A solution of $KMnO_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

A solution of KMnO₄ on reduction yields either a colourless solution or a brown precipitate or a green solutio

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KMnO_4

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Concept understanding — Inorganic Synthesis

Inorganic Synthesis – What It Really Means

Imagine you want to build a house. You need bricks, cement, steel, and a plan to put them together. Inorganic synthesis is exactly that — but for making chemical compounds that do not contain carbon-hydrogen bonds (the domain of organic chemistry). You take simple starting materials (elements or simple compounds) and, through a controlled chemical reaction, build a more complex inorganic product.

The intuition is simple: you are a chemist-craftsman. You decide what to make, choose the right ingredients, set the right conditions (temperature, pressure, solvent, time), and then isolate the pure product. The "synthesis" part is the entire journey from idea to pure substance.

The Precise Statement

Inorganic synthesis is the branch of chemistry concerned with the design, planning, and execution of chemical reactions to prepare inorganic compounds — including metals, alloys, coordination complexes, main-group compounds, solid-state materials, and nanomaterials — with controlled purity, structure, and properties.

It is not just "mixing chemicals." It involves:

Choosing the correct starting materials (precursors) — often simple salts, oxides, or elements.
Selecting a reaction method — solid-state heating, solution precipitation, electrochemical deposition, sol-gel, hydrothermal, etc.
Controlling reaction conditions — temperature, pressure, pH, concentration, atmosphere (inert gas, air, vacuum).
Purifying the product — recrystallization, distillation, sublimation, chromatography.
Characterising the product — proving you actually made what you intended (X-ray diffraction, spectroscopy, elemental analysis).

A Concrete Example: Making Copper(II) Sulfate Pentahydrate

You want to make the familiar blue crystal, $\ce{CuSO4·5H2O}$ .

Intuition: You have copper metal (a wire) and dilute sulfuric acid. Copper does not react with dilute acid directly — you need an oxidising agent. So you add nitric acid or simply heat copper with concentrated sulfuric acid.

Reaction:

\ce{Cu + 2H2SO4(conc.) -> CuSO4 + SO2 + 2H2O}

Then you evaporate the solution carefully. Blue crystals of $\ce{CuSO4·5H2O}$ appear.

What you did: You synthesised an inorganic compound from elemental copper and an acid. You controlled the concentration, temperature, and evaporation rate. You then filtered and dried the crystals.

Why It Matters

Inorganic synthesis is the foundation of:

Catalysts (e.g., $\ce{Pt}$ on alumina for car exhausts)
Electronic materials (silicon wafers, gallium arsenide for LEDs)
Medicinal compounds (cisplatin for cancer therapy)
Pigments (titanium dioxide white, Prussian blue)
Batteries (lithium cobalt oxide electrodes)

Without inorganic synthesis, modern technology would not exist.

A Common Misconception …

Why this formula?

Inorganic Synthesis: Why the Key Formulae Hold

Inorganic synthesis is the branch of chemistry concerned with the preparation of inorganic compounds — from simple salts to complex coordination compounds, organometallics, and solid-state materials. The key formulae in this field are not arbitrary; they arise from fundamental principles of stoichiometry, thermodynamics, kinetics, and coordination chemistry.

Let’s break down the reasoning behind the most important formulae.

1. The Yield Formula: Why It’s Not Just “Product/Reactant”

The most basic formula in any synthesis is:

\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

Why this holds:

Theoretical yield is calculated from the limiting reagent — the reactant that runs out first. This is based on the law of conservation of mass and the stoichiometric coefficients from the balanced chemical equation.
Actual yield is always less than theoretical because of:
- Side reactions (competing pathways)
- Incomplete reactions (equilibrium limitations)
- Loss during purification (filtration, crystallization, etc.)
The formula is a ratio because yield is a fractional measure of efficiency — it tells you how much of the maximum possible product you actually obtained.

