🧪 How to Calculate Oxidation Number (Oxidation State) – Easy Guide With Rules & Examples

The oxidation number (or oxidation state) tells us how many electrons an atom has gained, lost, or shared in a chemical compound.

It’s one of the most important concepts in chemistry, redox reactions, balancing equations, and inorganic chemistry.

The good news?

👉 Once you learn a few simple rules, calculating oxidation numbers becomes extremely easy.

Let’s break it down step-by-step with examples.

⭐ What is Oxidation Number?

The oxidation number represents the charge an atom would have if electrons were completely transferred.

Examples:

• Na in NaCl has an oxidation number of +1
• Cl in NaCl has an oxidation number of 1
• O in most compounds has –2

It helps in:

✔ Balancing redox reactions

✔ Identifying oxidation & reduction

✔ Predicting reaction products

✔ Understanding electron transfer

⭐ General Rules to Calculate Oxidation Number

Memorize these, and you can solve any question.

Rule 1: The oxidation number of any free element is 0

Examples:

• H₂ → 0
• O₂ → 0
• Fe (solid iron) → 0
• Cl₂ → 0

Rule 2: The oxidation number of a monoatomic ion = its charge

Examples:

• Na⁺ → +1
• Mg²⁺ → +2
• Cl⁻ → –1

Rule 3: Oxygen is usually –2

Examples:

• H₂O → O = –2
• CO₂ → O = –2

Exceptions:

• Peroxides (H₂O₂): O = –1
• OF₂: O = +2

Rule 4: Hydrogen is usually +1

Exceptions:

• In metal hydrides (NaH, CaH₂), H = –1

Rule 5: Fluorine is always –1

(Highest electronegativity)

Rule 6: The sum of oxidation numbers in a neutral compound = 0

Example:

H₂O → (2 × +1) + (–2) = 0

Rule 7: The sum of oxidation numbers in a polyatomic ion = ion charge

Example (SO₄²⁻):

Sum = –2

⭐ Step-by-Step: How to Calculate Oxidation Number

Let’s apply the rules.

🧮 Example 1: Find the oxidation number of Mn in KMnO₄

We know:

• K = +1
• O = –2

Let Mn = x

Equation:

(+1) + x + 4(–2) = 0

1 + x – 8 = 0

x – 7 = 0

x = +7

👉 Mn in KMnO₄ = +7

🧮 Example 2: Oxidation number of Cr in Cr₂O₇²⁻

Let Cr = x

Sum of oxidation numbers = –2

O = –2

Equation:

2x + 7(–2) = –2

2x – 14 = –2

2x = 12

x = +6

👉 Cr = +6

🧮 Example 3: Oxidation number of S in H₂SO₄

H = +1

O = –2

Let S = x

Equation:

2(+1) + x + 4(–2) = 0

2 + x – 8 = 0

x – 6 = 0

x = +6

👉 Sulfur = +6

🧮 Example 4: Oxidation number of N in NO₃⁻

Let N = x

O = –2

Charge = –1

Equation:

x + 3(–2) = –1

x – 6 = –1

x = +5

👉 Nitrogen = +5

🧮 Example 5: Oxidation number of Cl in HClO

H = +1

O = –2

Let Cl = x

Equation:

(+1) + x + (–2) = 0

x – 1 = 0

x = +1

👉 Cl = +1

⭐ Shortcut Trick to Save Time

Memorize these common oxidation states:

Element Usual Oxidation Number

H +1 (–1 in hydrides)

O –2 (–1 in peroxides)

F –1

Alkali metals (Group 1) +1

Alkaline earth metals (Group 2) +2

Al, Zn +3, +2

Halogens –1 (except with oxygen)

⭐ How to Identify Oxidation & Reduction

• Oxidation = Increase in oxidation number
• Reduction = Decrease in oxidation number

Example:

Fe²⁺ → Fe³⁺

(+2 → +3) = oxidation

Cl₂ → Cl⁻

(0 → –1) = reduction

⭐ Common Mistakes Students Make

❌ Confusing oxidation number with valency

❌ Forgetting the peroxide exception

❌ Not applying the overall charge rule

❌ Forgetting hydrogen is -1 in metal hydrides

⭐ Conclusion

Understanding oxidation numbers is straightforward with a few fixed rules. Practice with 7 to 10 compounds daily, and you’ll master redox reactions easily.