Oxidation Number Calculator
Calculate the oxidation number (oxidation state) of an element in a chemical compound.
Compound Information
0 for neutral molecules, or charge for ions
Quick Examples
Oxidation Number of Mn
+7
in KMnO4
Redox State
Oxidized state (lost electrons)
Common Oxidation States of Mn
+2, +4, +7
Calculation
Total charge: 0
Sum of known oxidation numbers: 0 - (-7) = 7
Per Mn atom: (7) / 1 = 7
Rules Used
- • Oxygen: -2 (usually)
- • Hydrogen: +1 (with nonmetals)
- • Alkali metals: +1
- • Alkaline earth metals: +2
- • Sum of oxidation numbers = total charge
About Oxidation Numbers
The oxidation number (or oxidation state) represents the hypothetical charge an atom would have if all bonds were completely ionic. It is useful for tracking electron transfer in redox reactions, balancing redox equations, and naming compounds. The sum of oxidation numbers in a neutral compound is zero, while for ions it equals the total charge.
What Is an Oxidation Number?
An oxidation number (also called oxidation state) is a hypothetical charge that an atom would have if all its bonds to other atoms were completely ionic — that is, if the shared electrons in each bond were assigned entirely to the more electronegative atom. While most chemical bonds are not purely ionic, the oxidation number is an extremely useful bookkeeping tool that allows chemists to track electron transfer in reactions, balance redox equations, and name compounds systematically.
The oxidation number of an atom in a compound is determined by a set of rules established by IUPAC. The most important rules are: the sum of oxidation numbers in a neutral compound equals zero; the sum in a polyatomic ion equals the ion's charge; hydrogen is usually +1 (except in metal hydrides where it is -1); oxygen is usually -2 (except in peroxides where it is -1 and in compounds with fluorine); fluorine is always -1 in its compounds; and alkali metals are always +1 while alkaline earth metals are always +2.
Oxidation numbers are essential for identifying redox (reduction-oxidation) reactions, where electrons are transferred between species. An increase in oxidation number indicates oxidation (loss of electrons), while a decrease indicates reduction (gain of electrons). This conceptual framework underlies everything from battery chemistry and corrosion to biological electron transport chains and photosynthesis.
Rules for Assigning Oxidation Numbers
The following rules are applied in order of priority. When a conflict arises, the earlier rule takes precedence:
- Free elements have an oxidation number of 0 (e.g., Fe, O₂, N₂, S₈).
- Monatomic ions have an oxidation number equal to their charge (e.g., Na⁺ = +1, Cl⁻ = -1).
- Fluorine is always -1 in its compounds (it is the most electronegative element).
- Hydrogen is +1 with nonmetals and -1 with metals (e.g., HCl: H = +1; NaH: H = -1).
- Oxygen is usually -2, except -1 in peroxides (H₂O₂) and +2 in OF₂.
- The sum of oxidation numbers in a neutral compound is 0; in an ion, it equals the ion charge.
These rules form a systematic framework that can be applied to any compound or ion. The key to using them effectively is to start with the elements that have fixed oxidation numbers (F, O, H, alkali metals, alkaline earth metals) and then solve for the unknown element using the charge constraint.
Oxidation Number Calculation
Where:
- ON= Oxidation number of each element
- n= Number of atoms of each element
- charge= Total charge (0 for neutral compounds)
How to Use This Calculator
This calculator determines the oxidation number of a target element in a chemical compound:
- Enter the Chemical Formula: Type the formula using standard notation (e.g., KMnO4, H2SO4, FeCl3). The calculator parses the formula to identify all elements and their counts.
- Specify the Target Element: Enter the element symbol whose oxidation number you want to find (e.g., Mn, S, Fe).
- Set the Total Charge: Enter 0 for neutral molecules, or the appropriate charge for polyatomic ions (e.g., -2 for CrO₄²⁻, -1 for NO₃⁻).
- View Results: The calculator displays the oxidation number, the redox state description, common oxidation states for the element, and a step-by-step calculation breakdown.
The quick example buttons provide instant access to common problems like KMnO₄ (permanganate), H₂SO₄ (sulfuric acid), Cr₂O₇²⁻ (dichromate), and NO₃⁻ (nitrate).
Understanding the Results
The calculator provides several pieces of information about the oxidation state of the target element:
Oxidation Number: The primary result is the oxidation number of the target element. Positive values indicate the element is in an oxidized state (has lost electrons relative to its elemental form). Negative values indicate a reduced state (has gained electrons). Zero means the element is in its elemental form.
Redox State Description: The calculator classifies the result as "Oxidized state (lost electrons)" for positive values, "Reduced state (gained electrons)" for negative values, or "Elemental/neutral state" for zero. This provides immediate chemical context for the result.
