Calculate oxidation number of Mn in KMnO4.

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Q. Calculate oxidation number of Mn in KMnO4.

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2023Subjective· 1mImportance★★★★★

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Concept understanding — Oxidation Number Calculation

Oxidation Number Calculation: From Intuition to Precision

Imagine you're watching a tug-of-war between two atoms in a molecule. Each atom has a certain "pull" on the shared electrons — chemists call this electronegativity. The oxidation number is like a scorecard that tells us: If the more electronegative atom took all the shared electrons, what charge would each atom end up with?

This isn't a real charge — it's a bookkeeping tool. Real molecules don't have these exact charges. But this imaginary scorecard helps us track where electrons go during chemical reactions, especially in redox (reduction-oxidation) processes.

The Core Idea

Oxidation number (also called oxidation state) is the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic — meaning the more electronegative atom keeps all the shared electrons.

Note

For an atom bonded to another atom of the same element (like O₂ or N₂), the electrons are shared equally. So the oxidation number is zero — no one "wins" the tug-of-war.

The Rules (Your Toolkit)

These rules are applied in order — rule 1 overrides rule 2, and so on. Memorise them in this sequence:

Free elements (uncombined, like Fe, O₂, H₂, S₈) have oxidation number = 0.
Monatomic ions have oxidation number = their charge.

Example: Na⁺ = +1, Cl⁻ = −1, Mg²⁺ = +2.
Fluorine is always −1 in compounds (it's the most electronegative element — it always "wins").
Oxygen is usually −2, except:
- In peroxides (like H₂O₂) it's −1
- In OF₂ (with fluorine) it's +2 (fluorine wins)
Hydrogen is usually +1 when bonded to non-metals, −1 when bonded to metals (like NaH, CaH₂).
The sum of oxidation numbers in a neutral compound = 0.

In a polyatomic ion, the sum = the ion's charge.

Watch out

Never apply rule 6 before rules 1–5. The sum rule is your check, not your starting point.

How to Calculate: A Step-by-Step Example

Let's find the oxidation number of sulphur in H₂SO₄ (sulphuric acid).

Step 1: Write the known oxidation numbers.

Hydrogen: +1 (rule 5, bonded to non-metal oxygen)

Oxygen: −2 (rule 4, not a peroxide)

Step 2: Let the unknown be $x$ (for sulphur).

Step 3: Apply the sum rule (rule 6). The compound is neutral, so:

2(+1) + x + 4(-2) = 0

Step 4: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

Important

Sulphur in H₂SO₄ has oxidation number +6.

Another Example: A Polyatomic Ion

Find the oxidation number of chromium in Cr₂O₇²⁻ (dichromate ion).

Oxygen: −2 (rule 4)

Let chromium = $x$

Sum of oxidation numbers = charge of ion (−2):

2x + 7(-2) = -2

2x - 14 = -2

2x = 12

x = +6

Tip

When you get a fractional oxidation number (like +2.5 in Fe₃O₄), it means the compound has two different oxidation states for the same element. Fe₃O₄ actually contains Fe²⁺ and Fe³⁺ in a 1:2 ratio.

Common Traps to Avoid

| Mistake | Why it's wrong |

|---------|----------------| …

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In KMnO4, K is +1 and each O is -2; balancing the neutral compound's charges gives Mn = +7.

Let the oxidation number of Mn be x.

Known oxidation numbers: K = +1 (alkali metal), O = -2 (usual value, no peroxide here).

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 49 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The oxidation numbers of boron in NaBH4 and Cr in K2Cr2O7 are$
(a) +4, +3
(b) +3, +6
(c) -3, +6
(d) -4, +12
›Reveal solutionSolution

Use the standard oxidation-number rules: Na = +1, H = -1 in metal hydrides, K = +1, O = -2, and the sum of oxidation numbers in a neutral compound equals zero.

For NaBH4:

Na = +1 (alkali metal, always +1).

H = -1 (H bonded to a less electronegative metal takes -1, as in metal hydrides).

Let oxidation number of B = x.

Sum = 0 (neutral compound):

(+1) + x + 4(-1) = 0

1 + x - 4 = 0

x = +3

So B is +3 in NaBH4.

For K2Cr2O7:

K = +1 each, so 2 K contribute +2. …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Oxidation number of oxygen in H2O2 is -$
(a) -1
(b) -2
(c) -1/2
(d) +1
›Reveal solutionSolution

Oxygen has oxidation number -1 in H2O2 because of the O-O peroxide bond.

To find oxidation number of O in H2O2: the molecule has formula H2O2, structure H-O-O-H. Each H contributes +1 (standard oxidation number of hydrogen bonded to a non-metal). Let oxidation number of each O be x. Since the molecule is neutral overall: 2(+1) + 2(x) = 0, so 2x = -2, x = -1.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Oxidation number of Fe in Fe2O3 is ______.$
›Reveal solutionSolution

Iron has oxidation number +3 in Fe2O3.

Let the oxidation number of Fe be x. Oxygen is assigned its standard oxidation number, -2 (no peroxide or superoxide linkage here). Fe2O3 is electrically neutral, so: 2(x) + 3 …

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CBSE 2026Set ANNUAL1 mark
$Q. Oxidation number of chlorine (Cl) in ClO3⁻ is .................$
›Reveal solutionSolution

Solving $x + 3(-2) = -1$ gives the oxidation number of chlorine as $+5$ .

