Match Column I with Column II for the oxidation states of the central atoms. Column I (i) Cr2O7²- (ii) MnO4^-

NCERT Exemplar · Q27

Q. Match Column I with Column II for the oxidation states of the central atoms. Column I

(i) Cr2O7^2-

(ii) MnO4^-

(iii) VO3^-

(iv) FeF6^3-
Column II

(a) +3

(b) +4

(d) +6

(e) +7

Rajasthan RbseShort· 2mImportance★★★★★est

86% · 67/78 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 |

|---------|----------------| …

Start your 14-day free trial to unlock the full solution →

Assign oxidation states by treating oxygen as $-2$ and fluorine as $-1$ , then balance the total charge. Cr₂O₇²⁻ → +6, MnO₄⁻ → +7, VO₃⁻ → +5, FeF₆³⁻ → +3.

The oxidation state of a central metal atom in a complex ion is found by accounting for the known oxidation states of the ligands and ensuring the algebraic sum equals the overall charge on the species. Oxygen almost always takes $-2$ (except in peroxides and superoxides, which are not present here), and fluorine invariably takes $-1$ as the most electronegative element.

1. Dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$

Let the oxidation state of each chromium atom be $x$ . The ion contains two chromium atoms and seven oxygen atoms.

2x + 7(-2) = -2

2x - 14 = -2

2x = +12 \implies x = +6

Each chromium is in the +6 oxidation state.

2. Permanganate ion, $\text{MnO}_4^{-}$

Let the oxidation state of manganese be $x$ . The ion has one manganese and four oxygen atoms.

x + 4(-2) = -1

x - 8 = -1

x = +7

Manganese is in the +7 oxidation state.

3. Vanadate ion, $\text{VO}_3^{-}$

Let the oxidation state of vanadium be $x$ . The ion contains one vanadium and three oxygen atoms.

x + 3(-2) = -1

x - 6 = -1

x = +5

Vanadium is in the +5 oxidation state.

4. Hexafluoroferrate(III) ion, $\text{FeF}_6^{3-}$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ26 — Balance the following ionic equations (i) Cr2O7²- + H^+ + I^- → Cr³+ +…→Next questionQ28 — Match the items in Column I with relevant items in Column II. Column I…

Q.Match Column I with Column II for the oxidation states of the central atoms. Column I

🎓Unlock everything free for 14 days

Q.Match Column I with Column II for the oxidation states of the central atoms.
Column I

Unlock everything free for 14 days