An oxidation state is a hypothetical charge of an atom in a compound, calculated assuming all bonds between atoms are ionic. It indicates how many electrons an atom has donated or attracted. On this page, you will find a clear definition, all the rules for determining oxidation states, a table of constant values, a step-by-step algorithm, and an interactive simulator to master the topic.
What is an oxidation state
Oxidation state (OS) is a hypothetical charge of an atom in a molecule that it would have if all shared electron pairs were completely transferred to the more electronegative atom. Simply put, it is an electron "counter": an atom that donates electrons gets a positive oxidation state, and one that attracts them gets a negative one.
Oxidation states are always written with the sign before the number and placed above the element symbol: for example, in water H₂O, hydrogen is +1 and oxygen is −2. This distinguishes it from an ion charge (S²⁻, Fe³⁺), where the sign is written after the number.
An oxidation state can be positive, negative, or zero, and for many elements, it is variable—changing from one compound to another. That is why you need to know how to determine it, rather than just memorizing it.
Oxidation state rules: table of constant values
To determine an oxidation state, you need to know a few firm rules. Some elements have a constant oxidation state in all compounds—their values are summarized in the reference table below. You should start your calculations with these.
In addition to constant values, there are three main rules:
- in a simple substance, the oxidation state of any atom is 0 (Fe, O₂, H₂, S₈);
- the sum of the oxidation states of all atoms in a molecule is 0—the molecule is electroneutral;
- in an ion, the sum of the oxidation states equals the ion charge (for example, in SO₄²⁻, the sum is −2).
| Element | Oxidation State | Exceptions |
|---|---|---|
| Simple substance O₂, H₂, Fe, Cu, S | 0 | no exceptions |
| Fluorine F | −1 | always −1, no other values |
| Group I metals Na, K, Li, Ag | +1 | constant |
| Group II metals Ca, Mg, Zn, Ba | +2 | constant |
| Aluminum Al | +3 | constant |
| Hydrogen H | +1 | −1 in metal hydrides: NaH, CaH₂ |
| Oxygen O | −2 | −1 in peroxides (H₂O₂); +2 in OF₂ |
How to determine oxidation state in a compound
When a formula contains an element with a variable oxidation state (sulfur, nitrogen, chlorine, manganese, phosphorus), it is found using an equation. The algorithm is universal: first, assign everything you know for sure, then denote the unknown as x and solve the sum equation. Five steps are below.
0, no need to calculate further+1, oxygen −2, fluorine −1 — according to the table above0 for a molecule or = charge for an ion+ or − signMaximum and minimum oxidation state
An element has a maximum and minimum oxidation state—the extreme values between which all others lie.
- The maximum oxidation state is usually equal to the group number in the Periodic Table (for main group elements). For example, sulfur in Group VI → max OS +6 (H₂SO₄), nitrogen in Group V → +5 (HNO₃), chlorine in Group VII → +7 (HClO₄). It appears in compounds with more electronegative elements—most often with oxygen.
- The minimum oxidation state for nonmetals is equal to the group number minus 8. For sulfur, this is 6 − 8 = −2 (H₂S), for nitrogen 5 − 8 = −3 (NH₃). It appears in compounds with hydrogen and metals.
There are exceptions: for example, fluorine does not have a maximum positive OS—it is the most electronegative and is always −1; for oxygen, a maximum OS of +2 occurs only in OF₂.
Examples of determining oxidation states
Example 1. Sulfur in H₂SO₄
Task: determine the oxidation state of sulfur in sulfuric acid H₂SO₄.
Step 1. This is not a simple substance—the formula contains three elements.
Step 2. Assign constant OS: hydrogen +1 (2 atoms), oxygen −2 (4 atoms).
Step 3. The oxidation state of sulfur is unknown—denote it as x.
Step 4. Set up the equation (sum of all OS in the molecule = 0):
Step 5. Solve: \(2 + x - 8 = 0 \Rightarrow x = +6\).
Answer: the oxidation state of sulfur in H₂SO₄ is +6.
Example 2. Manganese in KMnO₄
Task: determine the oxidation state of manganese in potassium permanganate KMnO₄.
Step 1–2. Constant OS: potassium (Group I metal) +1, oxygen −2 (4 atoms).
Step 3. Manganese is an element with a variable OS, denote as x.
Step 4. Sum equation:
Step 5. Solve: \(1 + x - 8 = 0 \Rightarrow x = +7\).