Key insight: The formula works only if you correctly identify the limiting reagent. For example, in the synthesis of $\ce{FeCl3}$ from $\ce{Fe}$ and $\ce{Cl2}$ , if you have 1 mol Fe and 2 mol $\ce{Cl2}$ , Fe is limiting (1:1.5 stoichiometry), so theoretical yield is based on Fe.

2. The Atom Economy Formula: Why It Measures “Greenness”

\text{Atom Economy} = \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Reactants}} \times 100\%

Why this holds:

This formula was introduced by Barry Trost (1991) to quantify how much of the starting materials ends up in the product.
It is not a yield — it’s a theoretical maximum based on the balanced equation. It assumes 100% yield.
The denominator includes all reactants (including solvents if they are consumed, but usually only stoichiometric reagents).
A high atom economy (e.g., 100% for addition reactions like $\ce{A + B -> C}$ ) means less waste. A low atom economy (e.g., substitution reactions with leaving groups) means more byproducts.

Example: In the synthesis of $\ce{NaCl}$ from $\ce{Na}$ and $\ce{Cl2}$ :

$2\ce{Na} + \ce{Cl2} \rightarrow 2\ce{NaCl}$

Atom economy = $\frac{2 \times 58.44}{2 \times 22.99 + 70.90} \times 100\% = 100\%$ — because all atoms end up in the product.

3. The Solubility Product and Precipitation: Why $K_{sp}$ Controls Synthesis

For a sparingly soluble salt like $\ce{AgCl}$ :

\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}

K_{sp} = [\ce{Ag+}][\ce{Cl-}]

Why this holds:

$K_{sp}$ is an equilibrium constant derived from the law of mass action. It applies only to saturated solutions.
In synthesis, you use $K_{sp}$ to predict whether a precipitate will form when mixing solutions. If the ion product $Q = [\ce{Ag+}][\ce{Cl-}]$ exceeds $K_{sp}$ , precipitation occurs.
The formula is temperature-dependent (because $\Delta G^\circ = -RT \ln K_{sp}$ ). So you must control temperature to control precipitation.

Reasoning: The equilibrium constant arises from the balance between the lattice energy (holding the solid together) and the hydration energy (stabilizing ions in solution). A very small $K_{sp}$ means the solid is very stable — useful for gravimetric synthesis.

4. The Coordination Number and Ligand Field Stabilization Energy (LFSE)

For an octahedral complex, the LFSE is:

\text{LFSE} = \left( -0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g} \right) \Delta_o

Why this holds:

This formula comes from crystal field theory (CFT). In an octahedral field, the five $d$ orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals).
The splitting energy $\Delta_o$ is the energy difference between these sets.
Electrons fill the $t_{2g}$ orbitals first (Hund’s rule), and each electron in $t_{2g}$ stabilizes the complex by $-0.4\Delta_o$ relative to the barycenter (average energy). Each electron in $e_g$ destabilizes by $+0.6\Delta_o$ .
The formula explains why certain coordination numbers are preferred: for example, $\ce{[Co(H2O)6]^{2+}}$ (high-spin $d^7$ ) has LFSE = $-0.8\Delta_o$ , while $\ce{[CoCl4]^{2-}}$ (tetrahedral) has a smaller LFSE — so the octahedral form is more stable. …

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The reduction of $KMnO_4$ proceeds through distinct colour changes depending on pH: in acidic medium it gives colourless $Mn^{2+}$ , in neutral/weakly alkaline it gives brown $MnO_2$ precipitate, and in strongly alkaline it gives green $MnO_4^{2-}$ . These represent successive stages of manganese reduction from +7 to +2.

The key to understanding this lies in the variable oxidation states of manganese and how pH controls the stability of the intermediate species. Permanganate ion ( $MnO_4^-$ ) is a powerful oxidising agent in all media, but the products differ because the reduction potential and the stability of manganese species change dramatically with pH.

Let me walk through each case systematically.

