$KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with KI, iodide ion is oxidised to ____________. (i) $I_2$ (ii) $IO^-$ (iii) $IO_3^-$ (iv) $IO_4^-$

KMnO₄ acts as an oxidising agent in alkaline medium. When alkaline KMnO₄ is treated with KI, iodide ion is oxi

NCERT Exemplar · Q16

KMnO_4

(i)

I_2

(ii)

IO^-

(iii)

IO_3^-

(iv)

IO_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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In alkaline medium, $KMnO_4$ is reduced to $MnO_2$ (or $MnO_4^{2-}$ ) and oxidises $\ce{I-}$ all the way to iodate ( $IO_3^-$ ), not just iodine. The correct option is (iii).

The key to this question lies in understanding how the oxidising power of $KMnO_4$ changes with pH. In acidic medium, $MnO_4^-$ is reduced to $\ce{Mn^{2+}}$ (a 5-electron change) and is a very strong oxidant — it can oxidise $\ce{I-}$ to $I_2$ easily. But in alkaline medium, the reduction product is different, and so is the extent of oxidation it can achieve.

When the medium is alkaline, $MnO_4^-$ typically reduces to $MnO_2$ (manganese dioxide, oxidation state +4) or, in strongly alkaline conditions, to manganate ion $MnO_4^{2-}$ (oxidation state +6). The number of electrons gained per $MnO_4^-$ is smaller (3 electrons for $MnO_2$ , 1 electron for $MnO_4^{2-}$ ), so the oxidising power per mole is less intense. However, the reaction is still vigorous enough to push iodide beyond elemental iodine.

Iodide ion $\ce{I-}$ (oxidation state -1) can be oxidised stepwise: first to $I_2$ (0), then to hypoiodite $IO^-$ (+1), then to iodite $IO_2^-$ (+3), then to iodate $IO_3^-$ (+5), and finally to periodate $IO_4^-$ (+7). In alkaline medium, $KMnO_4$ is strong enough to take it to the +5 state — iodate — but not to periodate (which requires even stronger oxidants or specific conditions like hot alkaline $KMnO_4$ with a catalyst).

Let’s walk through the actual reaction.

Identify the half-reactions. In alkaline medium, the reduction half-reaction for permanganate is:

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

(This is the most common version; in very concentrated alkali, $MnO_4^{2-}$ forms instead, but the principle is the same.)

Oxidation half-reaction for iodide. Iodide is oxidised to iodate:

I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-

Balance the electrons. The reduction consumes 3 electrons per $MnO_4^-$ , the oxidation produces 6 electrons per $I^-$ . To balance, we need 2 $MnO_4^-$ for every 1 $I^-$ :

2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-

I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-

Add the two half-reactions. Cancel water and hydroxide where possible: …

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Common Mistakes & How to Avoid Them

Mistake 1: Assuming the product is always $I_2$

Why students make this mistake:

Students often remember that $KI$ reacts with oxidising agents to give iodine ( $I_2$ ), especially from reactions in acidic medium. They apply this memory without checking the medium.

How to avoid:

Always check the medium (acidic, alkaline, neutral) before deciding the product. In alkaline medium, $KMnO_4$ is a stronger oxidising agent and can oxidise $I^-$ beyond $I_2$ .

Mistake 2: Forgetting the oxidation state change of Mn in alkaline medium

Why students make this mistake:

Students memorise that $KMnO_4$ in acidic medium gives $Mn^{2+}$ (change from +7 to +2). They incorrectly assume the same change in alkaline medium.

How to avoid:

Remember the three different reductions of $KMnO_4$ :

Medium	Product of Mn	Colour change
Acidic	$Mn^{2+}$ (colourless)	Purple → colourless
Neutral/Weakly alkaline	$MnO_2$ (brown)	Purple → brown
Strongly alkaline	$MnO_4^{2-}$ (green)	Purple → green

In alkaline medium, $KMnO_4$ is reduced to $K_2MnO_4$ (manganate, green), gaining only 1 electron per $MnO_4^-$ ion.

