ChemistryCBSE Class 12 · The d- and f-Block Elements

Potassium Dichromate and Potassium Permanganate: Class 12

Updated 15 October 2026 · Fact-checked

Potassium dichromate ($K_2Cr_2O_7$, orange, Cr +6) is made from chromite ore via sodium chromate; potassium permanganate ($KMnO_4$, purple, Mn +7) is made from pyrolusite via green manganate. Both are strong oxidising agents in acid: $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$ (6 electrons) and $MnO_4^-$ to $Mn^{2+}$ (5 electrons).

In acid, dichromate gains 6 electrons (orange to green) and permanganate gains 5 electrons (purple to pale pink), so both oxidise other substances.

Illustration drawn for QuestionBuddy.

Understand Potassium Dichromate and Potassium Permanganate

These two oxoanion salts are the most important compounds of transition metals in the syllabus. Questions ask for preparation equations, the effect of pH, structures, and balanced ionic equations for their oxidising reactions.

Potassium dichromate: preparation. Chromite ore, $F e C r_{2} O_{4}$ , is fused with sodium carbonate in plenty of air, giving sodium chromate: $4 F e C r_{2} O_{4} + 8 N a_{2} C O_{3} + 7 O_{2} \to 8 N a_{2} C r O_{4} + 2 F e_{2} O_{3} + 8 C O_{2}$ . The yellow chromate solution is filtered and acidified with sulphuric acid to give orange sodium dichromate: $2 N a_{2} C r O_{4} + 2 H^{+} \to N a_{2} C r_{2} O_{7} + 2 N a^{+} + H_{2} O$ . Sodium dichromate is more soluble than the potassium salt, so treating it with KCl crystallises orange $K_{2} C r_{2} O_{7}$ : $N a_{2} C r_{2} O_{7} + 2 K C l \to K_{2} C r_{2} O_{7} + 2 N a C l$ .

Chromate–dichromate equilibrium. Chromate and dichromate interconvert with pH: in acid, $2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ (yellow to orange); in alkali, $C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ (orange to yellow). Chromium stays +6 throughout; only the structure changes. Chromate is tetrahedral; dichromate is two tetrahedra sharing one corner oxygen, with a Cr–O–Cr angle of about 126°.

Oxidising action of dichromate. In acid, $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ ( $E^{\circ} = 1.33$ V). The orange solution turns green ( $C r^{3 +}$ ). It oxidises iodide to iodine, iron(II) to iron(III), $H_{2} S$ to sulphur and tin(II) to tin(IV). It is used as a primary standard in volumetric analysis, in the leather industry and as an oxidant in organic chemistry.

Potassium permanganate: preparation. Pyrolusite, $M n O_{2}$ , is fused with KOH and an oxidising agent such as $K N O_{3}$ (or air): $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ , giving dark green potassium manganate. In neutral or acidic solution manganate disproportionates: $3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$ . Commercially, the manganate is oxidised electrolytically in alkaline solution: $M n O_{4}^{2 -} \to M n O_{4}^{-} + e^{-}$ . In the lab, $M n^{2 +}$ is oxidised by peroxodisulphate: $2 M n^{2 +} + 5 S_{2} O_{8}^{2 -} + 8 H_{2} O \to 2 M n O_{4}^{-} + 10 S O_{4}^{2 -} + 16 H^{+}$ .

Properties and structure. $K M n O_{4}$ forms dark purple crystals; on heating to 513 K it decomposes: $2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$ . Both manganate (green, Mn +6, $d^{1}$ , paramagnetic) and permanganate (purple, Mn +7, $d^{0}$ , diamagnetic) are tetrahedral.

Oxidising action of permanganate depends on the medium. In acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ ( $E^{\circ} = 1.52$ V); the purple colour disappears, so $K M n O_{4}$ acts as its own indicator in titrations. It oxidises $I^{-}$ to $I_{2}$ , $F e^{2 +}$ to $F e^{3 +}$ , oxalate to $C O_{2}$ (at about 333 K), $H_{2} S$ (sulphide) to S, sulphite to sulphate and nitrite to nitrate. Hydrochloric acid is not used to acidify it, because $K M n O_{4}$ would oxidise $C l^{-}$ to chlorine. In neutral or faintly alkaline solution: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ ; brown $M n O_{2}$ forms. Here iodide is oxidised to iodate, thiosulphate to sulphate, and $M n^{2 +}$ to $M n O_{2}$ . Uses: analytical reagent, bleaching of wool, cotton and silk, and decolourising oils.