Acidic medium (pH < 1–2) In strong acid, the reduction goes all the way to $Mn^{2+}$ , which is colourless in dilute solution. The half-reaction is:

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

The $E^\circ$ is +1.51 V, making it the most powerful oxidising condition.

How to carry it out: Add dilute $H_2SO_4$ to the $KMnO_4$ solution, then add a reducing agent like oxalic acid, $FeSO_4$ , or $H_2O_2$ . The purple colour fades to colourless as $Mn^{2+}$ forms.

Neutral or weakly alkaline medium (pH ~7–9) Here the reduction stops at $MnO_2$ , a brown insoluble precipitate. The half-reaction is:

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

Notice that water provides the oxygen, and hydroxide ions are produced — so the solution becomes alkaline as the reaction proceeds.

How to carry it out: Simply add a reducing agent (like $Na_2SO_3$ or $KI$ ) to a neutral $KMnO_4$ solution. No acid or strong base is added. The purple colour turns brown as $MnO_2$ precipitates.

Strongly alkaline medium (pH > 12) In concentrated alkali, the reduction yields the green manganate ion $MnO_4^{2-}$ (oxidation state +6). The half-reaction is:

MnO_4^- + e^- \rightarrow MnO_4^{2-}

This is a one-electron reduction. The green colour is characteristic of $MnO_4^{2-}$ .

How to carry it out: Add excess $KOH$ or $NaOH$ to $KMnO_4$ solution (making it strongly alkaline), then add a mild reducing agent like $KI$ or $Na_2SO_3$ in small amounts. Alternatively, you can heat solid $KMnO_4$ with $KOH$ — but that's a different method. …

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Method: pH-Dependent Reduction of Permanganate

This problem is solved using the Redox Speciation Method — tracking how the oxidation state of manganese changes with pH.

Concept First (Why this happens)

$KMnO_4$ contains manganese in its +7 oxidation state. The reduction product depends on the $H^+$ concentration because:

In acidic medium, $H^+$ ions are available to stabilize lower oxidation states as cations.
In neutral/weakly basic medium, $MnO_2$ (insoluble) forms.
In strongly basic medium, the manganate ion ( $MnO_4^{2-}$ ) is stable.

Steps of the Method

Step 1: Identify the three reduction stages

pH condition	Product	Colour	Mn oxidation state
Acidic	$Mn^{2+}$ (aq)	Colourless	+2
Neutral/weakly basic	$MnO_2$ (s)	Brown precipitate	+4
Strongly basic	$MnO_4^{2-}$ (aq)	Green solution	+6

Step 2: Write the half-reactions for each case

Acidic medium (colourless $Mn^{2+}$ ):

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

Neutral/weakly basic (brown $MnO_2$ ):

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

Strongly basic (green $MnO_4^{2-}$ ):

MnO_4^- + e^- \rightarrow MnO_4^{2-}

Step 3: How to carry out each reduction

| Desired product | Reducing agent | Conditions |

|----------------|----------------|------------| …

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Here is a breakdown of the common mistakes students make on this classic inorganic synthesis question, along with how to avoid each.

The Core Concept (The "Why")

The key is that $KMnO_4$ (manganese in +7 oxidation state) is a powerful oxidising agent. The pH of the solution dictates the final reduction product of manganese because the reduction half-reaction involves $H^+$ ions.

In Acidic Medium ( $pH < 7$ ): $MnO_4^-$ is reduced to the colourless $Mn^{2+}$ ion.
In Neutral/Faintly Alkaline Medium ( $pH \approx 7-9$ ): $MnO_4^-$ is reduced to a brown precipitate of $MnO_2$ .
In Strongly Alkaline Medium ( $pH > 10$ ): $MnO_4^-$ is reduced to a green solution of $MnO_4^{2-}$ (manganate ion).

Common Mistake #1: Confusing the Colour of the Products

The Mistake: Students often mix up which product is formed in which medium. For example, they might say "green solution in acidic medium" or "colourless solution in alkaline medium."