Common Oxidation States: For many elements, the calculator lists their common oxidation states. For example, iron commonly exhibits +2 and +3 states, while manganese can be +2, +4, or +7. Comparing your calculated value to these common states helps verify the result is chemically reasonable.
Calculation Breakdown: The step-by-step calculation shows how the total charge is distributed among all atoms, making it easy to verify the arithmetic and understand the logic behind the result.
Real-World Applications
Oxidation numbers are fundamental to understanding and predicting chemical behavior across many fields:
Balancing Redox Reactions: The oxidation number method is one of the most systematic approaches for balancing complex redox equations. By identifying which elements are oxidized and which are reduced, and tracking the change in oxidation numbers, chemists can balance equations that would be nearly impossible to balance by inspection alone. This is essential in analytical chemistry, electrochemistry, and industrial chemistry.
Naming Chemical Compounds: The IUPAC naming system for transition metal compounds uses oxidation numbers to distinguish between different forms of the same element. For example, iron(II) chloride (FeCl₂) contains Fe²⁺, while iron(III) chloride (FeCl₃) contains Fe³⁺. Without knowing the oxidation number, the compound name would be ambiguous.
Corrosion Science: Understanding the oxidation states of metals in different environments helps predict and prevent corrosion. The passivation of stainless steel, for example, depends on chromium forming a stable Cr₂O₃ layer where chromium is in the +3 oxidation state. This protective oxide layer is what makes stainless steel resistant to rust.
Biological Redox Chemistry: Many biological processes involve electron transfer between different oxidation states of metal ions. The iron in hemoglobin cycles between Fe²⁺ and Fe³⁺ during oxygen transport, and the copper in cytochrome c oxidase alternates between Cu⁺ and Cu²⁺ in the electron transport chain. Understanding these oxidation state changes is crucial for biochemistry and medicine.
Worked Examples
Permanganate Ion (KMnO₄)
Problem:
Find the oxidation number of manganese in KMnO₄.
Solution Steps:
- 1KMnO₄ is a neutral compound, so the sum of oxidation numbers = 0
- 2K is an alkali metal: ON(K) = +1
- 3Oxygen is usually -2: ON(O) = -2, and there are 4 oxygen atoms
- 4Set up equation: (+1) + ON(Mn) + 4(-2) = 0
- 5Solve: ON(Mn) = 0 - 1 + 8 = +7
Result:
Mn = +7 in KMnO₄ (the highest oxidation state of manganese, making it a powerful oxidizing agent)
Sulfuric Acid (H₂SO₄)
Problem:
Find the oxidation number of sulfur in H₂SO₄.
Solution Steps:
- 1H₂SO₄ is neutral: sum = 0
- 2Hydrogen: ON(H) = +1, and there are 2 hydrogen atoms
- 3Oxygen: ON(O) = -2, and there are 4 oxygen atoms
- 4Set up equation: 2(+1) + ON(S) + 4(-2) = 0
- 5Solve: ON(S) = 0 - 2 + 8 = +6
Result:
S = +6 in H₂SO₄ (the highest oxidation state of sulfur, consistent with sulfuric acid being a strong oxidizing agent in concentrated form)
Dichromate Ion (Cr₂O₇²⁻)
Problem:
Find the oxidation number of chromium in Cr₂O₇²⁻.
Solution Steps:
- 1Cr₂O₇²⁻ has a total charge of -2: sum = -2
- 2Oxygen: ON(O) = -2, and there are 7 oxygen atoms
- 3Set up equation: 2(ON(Cr)) + 7(-2) = -2
- 4Solve: 2(ON(Cr)) = -2 + 14 = 12
- 5ON(Cr) = +6
Result:
Cr = +6 in Cr₂O₇²⁻ (a strong oxidizing agent used in organic chemistry and analytical methods)
Tips & Best Practices
- ✓Start with elements that have fixed oxidation numbers (F = -1, O = -2, H = +1, alkali metals = +1).
- ✓The sum of oxidation numbers in a neutral compound must equal zero.
- ✓For polyatomic ions, the sum of oxidation numbers must equal the ion's charge.
- ✓A positive oxidation number means the element is oxidized relative to its elemental form.
- ✓Transition metals often have multiple possible oxidation states — check common values to verify your answer.
- ✓Use oxidation numbers to balance redox equations by equating the total increase and decrease in oxidation numbers.
Frequently Asked Questions
Sources & References
Last updated: 2026-06-06
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Editorial Note
MyCalcBuddy Editorial Team
This page is maintained as an educational calculator reference.
Formula Source: Chemistry: The Central Science
by Brown, LeMay, Bursten