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

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