JEE Mains Chemistry · The d- and f-Block Elements

Potassium Dichromate and Chromium Compounds

Chromite ore is roasted to sodium chromate and turned into K₂Cr₂O₇, whose orange Cr₂O₇²⁻ and yellow CrO₄²⁻ interconvert with pH at a constant +6; in acid dichromate is a six-electron oxidant, and with a chloride it gives red CrO₂Cl₂, the start of the blue CrO₅ test.

All The d- and f-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-six PYQs, seventeen of them multiple choice, and seven from 2026, more than any other page. Eleven follow chromite ore to K₂Cr₂O₇ and the chromate–dichromate balance with pH; five use dichromate as an oxidant in acid; ten are the chromyl chloride test and the blue CrO₅ that ends it. Nine are numeric, and most of those are an oxidation state or a count of oxygen atoms.

Concept 1 of 3: From chromite ore to K₂Cr₂O₇, and chromate against dichromate

Chromium starts as Cr(III) in chromite,

\mathrm{FeCr_2O_4}

. Roasting with sodium carbonate in air oxidises it to yellow sodium chromate, Cr(VI). After that chromium never changes its oxidation state again: acid simply joins two chromate ions into one orange dichromate ion and lets out a water molecule, and base splits it back.

Definition

Roasting: $\mathrm{4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2}$ .
Acidify: $\mathrm{2Na_2CrO_4 + 2H^{+} \rightarrow Na_2Cr_2O_7 + 2Na^{+} + H_2O}$ .
Exchange: $\mathrm{Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl}$ . The potassium salt is less soluble and crystallises as orange crystals.
pH balance: $\mathrm{2CrO_4^{2-} + 2H^{+} \rightleftharpoons Cr_2O_7^{2-} + H_2O}$ . Acid favours orange dichromate; base favours yellow chromate. Cr is +6 in both.
Shapes: $\mathrm{CrO_4^{2-}}$ is tetrahedral. $\mathrm{Cr_2O_7^{2-}}$ is two tetrahedra sharing one corner: Cr–O–Cr angle 126°, a bent, symmetrical bridge, 6 terminal oxygens and 1 bridging.
$\mathrm{K_2Cr_2O_7}$ is a primary standard in volumetric analysis. $\mathrm{Na_2Cr_2O_7}$ is not: it is hygroscopic, so it cannot be weighed accurately.

The chromate–dichromate equilibrium

\mathrm{2CrO_4^{2-} + 2H^{+} \rightleftharpoons Cr_2O_7^{2-} + H_2O} \qquad \text{Cr stays } +6

Worked example

0.4 mol of chromite,

\mathrm{FeCr_2O_4}

, is roasted with sodium carbonate in air. How many moles of

\mathrm{O_2}

are used and of

\mathrm{Na_2CrO_4}

formed, and how many moles of

\mathrm{K_2Cr_2O_7}

can finally be made?

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q120Moderate

Example 1 · The d- and f-Block Elements · Potassium Dichromate and Chromium Compounds

Consider the following reactions

K_{2}Cr_{2}O_{7}\underset{-H_{2}O}{\overset{KOH}{\rightarrow}}\lbrack\text{ }A\rbrack\underset{-H_{2}O}{\overset{H_{2}SO_{4}}{\rightarrow}}\lbrack\text{ }B\rbrack +K_{2}SO_{4}

The products

\lbrack A\rbrack

and

\lbrack B\rbrack

, respectively are :

Chromate to dichromate is not a redox change

Colour changes from yellow to orange, but Cr is +6 on both sides. A question on 'the change in oxidation state' of Cr between chromate and dichromate has the answer 0.

Only the potassium salt is a primary standard

Sodium dichromate is hygroscopic, so a weighed sample is not pure. Potassium dichromate is less soluble, crystallises pure and is the primary standard.

Drill 10 more on from chromite ore to k₂cr₂o₇, and chromate against dichromate

Concept 2 of 3: Acidified dichromate as an oxidising agent

In acid, each chromium in dichromate drops from +6 to +3. Two chromiums per ion means six electrons per dichromate. The orange solution turns green as

\mathrm{Cr^{3+}}

forms, so anything that turns acidified dichromate green is a reducing agent.

Definition

Half-reaction: $\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O}$ , $E^\circ = +1.33$ V.
n-factor of $\mathrm{K_2Cr_2O_7}$ in acid = 6.
Iodide: $\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6I^{-} \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O}$ .
Iron(II): $\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$ .
Tin(II): $\mathrm{Cr_2O_7^{2-} + 14H^{+} + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O}$ .
Hydrogen sulphide: $\mathrm{Cr_2O_7^{2-} + 8H^{+} + 3H_2S \rightarrow 2Cr^{3+} + 3S + 7H_2O}$ .
Sulphur dioxide: $\mathrm{Cr_2O_7^{2-} + 2H^{+} + 3SO_2 \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O}$ . This is the dichromate-paper test for $\mathrm{SO_2}$ .
Species already in their higher state ( $\mathrm{Fe^{3+}}$ , $\mathrm{Sn^{4+}}$ , $\mathrm{SO_4^{2-}}$ ) cannot reduce it.