How to Avoid It:

Memorise the "pH-Colour" Triad: Create a simple mental map.
- Acid → Colourless ( $Mn^{2+}$ )
- Neutral → Brown ( $MnO_2$ )
- Alkaline → Green ( $MnO_4^{2-}$ )
Use a Mnemonic: "Acid gives Clear, Neutral gives Brown, Alkaline gives Green." (A-C, N-B, A-G).
Visualise the Ions: Remember that $Mn^{2+}$ is a very pale pink (appears colourless in dilute solution), $MnO_2$ is a solid brown precipitate, and $MnO_4^{2-}$ is a distinct green colour in solution.

Common Mistake #2: Writing the Wrong Half-Reactions

The Mistake: Students write the reduction half-reaction incorrectly, especially forgetting to balance $H^+$ and $H_2O$ or using the wrong number of electrons.

How to Avoid It:

Always Balance by the "ION-ELECTRON" Method: For each medium, write the balanced half-reaction. This is non-negotiable for exam accuracy.

1. Acidic Medium (to $Mn^{2+}$ ):

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

*Note: 5 electrons are gained. The solution becomes colourless.*

**2. Neutral/Faintly Alkaline Medium (to $MnO_2$):**

MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

*Note: 3 electrons are gained. The brown precipitate is $MnO_2$.*

**3. Strongly Alkaline Medium (to $MnO_4^{2-}$):**

MnO_4^- + e^- \rightarrow MnO_4^{2-}

*Note: Only 1 electron is gained. The green colour is due to the manganate ion.*

Check the Number of Electrons: The number of electrons gained decreases as the pH increases (5 → 3 → 1). This is a good sanity check.

Common Mistake #3: Forgetting the "How" (The Reagents)

The Mistake: Students can state the products but cannot describe how to carry out the reduction (e.g., what reagent to add).

How to Avoid It:

Learn the Specific Reducing Agents: The question asks "how are they carried out?" You must know the common reagents.
- For Colourless $Mn^{2+}$ (Acidic): Add a reducing agent like oxalic acid ( $H_2C_2O_4$ ) or ferrous sulphate ( $FeSO_4$ ) in the presence of dilute $H_2SO_4$ .
  - Example: $2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
- For Brown $MnO_2$ (Neutral): Add a reducing agent like sodium sulphite ( $Na_2SO_3$ ) or hydrogen peroxide ( $H_2O_2$ ) in neutral or faintly alkaline conditions.
  - Example: $2KMnO_4 + 3Na_2SO_3 + H_2O \rightarrow 2MnO_2 + 3Na_2SO_4 + 2KOH$
- For Green $MnO_4^{2-}$ (Strongly Alkaline): Add a reducing agent like potassium sulphite ( $K_2SO_3$ ) or potassium iodide ( $KI$ ) in a concentrated solution of KOH or NaOH. …

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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 23 on this concept.

CBSE 2025Set ANNUAL1 mark
$Q. How will you prepare K2MnO4 from pyrolusite? (Give chemical equation only)$
›Reveal solutionSolution

Fusion of pyrolusite (MnO2) with KOH in the presence of an oxidising agent (air/O2 or KNO3) gives potassium manganate.

Pyrolusite ( $MnO_2$ ) is fused with $KOH$ in presence of air (or an oxidising agent like $KNO_3$ ):

$2MnO_2 + 4KOH + O_2 \xrightarrow{\text{fuse}} 2K_2MnO_4 + 2H_2O$

…

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CBSE 2025Set ANNUAL1 mark
$Q. How will you prepare Potassium dichromate from Sodium dichromate? (Give chemical equation only)$
›Reveal solutionSolution

KCl is added to a solution of sodium dichromate; the less soluble potassium dichromate crystallises out.