Mistake 3: Not balancing the half-reactions to find the correct product

Why students make this mistake:

Students guess the product without writing balanced equations. They may pick $IO_4^-$ thinking "more oxidation is better" without checking electron balance.

How to avoid:

Write and balance the half-reactions:

Reduction half (alkaline medium):

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

Oxidation half:

Iodide ( $I^-$ , oxidation state -1) is oxidised. Check possible products:

$I_2$ (0): loses 1 electron per I atom → 2 electrons per $I_2$ molecule
$IO^-$ (+1): loses 2 electrons per I atom
$IO_3^-$ (+5): loses 6 electrons per I atom
$IO_4^-$ (+7): loses 8 electrons per I atom

Balance electrons:

Each $MnO_4^-$ gains only 1 electron (→ $MnO_4^{2-}$ ), so the electrons lost in the oxidation half must be supplied by the right number of permanganate ions. In cold alkaline medium the iodide is oxidised all the way to iodate, $IO_3^-$ (oxidation state $+5$ ), which means each iodine loses 6 electrons:

$I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{MnO}_2$
›Reveal solutionSolution
When MnO₂ is fused with KOH in the presence of O₂, it is oxidised to the manganate(VI) ion, giving a dark green melt. The product is K₂MnO₄, not KMnO₄, because the fusion conditions (strong alkali, limited oxidising power) stop at the +6 state. The correct option is (D).

The key concept here is oxidation state control by reaction conditions. Manganese can exist in many oxidation states, and the product formed depends on the strength of the oxidising agent and the medium (acidic vs. alkaline). In a strongly alkaline fusion with a moderate oxidant (O₂), manganese(IV) in MnO₂ is oxidised to manganese(VI), not to manganese(VII). The deep green colour of the manganate(VI) ion is a classic identifying feature.

Let’s walk through the reasoning step by step.

Identify the starting material and conditions.

We begin with $\mathrm{MnO}_2$ , where manganese is in the +4 oxidation state. It is fused (heated strongly) with KOH (a strong base) in the presence of oxygen gas ( $\mathrm{O}_2$ ). The fusion creates a molten, highly alkaline environment.

Determine the possible oxidation products.

In alkaline conditions, manganese can be oxidised to:

$\mathrm{Mn}_2\mathrm{O}_3$ (Mn in +3) — but this is less oxidised than the starting +4, so it would require reduction, not oxidation. Not possible here.

$\mathrm{MnO}_2$ (Mn in +4) — unchanged, but the reaction explicitly includes O₂, so oxidation is expected.

$\mathrm{K}_2\mathrm{MnO}_4$ (manganate, Mn in +6) — a dark green compound.

$\mathrm{KMnO}_4$ (permanganate, Mn in +7) — a purple compound.

Recall the classic fusion reaction.

The standard laboratory preparation of potassium manganate is exactly this:

$2\,\mathrm{MnO}_2 + 4\,\mathrm{KOH} + \mathrm{O}_2 \xrightarrow{\text{fusion}} 2\,\mathrm{K}_2\mathrm{MnO}_4 + 2\,\mathrm{H}_2\mathrm{O}$
The product is potassium manganate(VI), which is dark green. This is a well-known reaction in inorganic chemistry.

Why not KMnO₄? …
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COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
›Reveal solutionSolution
The key idea is that dichromate in basic medium converts to chromate, not to a hydroxide or chromium trioxide. The final answer is chromate ion, $\mathrm{CrO}_4^{2-}$ , so option (C) is correct.

Concept & Intuition

The chemistry here hinges on the acid–base equilibrium between dichromate ( $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ ) and chromate ( $\mathrm{CrO}_4^{2-}$ ). In acidic solution, dichromate is stable; in basic solution, it shifts to chromate. Heating with aqueous NaOH provides a strongly basic environment, so the reaction is simply a deprotonation/condensation reversal — no redox occurs, and no insoluble hydroxide forms because chromate is soluble.

Recognize the species involved

Potassium dichromate ( $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ ) in water gives $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ ions. Aqueous NaOH supplies $\mathrm{OH}^-$ ions, making the solution basic.

