr₂O₇.
4. N in NH₄⁺?
5. In 2Al + 3CuCl₂ → 2AlCl₃ + 3Cu, identify oxidizing and reducing agents.