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Oxidation Number Calculator — Solve for Unknown Oxidation States

Find the oxidation number of any element in a compound instantly. Enter the known oxidation states and atom counts, get the unknown — with step-by-step verification. Works for sulfuric acid, permanganate, dichromate, and any compound.

  • Data verified · June 2026
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The oxidation number (or oxidation state) is the formal charge an atom would carry if all bonds in the compound were ionic. It is the foundation of redox chemistry — you need it to identify oxidizing and reducing agents, balance half-reactions, and name inorganic compounds systematically. This calculator solves for the unknown oxidation state of one element when you know the rest.

When to use this calculator

  • Find the oxidation state of S in H₂SO₄, SO₃, SO₄²⁻, and other sulfur compounds.
  • Determine Mn in permanganate (MnO₄⁻), MnO₂, and Mn₂O₇.
  • Solve for Cr in dichromate (Cr₂O₇²⁻), CrO₄²⁻, and chromium oxides.
  • Identify oxidizing and reducing agents in redox reactions.
  • Balance redox equations using the half-reaction method.
  • Assign systematic IUPAC names to inorganic compounds.

Fundamental Oxidation Number Rules

Element / SituationOxidation NumberExample
Free element (O₂, Fe, Cl₂)0Cl₂ = 0
Monatomic ion= ionic chargeNa⁺ = +1
Fluorine (always)−1HF: F = −1
Oxygen (most compounds)−2H₂O: O = −2
Oxygen in peroxides−1H₂O₂: O = −1
Oxygen bonded to F+2OF₂: O = +2
Hydrogen (most compounds)+ 1HCl: H = +1
Hydrogen (metal hydrides)−1NaH: H = −1
Alkali metals (Group 1)+ 1Na = +1
Alkaline earth metals (Group 2)+ 2Ca = +2

Fuente: IUPAC Recommendations on Oxidation States — Pure and Applied Chemistry (iupac.org)

How it works

What is an Oxidation Number?

The oxidation number (also called oxidation state) is a formalism that assigns each atom a hypothetical charge based on electronegativity. In any compound:

  • Σ (oxidation state × atom count) = total charge of the species
  • This is the master equation used to find unknown oxidation states.

    Fundamental Rules at a Glance

    Element / SituationOxidation Number
    Free element (O₂, Fe, Cl₂)0
    Monatomic ion (Na⁺, Ca²⁺)= ionic charge
    Hydrogen (most compounds)+1
    Hydrogen (metal hydrides: NaH)−1
    Oxygen (most compounds)−2
    Oxygen in peroxides (H₂O₂)−1
    Oxygen bonded to F (OF₂)+2
    Fluorine (always)−1
    Alkali metals (Group 1)+1
    Alkaline earth metals (Group 2)+2

    Common Elements and Their Possible Oxidation States

    ElementTypical statesHighestLowest
    Hydrogen (H)+1, −1+1−1
    Carbon (C)−4 to +4+4−4
    Nitrogen (N)−3 to +5+5 (HNO₃)−3 (NH₃)
    Sulfur (S)−2, +4, +6+6 (H₂SO₄)−2 (H₂S)
    Manganese (Mn)+2, +3, +4, +7+7 (MnO₄⁻)+2
    Iron (Fe)+2, +3+3+2
    Chromium (Cr)+2, +3, +6+6 (Cr₂O₇²⁻)+2
    Chlorine (Cl)−1, +1, +3, +5, +7+7 (HClO₄)−1 (HCl)
    Phosphorus (P)−3, +3, +5+5 (H₃PO₄)−3 (PH₃)
    Copper (Cu)+1, +2+2+1

    Step-by-Step Method

    1. Write out all known oxidation states for each element.
    2. Set up the equation: n₁·ox₁ + n₂·ox₂ + ... + nₓ·x = total charge.
    3. Solve for x algebraically.
    4. Verify: substitute back and confirm the sum matches the total charge.

    Real Examples

    CompoundEquationUnknown
    H₂SO₄2(+1) + x + 4(−2) = 0S = +6
    MnO₄⁻x + 4(−2) = −1Mn = +7
    Cr₂O₇²⁻2x + 7(−2) = −2Cr = +6
    H₂O₂2(+1) + 2x = 0O = −1
    FeO·Fe₂O₃ (Fe₃O₄)3x + 4(−2) = 0Fe = +8/3

    Worked Example: Chromium in Dichromate (Cr₂O₇²⁻)

    Given: 2 Cr atoms (unknown), 7 O atoms at −2, total ion charge = −2.
    Equation: 2(x) + 7(−2) = −2 → 2x − 14 = −2.
    Solve: 2x = 12 → x = +6.
    Verify: 2(+6) + 7(−2) = 12 − 14 = −2 ✓
    Interpretation: Chromium is +6 in dichromate (highest common state), making it a strong oxidizing agent.
    Chromium = +6 in Cr₂O₇²⁻. This is its maximum state and explains why dichromate is a powerful oxidizer.