Answer: the oxidation state of manganese is +7—this is its maximum oxidation state (Mn is in Group VII).
Example 3. Determining from the oxide formula Fe₂O₃
Task: determine the oxidation state of iron in the oxide Fe₂O₃.
Step 1–2. Oxygen is always −2, there are 3 oxygen atoms, so the total contribution of oxygen is: \(3 \cdot (-2) = -6\).
Step 3. Denote the oxidation state of iron as x, there are 2 iron atoms.
Step 4. Equation: \(2x + 3 \cdot (-2) = 0\).
Step 5. Solve: \(2x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = +3\).
Answer: the oxidation state of iron is +3. Check: two atoms at +3 give +6, three oxygen atoms give −6, sum is 0 — correct.
Common mistakes when determining oxidation states
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Confusing oxidation state with valency.
Valency is the number of chemical bonds an atom forms; it has no sign (always a positive number: I, II, III). Oxidation state is a hypothetical charge with a sign (+ or −) and can be zero. For example, in N₂, the valency of nitrogen is III, but the oxidation state is 0.
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Incorrectly placing the sign: writing the number then the plus or minus (2+ instead of +2).
For oxidation state, the sign is placed BEFORE the number and above the element symbol: +2, −3, +6. Writing the sign after the number (S²⁻) refers to an ion charge, not an oxidation state.
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Assuming oxygen is always −2 and hydrogen is always +1.
This works almost always, but there are exceptions. Oxygen in peroxides (H₂O₂) has an OS of −1, and in OF₂ it is even +2. Hydrogen in metal hydrides (NaH, CaH₂) has an OS of −1 because the metal is even more electropositive.
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Forgetting that the sum of oxidation states in a molecule equals zero.
A molecule is electroneutral, so the sum of (OS × number of atoms) for all elements is always = 0. For an ion, this sum equals the ion charge. This is the main equation used to find the unknown oxidation state.
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Not multiplying the oxidation state by the number of atoms of the element.
In the sum equation, each OS must be multiplied by the subscript (number of atoms). In H₂SO₄, hydrogen contributes not +1, but 2·(+1) = +2; oxygen contributes not −2, but 4·(−2) = −8. Skipping the subscript is the most common arithmetic error.
Questions and answers
How does oxidation state differ from valency?
Valency shows the number of chemical bonds an atom forms and has no sign (denoted by Roman numerals: I, II, III). Oxidation state is a hypothetical charge of an atom; it has a sign (+ or −) and can be zero or fractional. For many elements, these values coincide in magnitude, but not always: in the N₂ molecule, the valency of nitrogen is III, while the oxidation state is 0.
How to determine the maximum oxidation state of an element?
The maximum (highest positive) oxidation state of a main group element is equal to its group number in the Periodic Table. For example, sulfur in Group VI has a max OS of +6, nitrogen in Group V has +5, and chlorine in Group VII has +7. It appears in compounds with oxygen. Exceptions: oxygen and fluorine do not have a maximum positive OS equal to the group number.
How to determine the minimum oxidation state?
For nonmetals, the minimum (lowest) oxidation state is equal to the group number minus 8. For sulfur, this is 6 − 8 = −2 (in H₂S), for nitrogen 5 − 8 = −3 (in NH₃), for chlorine 7 − 8 = −1 (in HCl). The minimum OS appears in compounds with hydrogen and metals. For metals, the minimum oxidation state is 0 (in a simple substance).
What is the oxidation state of a simple substance?
For any atom in a simple substance, the oxidation state is 0. This applies to both metals (Fe, Cu, Na) and nonmetals (O₂, H₂, S₈, Cl₂, P₄). The reason is simple: atoms of the same element attract shared electrons equally, so the hypothetical charge is zero.
Is oxygen always −2?
Almost always, but there are two exceptions. In peroxides (H₂O₂, Na₂O₂), the oxidation state of oxygen is −1 because oxygen atoms are bonded to each other. In the compound with fluorine OF₂, oxygen has a positive OS of +2 because fluorine is even more electronegative.
Can an oxidation state be fractional?
Yes, but it is an average (formal) value. For example, in Fe₃O₄, the average oxidation state of iron is +8/3 because the formula contains a mixture of atoms with OS +2 and +3. In the school curriculum, such cases are rarely discussed—usually, the oxidation state is an integer.