Sodium dichromate solution is treated with potassium chloride:

$Na_2Cr_2O_7 + 2KCl \to K_2Cr_2O_7 + 2NaCl$

…

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CBSE 2024Set 56/2/11 markMCQ
$MnO_2$
›Reveal solutionSolution
Fusing $MnO_2$ with KOH in air oxidises Mn(IV) to Mn(VI), forming the green manganate ion $MnO_4^{2-}$ . The product is potassium manganate, $K_2MnO_4$ , option (B).

This is a classic example of an oxidation reaction in a fused alkaline medium. The key is to track the oxidation state of manganese and the role of the environment.

Why this approach works: In solid-state or fused-salt reactions, the strong alkaline medium (KOH) and the oxidising power of atmospheric oxygen work together. $MnO_2$ is already a common starting material for manganese chemistry. When you fuse it with KOH, you create a melt rich in $OH^-$ ions. Air ( $O_2$ ) acts as the oxidising agent, pulling electrons away from manganese. The Mn(IV) in $MnO_2$ cannot stay at +4 in such a strongly oxidising, basic melt — it gets pushed to a higher stable state. The +6 state (manganate) is particularly stable in alkaline conditions, while the +7 state (permanganate) requires even stronger oxidising conditions or a different workup.

Let’s walk through the reasoning step by step.

Identify the starting oxidation state. In $MnO_2$ , oxygen is $-2$ (usual for oxides). Let the Mn oxidation state be $x$ . Then $x + 2(-2) = 0$ , so $x = +4$ . Manganese is in the +4 oxidation state.

Recognise the reaction conditions. “Fused with KOH in air” means:

High temperature (fusion) — the mixture is molten.

Strongly basic medium — excess KOH provides $OH^-$ ions.

Presence of atmospheric oxygen ( $O_2$ ) — a good oxidising agent.

Predict the likely product. In alkaline conditions, manganese can exist in several oxidation states. The +6 state, as the manganate ion $MnO_4^{2-}$ , is well-known and stable in basic solution. The +7 state, as permanganate $MnO_4^-$ , is more stable in acidic or neutral conditions. Here, the basic melt favours the manganate. Also, $O_2$ is a moderately strong oxidiser — it can take Mn from +4 to +6, but not easily to +7 (that usually requires a stronger oxidant like $KNO_3$ or $KClO_3$ ).

Write the balanced chemical equation. The reaction is:

$2 MnO_2 + 4 KOH + O_2 \rightarrow 2 K_2MnO_4 + 2 H_2O$
Check: Mn goes from +4 to +6 (loss of 2 electrons per Mn). $O_2$ goes from 0 to $-2$ (gain of 4 electrons per $O_2$ ). Two Mn atoms lose 4 electrons total, exactly balancing the gain by one $O_2$ molecule. The KOH provides the potassium ions and the oxygen for the water. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The chemical formula of chromite ore is -$
(a) MnO2
(b) Na2Cr2O4
(c) FeCr2O4
(d) Na2CrO4
›Reveal solutionSolution

Chromite ore, the main source of chromium, has the formula FeCr2O4 (iron(II) chromite, a mixed oxide of iron and chromium).

Chromite crystallises in the spinel structure, in which Fe2+ ions occupy tetrahedral holes and Cr3+ ions occupy octahedral holes of a close-packed oxide lattice, giving the overall formula FeCr2O4 (equivalently FeO.Cr2O3). …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Process of commercial production of nitric acid is$
(a) Haber process
(b) Ostwald's process
(c) Contact process
(d) Deacon's process
›Reveal solutionSolution

Ostwald's process is named specifically for industrial nitric-acid manufacture, distinguishing it from Haber's (ammonia), Contact (sulphuric acid) and Deacon's (chlorine) processes.