    Frequently asked questions

    What is an oxidation number?
    An oxidation number is the hypothetical charge an atom would have if all bonds in a compound were completely ionic. It is positive when the atom gives up electrons and negative when it gains them. Unlike actual ionic charge, it can be a non-integer (e.g., Fe in Fe₃O₄ is +8/3).
    What are the six essential rules for assigning oxidation numbers?
    1. Free elements (O₂, H₂, Na, Fe) = 0. 2. Monatomic ions equal their charge (+2 for Ca²⁺). 3. Hydrogen = +1 in most compounds; −1 in metal hydrides (NaH, CaH₂). 4. Oxygen = −2 in most compounds; −1 in peroxides (H₂O₂, BaO₂); −½ in superoxides. 5. Fluorine = −1 always (most electronegative element). 6. The sum over all atoms equals the total charge (0 for neutral, ion charge for ions).
    How do you find the oxidation number of sulfur in H₂SO₄?
    Apply the rule: 2(+1) + 1(x) + 4(−2) = 0. Solve: 2 + x − 8 = 0 → x = +6. Sulfur is in its highest oxidation state (+6) in sulfuric acid. In SO₂ it is +4, in H₂S it is −2.
    What is the oxidation state of manganese in permanganate (MnO₄⁻)?
    The ion has charge −1. Set 1(x) + 4(−2) = −1 → x − 8 = −1 → x = +7. Manganese reaches its maximum oxidation state (+7) in permanganate. In MnO₂ it is +4, and as Mn²⁺ it is +2.
    What is the oxidation number of chromium in dichromate (Cr₂O₇²⁻)?
    Ion charge = −2, oxygen = −2 (×7 = −14). Equation: 2(x) + 7(−2) = −2 → 2x − 14 = −2 → 2x = 12 → x = +6. Chromium is +6 in all chromate/dichromate compounds. In Cr₂O₃ it is +3.
    What is the difference between oxidation number and ionic charge?
    Ionic charge is the actual charge on an isolated ion (measured). Oxidation number is a bookkeeping tool — a theoretical charge assuming ionic bonding even in covalent molecules. For NaCl, both agree: Na = +1. But in H₂O, oxygen's oxidation number is −2 while the actual partial charge is much smaller.
    Can an oxidation number be fractional?
    Yes. In Fe₃O₄ (magnetite), the average oxidation state of iron is +8/3 ≈ +2.67. This is because Fe₃O₄ contains both Fe²⁺ and Fe³⁺ ions. Fractional oxidation states appear in mixed-valence compounds and are chemically meaningful.
    How do oxidation numbers help identify redox reactions?
    In a redox reaction, at least one element's oxidation number increases (oxidation — the substance is an reducing agent) and at least one decreases (reduction — the substance is an oxidizing agent). If no oxidation numbers change, the reaction is not redox. For example, in 2H₂ + O₂ → 2H₂O: H goes 0→+1 (oxidized), O goes 0→−2 (reduced).
    What happens when oxygen is NOT −2?
    Three common exceptions: Peroxides (H₂O₂, Na₂O₂): O = −1. Superoxides (KO₂): O = −½. Bonded to fluorine (OF₂): O = +2 (fluorine is more electronegative and takes the negative oxidation state). Always check for these before applying the default −2 rule.

    Methodology & trust

    Editorial

    ciencia calculator with its formula verified automatically against IUPAC Recommendations on Oxidation States — Pure and Applied Chemistry, per our editorial policy and methodology.

    Updates

    Updated: June 2026. Parameters are verified periodically against the cited sources.

    Privacy

    Calculations run 100% in your browser. We do not store or transmit your data.

    Limitations

    Indicative results. For critical decisions, consult a professional.

    📌 How to cite this calculator
    APA format

    Rodríguez, M. (2026). Oxidation Number Calculator — Solve for Unknown Oxidation States. Hacé Cuentas. https://hacecuentas.com/en/oxidation-number-element

    BibTeX
    @misc{hacecuentas_oxidation_number_element_2026,
      author       = {Rodríguez, Martín},
      title        = {{Oxidation Number Calculator — Solve for Unknown Oxidation States}},
      year         = {2026},
      howpublished = {\url{https://hacecuentas.com/en/oxidation-number-element}},
      note         = {Hacé Cuentas}
    }

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