Potassium Dichromate and Potassium Permanganate: formulas and key facts

Dichromate in acid: $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$; $E^{\circ} = 1.33$ V; orange → green.
Permanganate in acid: $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$; $E^{\circ} = 1.52$ V; purple → colourless (self-indicator).
Permanganate in neutral or faintly alkaline medium: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$; Brown precipitate of $M n O_{2}$ .
Chromate–dichromate interconversion: $2 C r O_{4}^{2 -} + 2 H^{+} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O$; Acid: orange dichromate; alkali: yellow chromate. Cr stays +6.
Manganate to permanganate: $3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$; Disproportionation (Mn: +6 → +7 and +4).
Mole ratios in titrations (acid): $M n O_{4}^{-} : F e^{2 +} = 1 : 5; M n O_{4}^{-} : C_{2} O_{4}^{2 -} = 2 : 5; C r_{2} O_{7}^{2 -} : F e^{2 +} = 1 : 6$; From the balanced ionic equations.

How to solve Potassium Dichromate and Potassium Permanganate questions

Most questions ask you to 'complete and balance' an oxidation by $K M n O_{4}$ or $K_{2} C r_{2} O_{7}$ . Balance by the ion-electron method; it is faster and safer than trial.

1Write the reduction half: $M n O_{4}^{-} \to M n^{2 +}$ (acid, 5e⁻), $M n O_{4}^{-} \to M n O_{2}$ (neutral, 3e⁻) or $C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}$ (6e⁻), with $H^{+}$ and $H_{2} O$ already balanced (use the standard forms).
2Write the oxidation half for the reducing agent, e.g. $F e^{2 +} \to F e^{3 +} + e^{-}$ , $2 I^{-} \to I_{2} + 2 e^{-}$ , $C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-}$ .
3Multiply each half so the electrons are equal, then add and cancel.
4Check atoms and total charge on both sides.
5For titrations, read the mole ratio from the balanced equation, then use moles = molarity × volume (L).
6For preparation questions, write each step with its condition (fusion with alkali, air or $K N O_{3}$ ; acidification; electrolytic oxidation; KCl).
7State colour changes and oxidation states where asked: dichromate orange → $C r^{3 +}$ green; permanganate purple → colourless ( $M n^{2 +}$ ) or brown ( $M n O_{2}$ ).

Potassium Dichromate and Potassium Permanganate in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Permanganate in neutral solution gives brown MnO₂

Suppose a family washes vegetables in a dilute pink solution of potassium permanganate in tap water. After some time the water turns light brown and a faint brown deposit appears.

Explain the colour change, write the half-reaction responsible, and state the change in oxidation state of manganese.

Try it first, then show the solutionSolution

Tap water is roughly neutral, so permanganate acts by the neutral-medium route.
$M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ : the electrons come from organic matter and microbes it oxidises.
Mn changes from +7 (purple $M n O_{4}^{-}$ ) to +4 (brown $M n O_{2}$ ).
The brown colour and deposit are the insoluble $M n O_{2}$ .

Answer: Permanganate (Mn +7) is reduced to brown insoluble $M n O_{2}$ (Mn +4) in neutral water: $M n O_{4}^{-} + 2 H_{2} O + 3 e^{-} \to M n O_{2} + 4 O H^{-}$ .

Memory hook: Purple to brown in neutral water, purple to colourless in acid: the medium decides where manganese ends up.

Tricks to remember Potassium Dichromate and Potassium Permanganate

Short memory aids. Say each one aloud once, then use it on a question the same day.

8-5-4 and 14-6-7
Permanganate: 8 $H^{+}$ , 5 $e^{-}$ , 4 $H_{2} O$ . Dichromate: 14 $H^{+}$ , 6 $e^{-}$ , 7 $H_{2} O$ .
Use it when: Writing reduction half-reactions instantly.
Acid makes orange
Add acid, yellow chromate becomes orange dichromate; add alkali, it goes back to yellow. Oxidation state stays +6.
Use it when: pH-effect questions.
Green → purple + brown
Green manganate in acid splits into purple permanganate and brown $M n O_{2}$ in a 2 : 1 ratio.
Use it when: Disproportionation equation.

