Calculate oxidation numbers from formulas and charges
Apply oxidation-number rules to compounds and ions, then check each result against the total charge.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Check atom balance, total charge and oxidation-state changes separately. Written practice is self-checked
Before you begin
- Recognise common ions and write formulas for simple compounds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the oxidation number of a monatomic ion?
What you will learn
- Calculate oxidation numbers in compounds and ions
- Use total charge to solve for an unknown oxidation number
- Distinguish oxidation number from an atom's literal charge
Use known values and total charge
An element in its uncombined form has oxidation number 0. A monatomic ion's oxidation number equals its charge. In compounds, oxygen is usually −2 and hydrogen is usually +1; use any exceptions stated in a question.
The sum of oxidation numbers is 0 for a neutral compound and equals the overall charge for a polyatomic ion. Multiply an element's oxidation number by the number of its atoms before adding the values.
Solve algebraically and verify
Assign the unknown oxidation number a variable, form the total-charge equation, and solve. Then substitute the result to check the sum. An oxidation number is a bookkeeping quantity for electron distribution; it is not automatically a measured ionic charge on each atom in a covalent molecule.
Use the formula exactly as written, including subscripts and brackets around polyatomic ions. A charge written outside brackets belongs to the whole ion.
Pause and explain
What must the sum of oxidation numbers equal in Cr₂O₇²⁻?
Worked example
Find the oxidation number of sulfur in H₂SO₄.
Show the worked solution
- Hydrogen contributes 2×(+1)=+2 and oxygen contributes 4×(−2)=−8.
- The neutral molecule has total oxidation number 0.
- Let sulfur be x: x + 2 − 8 = 0, so x = +6.
Answer Sulfur is +6 in H₂SO₄.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the oxidation number of manganese in KMnO₄.
Give me a hint
Use K = +1, O = −2 and a neutral total.
Compare my reasoning
- K contributes +1.
- Four O atoms contribute −8.
- Set +1 + x − 8 = 0.
Mn = +7.
Look for these in your work
- Known values assigned
- Subscripts included
- Neutral total solved
Calculate the oxidation number of chromium in Cr₂O₇²⁻.
Give me a hint
Use seven oxygen atoms and set the sum to the ion's charge.
Compare my reasoning
- O contributes 7×(−2)=−14.
- Let each Cr be x: 2x − 14 = −2.
- Solve for x.
Cr = +6.
Look for these in your work
- Oxygen contribution correct
- Ion charge used
- Chromium value found
A student assigns sulfur +4 in SO₄²⁻. Check the answer and state the correct oxidation number.
Give me a hint
Sum four oxygen contributions and match the ion charge.
Compare my reasoning
- O contributes 4×(−2)=−8.
- Set x − 8 = −2.
- Solve for x and substitute to check.
Sulfur is +6 because +6 − 8 = −2.
Look for these in your work
- Ion charge respected
- Oxidation number solved
- Sum verified
Common mistakes
- Forgetting to multiply by subscripts
- Setting an ion's total to zero instead of its charge
- Using a neutral atom's value for a monatomic ion
- Treating oxidation number as always a literal ionic charge
What can you explain now?
What is the oxidation number of a monatomic ion?
Compare with the explanation
Its ionic charge
For a single-atom ion, the value equals its charge.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which values are known first, and what total must their sum match?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.