Let the oxidation number of Cl be $x$ . Oxygen is assigned $-2$ in this oxoanion. The sum of oxidation numbers equals the overall ionic charge: …

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CBSE 2026Set sz1 markMCQ
$Q. Select the correct one: The oxidation state of chromium in chromium trioxide is$
(a) +3
(b) +4
(c) +5
(d) +6
›Reveal solutionSolution

Using the standard oxidation-state rule for oxygen (-2) and balancing to zero net charge for the neutral molecule CrO3 gives Cr = +6.

Let the oxidation state of chromium in CrO3 be x.

Oxygen is assigned an oxidation state of -2 (standard rule, since CrO3 is not a peroxide or superoxide).

CrO3 is a neutral molecule, so the sum of oxidation states must equal zero:

x + 3(-2) = 0

x - 6 = 0

x = +6

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In which of the following oxidation number of oxygen is maximum ?$
(a) H2O2
(b) K2O
(c) KO2
(d) O2F2
›Reveal solutionSolution
Oxygen shows +1 in O2F2, its maximum among the given species.

Assign oxidation numbers of oxygen:

H2O2: O = −1 (peroxide).

K2O: O = −2 (normal oxide).

KO2: O = −1/2 (superoxide). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. In which of the following compounds oxidation number of Cl is + 5 ?$
(a) HClO4
(b) HClO2
(c) HClO3
(d) HClO
›Reveal solutionSolution
Cl is +5 in HClO3.

Using H = +1, O = −2 and net charge 0:

HClO: 1 + Cl − 2 = 0 → Cl = +1.

HClO2: 1 + Cl − 4 = 0 → Cl = +3.

HClO3: 1 + Cl − 6 = 0 → Cl = +5. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The oxidation number of Sulphur in Na2S4O6 is :$
(1) 2.5
(2) 1.5
(3) 2
(4) 3
›Reveal solutionSolution

The average oxidation number of sulphur in Na2S4O6 is +2.5 — option (1).

For the neutral compound Na2S4O6, the sum of all oxidation numbers is zero. Taking Na = +1 and O = -2, and letting the (average) oxidation number of S be x:

2(+1) + 4(x) + 6(-2) = 0

2 + 4x - 12 = 0

4x = 10

x = +2.5

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Oxidation state of Cl in CaOCl2 is$
(a) 0
(b) +1
(c) -1
(d) +1, -1
›Reveal solutionSolution

CaOCl2 (bleaching powder) contains chlorine in two oxidation states at once: +1 and -1.

Bleaching powder, CaOCl2, is best represented as the mixed salt Ca(OCl)Cl, containing both a hypochlorite ion (OCl-, in which Cl has oxidation state +1) and a chloride ion (Cl-, oxidation state -1). This dual oxidation state is a direct consequence of how bleaching powder is manufactured: by passing chlorine gas (oxidation state 0) over s …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the oxidation state of S in H2SO4?$
(a) +4
(b) +5
(c) +6
(d) +8
›Reveal solutionSolution
Using the standard oxidation-state rules (H = +1, O = -2) and the fact that the molecule is neutral overall, solving for S gives +6.

Oxidation state rules used here:

Hydrogen is almost always +1 in compounds (except metal hydrides)

Oxygen is almost always -2 in compounds (except peroxides/superoxides)

The sum of all oxidation states in a neutral molecule must equal zero

For H2SO4 (2 H atoms, 1 S atom, 4 O atoms), let x = oxidation state of S:

2(+1) + 1(x) + 4(-2) = 0

2 + x - 8 = 0

x - 6 = 0

x = +6

…
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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: Oxidation number of Sulphur (S) in H2S2O7 is ___________.$
›Reveal solutionSolution

Using the standard oxidation states H = +1, O = -2 and solving for S in the neutral molecule gives S = +6 (same as in H $_2$ SO $_4$ , since H $_2$ S $_2$ O $_7$ is oleum/pyrosulfuric acid built from two SO $_4$ units).

$H_2S_2O_7$ is a neutral molecule, so the sum of all oxidation numbers must equal 0.

Let the oxidation number of each S atom be $x$ (both S atoms are equivalent by symmetry in pyrosulfuric acid).

$2(+1) + 2(x) + 7(-2) = 0$

$2 + 2x - 14 = 0$

$2x = 12$

$x = +6$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When methane is burnt in O2 to produce CO2 and H2O the oxidation number of carbon changes by?$
(a) +4
(b) +8
(c) Zero
(d) +2
›Reveal solutionSolution

Carbon's oxidation state goes from -4 in methane to +4 in carbon dioxide during combustion, a change of +8.

In CH4, each H is assigned +1 (by convention). Let oxidation number of C = x.

x + 4(+1) = 0 (CH4 is neutral)

x = -4

In CO2, oxygen is assigned -2 (standard). Let oxidation number of C = y.

y + 2(-2) = 0

y = +4

…

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Q.Calculate oxidation number of Mn in KMnO4.

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