Q.Describe the oxidation number of the starred element in each of the following:
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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.
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:
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Free elements (uncombined, like Fe, O₂, H₂, S₈) have oxidation number = 0.
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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").
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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).
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The sum of oxidation numbers in a neutral compound = 0.
In a polyatomic ion, the sum = the ion's charge.
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
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
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 |
|---------|----------------| …
Each starred atom's oxidation number is found by using the known, fixed oxidation numbers of the other atoms in the formula (H = +1, O = -2 usually, alkali metals = +1) and the overall charge, then solving for the unknown. …
Using O = -2 (or -1 for peroxide oxygens), H = +1, and matching the sum of oxidation numbers to the overall charge of each species, the starred elements' oxidation numbers work out to: Mn = +6, S = +6, N = -3, Cr = +6, Na = +1.
- K2Mn*O4 (potassium manganate): let oxidation number of Mn = x. K = +1 each (2 K = +2), O = -2 each (4 O = -8). The compound is neutral: 2(+1) + x + 4(-2) = 0 2 + x - 8 = 0 x = +6
- H2S2*O8 (peroxodisulfuric/Marshall's acid): this molecule contains one peroxide (-O-O-) linkage, so 2 of its 8 oxygens are peroxide oxygens (oxidation number -1 each) and the remaining 6 are normal oxygens (-2 each). Let oxidation number of each S = x (2 S atoms): 2(+1) + 2x + 6(-2) + 2(-1) = 0 2 + 2x - 12 - 2 = 0 2x = 12 x = +6 (per S atom)
- N*H4- (this is the ammonium ion, NH4+, overall charge +1; treating the printed superscript as the ion's charge notation): let oxidation number of N = x. H = +1 each (4 H = +4): x + 4(+1) = +1 x + 4 = 1 x = -3 …
Showing the 12 most recent of 49 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.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. …
- CBSE 2026Set ANNUAL1 markMCQQ.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.
…
- CBSE 2026Set ANNUAL1 markQ.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 …
- CBSE 2026Set ANNUAL1 markQ.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: …
- CBSE 2026Set sz1 markMCQQ.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
…
- CBSE 2026Set ANNUAL1 markMCQQ.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). …
- CBSE 2026Set ANNUAL1 markMCQQ.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. …
- CBSE 2026Set ANNUAL1 markMCQQ.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
…
- CBSE 2025Set ANNUAL1 markMCQQ.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 …
- CBSE 2025Set ANNUAL1 markMCQQ.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
…
- CBSE 2025Set ANNUAL1 markQ.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 H2SO4, since H2S2O7 is oleum/pyrosulfuric acid built from two SO4 units).
H2S2O7 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
…
- CBSE 2025Set ANNUAL1 markMCQQ.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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