Which of the following arrangements represent increasing oxidation number of the central atom? (i) CrO2^- , Cl

NCERT Exemplar · Q8

Q. Which of the following arrangements represent increasing oxidation number of the central atom?

(i) CrO2^- , ClO3^- , CrO4^2- , MnO4^-

(ii) ClO3^- , CrO4^2- , MnO4^- , CrO2^-

(iii) CrO2^- , ClO3^- , MnO4^- , CrO4^2-

(iv) CrO4^2- , MnO4^- , CrO2^- , ClO3^-

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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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To arrange the given species by increasing oxidation number of their central atoms, we calculate the oxidation state for each: Cr in $\text{CrO}_2^-$ is +3, Cl in $\text{ClO}_3^-$ is +5, Cr in $\text{CrO}_4^{2-}$ is +6, and Mn in $\text{MnO}_4^-$ is +7. The increasing order is $\text{CrO}_2^- < \text{ClO}_3^- < \text{CrO}_4^{2-} < \text{MnO}_4^-$ .

The oxidation number (or oxidation state) of an atom in a compound represents the hypothetical charge it would have if all bonds were purely ionic. It's a useful concept for tracking electron transfer in redox reactions and understanding the chemical behaviour of elements. To determine the oxidation number of a central atom in a polyatomic ion, we use a set of rules:

The oxidation number of oxygen is typically -2, except in peroxides (like $\text{H}_2\text{O}_2$ ) where it is -1, and in superoxides (like $\text{KO}_2$ ) where it is -1/2, or when bonded to fluorine (like $\text{OF}_2$ ) where it is +2. For this problem, oxygen will be -2.
The sum of the oxidation numbers of all atoms in a neutral compound is zero.
The sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

We will apply these rules to each given species to find the oxidation number of its central atom.

Calculate the oxidation number of Cr in $\text{CrO}_2^-$ :

Let the oxidation number of Cr be $x$ .

The oxidation number of oxygen is -2.

The overall charge of the ion is -1.

So, we set up the equation:

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

$x - 4 = -1$

$x = -1 + 4$

$x = +3$

The oxidation number of Cr in $\text{CrO}_2^-$ is +3.
Calculate the oxidation number of Cl in $\text{ClO}_3^-$ :

Let the oxidation number of Cl be $x$ .

The oxidation number of oxygen is -2.

The overall charge of the ion is -1.

So, we set up the equation:

$x + 3(-2) = -1$

$x - 6 = -1$

$x = -1 + 6$

$x = +5$

The oxidation number of Cl in $\text{ClO}_3^-$ is +5.
Calculate the oxidation number of Cr in $\text{CrO}_4^{2-}$ :

Let the oxidation number of Cr be $x$ .

The oxidation number of oxygen is -2.

The overall charge of the ion is -2.

So, we set up the equation:

$x + 4(-2) = -2$

$x - 8 = -2$

$x = -2 + 8$

$x = +6$

The oxidation number of Cr in $\text{CrO}_4^{2-}$ is +6.
Calculate the oxidation number of Mn in $\text{MnO}_4^-$ :

Let the oxidation number of Mn be $x$ .

The oxidation number of oxygen is -2.

The overall charge of the ion is -1.

So, we set up the equation: …

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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.Which of the following arrangements represent increasing oxidation number of the central atom?

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