Potassium Dichromate and Potassium Permanganate: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Chromate in acid · 1 mark · MCQBoard level

When dilute sulphuric acid is added to a yellow solution of potassium chromate, the solution:
(a) turns green because Cr is reduced
(b) turns orange because dichromate forms
(c) turns colourless
(d) turns purple

Try it first, then show the solutionSolution

$2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$ : no change in oxidation state; orange dichromate forms.

Answer: (b).

Example 2: Two classic ionic equations · 2 marks · Very short answerBoard level

Complete and balance: (a) $C r_{2} O_{7}^{2 -} + I^{-} + H^{+} \to$ (b) $M n O_{4}^{-} + F e^{2 +} + H^{+} \to$

Try it first, then show the solutionSolution

(a) $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ and $2 I^{-} \to I_{2} + 2 e^{-}$ (×3): $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$ (1 mark).
(b) $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ and $F e^{2 +} \to F e^{3 +} + e^{-}$ (×5): $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ (1 mark).

Answer: (a) $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$ . (b) $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ .

Example 3: Preparing KMnO₄ from pyrolusite · 3 marks · Short answerHarder board level

Describe, with balanced equations, how potassium permanganate is prepared from pyrolusite ore. Give the colour and oxidation state of manganese at each stage.

Try it first, then show the solutionSolution

Fuse $M n O_{2}$ (black, +4) with KOH in the presence of air or $K N O_{3}$ : $2 M n O_{2} + 4 K O H + O_{2} \to 2 K_{2} M n O_{4} + 2 H_{2} O$ ; dark green manganate, Mn +6 (1 mark).
In neutral or acidic solution manganate disproportionates: $3 M n O_{4}^{2 -} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O$ ; purple permanganate (+7) and brown $M n O_{2}$ (+4) (1 mark).
Commercially, alkaline manganate is oxidised electrolytically: $M n O_{4}^{2 -} \to M n O_{4}^{-} + e^{-}$ (1 mark).

Answer: $M n O_{2}$ (+4, black) → $K_{2} M n O_{4}$ (+6, green) → $K M n O_{4}$ (+7, purple), by alkaline fusion then disproportionation in acid or electrolytic oxidation.

Example 4: A permanganate titration · 3 marks · Short answerHarder board level

What volume of 0.020 M $K M n O_{4}$ is needed to oxidise 20.0 mL of 0.10 M $F e S O_{4}$ solution in dilute sulphuric acid? Why is no separate indicator needed, and why is HCl not used to acidify?

Try it first, then show the solutionSolution

$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , so 1 mol $M n O_{4}^{-}$ reacts with 5 mol $F e^{2 +}$ (½ mark).
Moles $F e^{2 +} = 0.10 \times 0.0200 = 2.0 \times 1 0^{- 3}$ ; moles $M n O_{4}^{-} = 4.0 \times 1 0^{- 4}$ (1 mark).
Volume $= \frac{4.0 \times 1 0 ^{- 4}}{0.020} = 0.020$ L $= 20.0$ mL (½ mark).
The purple $M n O_{4}^{-}$ becomes nearly colourless $M n^{2 +}$ ; the first permanent pink shows the end point (self-indicator). HCl is avoided because $K M n O_{4}$ would oxidise $C l^{-}$ to $C l_{2}$ , using up extra permanganate (1 mark).

Answer: 20.0 mL; $K M n O_{4}$ is a self-indicator; HCl would be oxidised to chlorine.

Example 5: Same reagent, different medium · 5 marks · Long answerHOTS / competency

(a) Write balanced ionic equations for the reaction of $K M n O_{4}$ with iodide ions in (i) acidic and (ii) neutral or faintly alkaline solution. (b) How many electrons does each $M n O_{4}^{-}$ gain in each case? (c) Compare the structure, colour and magnetic behaviour of $M n O_{4}^{2 -}$ and $M n O_{4}^{-}$ . (d) Why is $K_{2} C r_{2} O_{7}$ , rather than $K M n O_{4}$ , used as a primary standard?

Try it first, then show the solutionSolution

(a)(i) $2 M n O_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 I_{2} + 8 H_{2} O$ (1 mark). (ii) $2 M n O_{4}^{-} + H_{2} O + I^{-} \to 2 M n O_{2} + 2 O H^{-} + I O_{3}^{-}$ (1 mark).
(b) Acidic: Mn +7 → +2, 5 electrons. Neutral: +7 → +4, 3 electrons (1 mark).
(c) Both are tetrahedral. Manganate: green, Mn +6 ( $d^{1}$ ), one unpaired electron, paramagnetic. Permanganate: purple, Mn +7 ( $d^{0}$ ), diamagnetic (1 mark).
(d) $K_{2} C r_{2} O_{7}$ can be obtained very pure, is stable and not hygroscopic, so a solution of exactly known concentration can be made by weighing; $K M n O_{4}$ solutions slowly decompose and must be standardised (1 mark).