In Ostwald's process, ammonia is catalytically oxidised over a Pt-Rh catalyst to nitric oxide, which is further oxidised to NO2 and then absorbed in water to give nitric acid:

4NH3 + 5O2 --(Pt/Rh, 500 K, 9 bar)--> 4NO + 6H2O …

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CBSE 2022Set M1 mark
$Q. Name the method used for concentration of sulphide ore.$
›Reveal solutionSolution

Sulphide ores are concentrated by the froth flotation process.

The froth flotation process is used to concentrate sulphide ores. The powdered ore is mixed with water and a collector/frother (e.g. pine oil); air is blown through. The sulphide ore particles are preferentially wetted by the oil and rise with t …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. Zone refining is used for obtaining ultra pure sample of$
(a) copper
(b) sodium
(c) germanium
(d) zinc
›Reveal solutionSolution

Zone refining purifies a metal based on the difference in solubility of impurities in the molten vs solid state of the metal.

In zone refining, a mobile induction heater melts a narrow zone of an impure metal rod at one end and moves slowly to the other end. Impurities are more soluble in the molten zone than in the solid, so they get swept along with the moving molten zone and concentrate at one end, which is then cut off. This te …

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CBSE 2020Set ANNUAL1 mark
$Q. Iron scraps are advisable and advantageous than zinc scraps for reducing the low grade copper ores. Why?$
›Reveal solutionSolution

Iron and zinc both lie above copper in the reactivity series and can reduce $Cu^{2+}$ , but iron scrap is far cheaper and more abundant, so it is the economical choice.

Concept. In hydrometallurgy of copper, a low-grade ore is leached and the copper in solution is displaced by a more reactive metal:

$Cu^{2+}(aq) + M \rightarrow Cu + M^{2+}(aq)$

where $M$ must lie above copper in the activity series.

Reason. Both $Fe$ and $Zn$ are more reactive than $Cu$ , so either can reduce $Cu^{2+}$ to $Cu$ :

$Cu^{2+} + Fe \rightarrow Cu + Fe^{2+}$ …

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CBSE 2020Set ANNUAL1 mark
$Q. Complete the reaction XeF₆ + H₂O ⟶ ? + 2HF .$
›Reveal solutionSolution

One molecule of water partially hydrolyses $XeF_6$ to $XeOF_4$ , liberating $2HF$ .

Concept. Xenon hexafluoride is readily hydrolysed. The extent of hydrolysis depends on the amount of water. With a limited amount (1 mole of water), only partial hydrolysis occurs.

Reaction (partial hydrolysis).

$XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$

Here one $O$ atom replaces two $F$ atoms, and the two displaced $F$ combine with the two $H$ of water to give $2HF$ .

…

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CBSE 2019Set ANNUAL1 mark
$Q. What is the role of depressant (NaCN) in Froth-Flotation method?$
›Reveal solutionSolution

NaCN selectively prevents ZnS from being wetted by the collector oil (by forming a complex on its surface), so ZnS sinks while PbS floats — separating a mixed Pb–Zn sulphide ore.

Concept: Froth flotation concentrates sulphide ores: pine-oil collectors make the mineral surface hydrophobic so it rises with the froth. When two sulphides are present, a depressant is used to keep one down.

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Which of the following noble gases is abundant in air?$
(i) He
(ii) Ne
(iii) Ar
(iv) Kr
›Reveal solutionSolution

Argon is the most abundant noble gas in air.

Dry air contains about 0.93% argon by volume, whereas neon, helium and krypton are present only in trace amounts (of the order of parts per million). Hen …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Which one is the ore of copper?$
(i) Haematite
(ii) Chalcopyrite
(iii) Dolomite
(iv) Bauxite
›Reveal solutionSolution

Chalcopyrite (CuFeS2) is the ore of copper.

An ore is a mineral from which a metal is extracted profitably. Chalcopyrite (copper pyrites), CuFeS2, is the principal ore of copper. Haematite (Fe2O3) is an iron ore, dolomite (CaCO3·MgCO3) is a c …

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Q.A solution of KMnO4KMnO_4KMnO4​ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

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Q.A solution of $KMnO_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

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