Electron balance with dichromate

6 \times n\left(\mathrm{Cr_2O_7^{2-}}\right) = (\text{electrons lost per reductant}) \times n(\text{reductant})

Worked example

How many moles of

\mathrm{SO_2}

are oxidised to sulphate by 0.02 mol of acidified

\mathrm{K_2Cr_2O_7}

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q36Moderate

Example 2 · The d- and f-Block Elements · Potassium Dichromate and Chromium Compounds

Which of the following sets includes all the species that will change the orange colour of

K_{2}Cr_{2}O_{7}

in acidic medium?

Electrons per what?

One dichromate takes 6 electrons. But forming one

\mathrm{I_2}

from iodide involves 2 electrons, and forming one S from sulphide also 2. Read whether a question counts per dichromate, per product molecule or per balanced equation before you add.

The green paper is not proof of SO₂ alone

Acidified dichromate paper turning green is the standard test for

\mathrm{SO_2}

, but

\mathrm{H_2S}

also reduces dichromate and turns it green. The key follows the named test; the chemistry allows both.

Drill 4 more on acidified dichromate as an oxidising agent

Concept 3 of 3: The chromyl chloride test and blue CrO₅

Heat a chloride with dichromate and concentrated sulphuric acid, and orange-red vapours of chromyl chloride come off. Chromium is still +6: nothing is reduced. The vapour dissolves in NaOH as yellow chromate, and chromate with hydrogen peroxide in acid gives a deep-blue peroxide,

\mathrm{CrO_5}

, that dissolves in amyl alcohol. Bromides and iodides give no chromyl compound, so the test picks out chloride.

Definition

Step 1: $\mathrm{4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O}$ .
$\mathrm{CrO_2Cl_2}$ : orange-red vapour, Cr +6, d⁰ (the same d count as Ti(IV), V(V) and Mn(VII)).
Step 2: $\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O}$ (yellow solution).
Step 3: $\mathrm{CrO_4^{2-} + 2H^{+} + 2H_2O_2 \rightarrow CrO_5 + 3H_2O}$ . The blue $\mathrm{CrO_5}$ is extracted into amyl alcohol or ether.
CrO₅ structure:one Cr=O and two peroxo (O–O) groups, a butterfly shape. Cr is +6, not +10.
With KBr instead of a chloride, brown bromine vapour comes off and the NaOH solution stays colourless: no chromate, no blue layer.

Oxidation state of Cr in CrO₅

x + 1(-2) + 4(-1) = 0 \;\Rightarrow\; x = +6

Worked example

Show that chromium in

\mathrm{CrO_5}

is +6 and not +10, and count the peroxide oxygens.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q45Moderate

Example 3 · The d- and f-Block Elements · Potassium Dichromate and Chromium Compounds

On heating a mixture of common salt and

K_{2}Cr_{2}O_{7}

in equal amount along with concentrated

H_{2}SO_{4}

in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are :

CrO₅ is +6, not +10

Four of its five oxygens are in peroxo groups at −1. Treating all five as −2 gives an impossible +10.

The formula is CrO₂Cl₂

One chromium, two oxygens, two chlorines. Options such as

\mathrm{Cr_2O_2Cl_2}

or a +5 or +3 state are the distractors; there is no redox for chromium in this step.

Drill 9 more on the chromyl chloride test and blue cro₅

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

From chromite ore to K₂Cr₂O₇, and chromate against dichromate
The chromate–dichromate equilibrium
$\mathrm{2CrO_4^{2-} + 2H^{+} \rightleftharpoons Cr_2O_7^{2-} + H_2O} \qquad \text{Cr stays } +6$
Acidified dichromate as an oxidising agent
Electron balance with dichromate
$6 \times n\left(\mathrm{Cr_2O_7^{2-}}\right) = (\text{electrons lost per reductant}) \times n(\text{reductant})$
The chromyl chloride test and blue CrO₅
Oxidation state of Cr in CrO₅
$x + 1(-2) + 4(-1) = 0 \;\Rightarrow\; x = +6$

Watch out for (6)

Chromate to dichromate is not a redox change
→ From chromite ore to K₂Cr₂O₇, and chromate against dichromate
Only the potassium salt is a primary standard
→ From chromite ore to K₂Cr₂O₇, and chromate against dichromate
Electrons per what?
→ Acidified dichromate as an oxidising agent
The green paper is not proof of SO₂ alone
→ Acidified dichromate as an oxidising agent
CrO₅ is +6, not +10
→ The chromyl chloride test and blue CrO₅
The formula is CrO₂Cl₂
→ The chromyl chloride test and blue CrO₅

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