Answer: (a) Acid: $I^{-} \to I_{2}$ ; neutral: $I^{-} \to I O_{3}^{-}$ . (b) 5 and 3. (c) Both tetrahedral; $M n O_{4}^{2 -}$ green, paramagnetic; $M n O_{4}^{-}$ purple, diamagnetic. (d) Dichromate is pure and stable.

Where students lose marks in Potassium Dichromate and Potassium Permanganate

The slips examiners see most often, and what to write instead.

Mistake: Wrong number of H⁺ or water in the half-reaction
Wrong working
$M n O_{4}^{-} + 4 H^{+} + 5 e^{-} \to M n^{2 +} + 2 H_{2} O$
Right working
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Do this: Balance O with $H_{2} O$ , then H with $H^{+}$ , then charge with electrons.
Usually costs: 1 mark per equation.
Mistake: Treating the chromate–dichromate change as redox
Do this: Chromium is +6 in both; it is an acid–base (condensation) equilibrium.
Usually costs: ½ to 1 mark.
Mistake: Giving $M n^{2 +}$ as the product in neutral medium
Do this: Neutral or faintly alkaline: $M n O_{2}$ (brown, 3 electrons). Acid: $M n^{2 +}$ (5 electrons).
Usually costs: 1 mark.
Mistake: Unbalanced charge in 'complete and balance' answers
Atoms are checked but charge is not.
Do this: Add up ionic charges on each side as the last step.
Usually costs: ½ to 1 mark.

Exam tips

Memorise the two reduction half-reactions; every ionic equation is built from them.
Write states or colours if asked: 'orange to green', 'purple to colourless', 'brown precipitate'.
In preparation answers, name the ore (chromite $F e C r_{2} O_{4}$ , pyrolusite $M n O_{2}$ ) and each intermediate.
For structure questions, describe dichromate as two tetrahedra sharing a corner oxygen.
In titration numericals, show the mole ratio from the balanced equation before calculating.

Practice questions on Potassium Dichromate and Potassium Permanganate

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The oxidation state of chromium in potassium dichromate is:
(a) +3 (b) +4 (c) +6 (d) +7
Hint
$2 (+ 1) + 2 x + 7 (- 2) = 0$ .
Answer
(c) +6.
Q2 · 2 marks · Very short answerBoard level
What is the effect of increasing pH on a solution of potassium dichromate? Write the equation.
Hint
Higher pH means more $O H^{-}$ .
Answer
Orange dichromate changes to yellow chromate: $C r_{2} O_{7}^{2 -} + 2 O H^{-} \to 2 C r O_{4}^{2 -} + H_{2} O$ . Chromium stays +6.
Q3 · 2 marks · Very short answerHarder board level
Complete and balance the following ionic equations: (a) $M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to$ (b) $M n O_{4}^{-} + N O_{2}^{-} + H^{+} \to$
Hint
Oxalate gives $C O_{2}$ (2e⁻ per $C_{2} O_{4}^{2 -}$ ); nitrite gives nitrate (2e⁻).
Answer
(a) $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ . (b) $2 M n O_{4}^{-} + 5 N O_{2}^{-} + 6 H^{+} \to 2 M n^{2 +} + 5 N O_{3}^{-} + 3 H_{2} O$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): No external indicator is needed in titrations with acidified $K M n O_{4}$ .
Reason (R): $M n O_{4}^{-}$ is intensely purple while $M n^{2 +}$ is almost colourless.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Self-indicator.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
25.0 mL of 0.050 M sodium oxalate is titrated with 0.020 M $K M n O_{4}$ in warm dilute $H_{2} S O_{4}$ . Calculate the volume of $K M n O_{4}$ required.
Hint
$M n O_{4}^{-} : C_{2} O_{4}^{2 -} = 2 : 5$ .
Answer
Moles oxalate $= 0.050 \times 0.0250 = 1.25 \times 1 0^{- 3}$ ; moles $M n O_{4}^{-} = \frac{2}{5} \times 1.25 \times 1 0^{- 3} = 5.0 \times 1 0^{- 4}$ ; volume $= \frac{5.0 \times 1 0 ^{- 4}}{0.020} = 0.025$ L = 25.0 mL.
Q6 · 2 marks · Very short answerHarder board level
Calculate the mass of $K_{2} C r_{2} O_{7}$ (294 g mol⁻¹) needed to oxidise 0.060 mol of $F e^{2 +}$ in acidic solution. Write the balanced ionic equation.
Hint
1 : 6 ratio.
Answer
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 F e^{2 +} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O$ . Moles dichromate $= 0.060/6 = 0.010$ ; mass $= 0.010 \times 294 = 2.94$ g.
Q7 · 3 marks · Short answerHOTS / competency
Write the equation for the action of heat on $K M n O_{4}$ . Identify the oxidation states of Mn in all manganese species and say which product is paramagnetic.
Hint
Three products, one of them a gas.
Answer
$2 K M n O_{4} 513 K K_{2} M n O_{4} + M n O_{2} + O_{2}$ . Mn: +7 in $K M n O_{4}$ , +6 in $K_{2} M n O_{4}$ , +4 in $M n O_{2}$ . $K_{2} M n O_{4}$ ( $d^{1}$ ) and $M n O_{2}$ ( $d^{3}$ ) are paramagnetic; $K M n O_{4}$ ( $d^{0}$ ) is diamagnetic.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student adds a few drops of acidified $K M n O_{4}$ to three test tubes containing (A) sodium sulphite, (B) sodium nitrite and (C) hydrogen sulphide water. In each tube the purple colour disappears; tube C also turns milky.
(a) Write balanced ionic equations for the reactions in A and B. (2 marks)
(b) What causes the milkiness in C? Write the equation using sulphide ions. (1 mark)
(c) Why is dilute $H_{2} S O_{4}$ , not HCl, used to acidify the permanganate? (1 mark)
Hint
Sulphite → sulphate, nitrite → nitrate, sulphide → sulphur.
Answer
(a) A: $5 S O_{3}^{2 -} + 2 M n O_{4}^{-} + 6 H^{+} \to 2 M n^{2 +} + 5 S O_{4}^{2 -} + 3 H_{2} O$ . B: $5 N O_{2}^{-} + 2 M n O_{4}^{-} + 6 H^{+} \to 2 M n^{2 +} + 5 N O_{3}^{-} + 3 H_{2} O$ . (b) Colloidal sulphur: $5 S^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 2 M n^{2 +} + 8 H_{2} O + 5 S$ . (c) Permanganate would oxidise chloride ions to chlorine, consuming extra $K M n O_{4}$ and giving wrong results.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Ore used to make $K_{2} C r_{2} O_{7}$ ?Show answer
Chromite, $F e C r_{2} O_{4}$ .
2.Colour of manganate ion and oxidation state of Mn?Show answer
Green; +6.
3.Electrons gained by $C r_{2} O_{7}^{2 -}$ in acid?Show answer
6.
4.Product of $M n O_{4}^{-}$ with $I^{-}$ in neutral medium?Show answer
$I O_{3}^{-}$ (and $M n O_{2}$ ).
5.Why is KCl added to sodium dichromate?Show answer
To crystallise the less soluble $K_{2} C r_{2} O_{7}$ .
6.Shape of $M n O_{4}^{-}$ ?Show answer
Tetrahedral.

Potassium Dichromate and Potassium Permanganate: frequently asked questions

How is potassium dichromate prepared in Class 12?

Chromite ore is fused with sodium carbonate in air to give sodium chromate, which is acidified to sodium dichromate and then treated with KCl to crystallise potassium dichromate.

Why is KMnO₄ called a self-indicator?

Its purple colour disappears as it is reduced to nearly colourless Mn²⁺, so the first lasting pink colour marks the end point.

What happens when acid is added to a chromate solution?

Yellow chromate changes to orange dichromate: 2CrO_4^{2-} + 2H^+ \to Cr_2O_7^{2-} + H_2O.

Why is permanganate a stronger oxidising agent in acid?

In acid each MnO_4^- gains 5 electrons and E^\circ is 1.52 V; in neutral solution it gains only 3 to form MnO_2.

Is manganate paramagnetic?

Yes. MnO_4^{2-} has Mn in +6 (d^1) with one unpaired electron; permanganate (d^0) is diamagnetic.