Write the equilibrium

The dichromate–chromate equilibrium is:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons 2\mathrm{CrO}_4^{2-} + 2\mathrm{H}^+$
In basic solution, $\mathrm{OH}^-$ consumes $\mathrm{H}^+$ , pulling the equilibrium to the right.

Apply Le Chatelier’s principle

Adding $\mathrm{OH}^-$ removes $\mathrm{H}^+$ (forming water), so the system shifts to produce more $\mathrm{CrO}_4^{2-}$ . Heating accelerates the reaction but does not change the product.

Check the options

(A) $\mathrm{Cr}(\mathrm{OH})_2$ : This would require reduction of Cr(VI) to Cr(II), which does not occur here (no reducing agent). …
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KCET 2025Set D-41 markMCQ
$\ce{H2S}$
›Reveal solutionSolution
Permanganate is the oxidant (Mn $+7 \to +2$ , so it is reduced) and sulphide is the reductant (S $-2 \to 0$ , so it is oxidised to free sulphur) — the only option that gets both directions right.

Step 1 — The concept: oxidant is reduced, reductant is oxidised

Half the options here are traps built on confusing these two words. Fix them first:

Oxidising agent — accepts electrons ⇒ its own oxidation number falls ⇒ it is reduced.

Reducing agent — donates electrons ⇒ its own oxidation number rises ⇒ it is oxidised.

So an option saying " $\ce{MnO4-}$ is oxidised to $\ce{Mn^{2+}}$ " is self-contradictory on its face: $+7 \to +2$ is a fall, which is reduction. That immediately kills options (A) and (C).

Step 2 — What must $\ce{MnO4-}$ do?

Oxidation number of Mn in $\ce{MnO4-}$ :
$x + 4(-2) = -1 \;\Rightarrow\; x = +7$
$+7$ is manganese's highest possible oxidation state — it has no electrons left to lose, so it cannot be oxidised. It can only gain electrons. In acidic medium (as stated) the standard 5-electron reduction is:
$\ce{MnO4^-_{(aq)} + 8H^+_{(aq)} + 5e^- -> Mn^{2+}_{(aq)} + 4H2O_{(l)}} \qquad E^\circ = +1.51~\text{V}$
Mn: $\;+7 \to +2$ . Reduced. ✓ (The purple colour discharges to the almost-colourless $\ce{Mn^{2+}}$ — the basis of permanganate titrations being self-indicating.)

Step 3 — What must $\ce{H2S}$ do?

Oxidation number of S in $\ce{H2S}$ :
$2(+1) + x = 0 \;\Rightarrow\; x = -2$
$-2$ is sulphur's lowest oxidation state — it has a full octet and no electrons left to gain, so it cannot be reduced. It can only lose electrons. So option (C)'s " $\ce{H2S}$ is reduced to $\ce{SO2}$ " is doubly absurd: it is not reduction, and $-2 \to +4$ is a rise.

With permanganate, sulphide is oxidised to free sulphur:
$\ce{H2S -> S + 2H^+ + 2e^-} \qquad \text{S: } -2 \to 0$
Step 4 — Balance the overall reaction

Multiply the Mn half-reaction by 2 (10 e⁻) and the S half-reaction by 5 (10 e⁻) so the electrons cancel: …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{W}$
›Reveal solutionSolution

The orange oxidiser Z is $\text{K}_2\text{Cr}_2\text{O}_7$ ; tracing its manufacture backwards, the starting ore W is chromite, $\text{FeCr}_2\text{O}_4$ - option (B).

Z is described as orange crystals used as an oxidising agent in acid medium - this is potassium dichromate, $\text{K}_2\text{Cr}_2\text{O}_7$ . Reading the standard industrial preparation backwards:

Roasting with $\text{Na}_2\text{CO}_3$ / $\text{O}_2$ : the chromite ore is fused in air, oxidising Cr(III) to Cr(VI) as sodium chromate (X):
$4\,\text{FeCr}_2\text{O}_4 + 8\,\text{Na}_2\text{CO}_3 + 7\,\text{O}_2 \rightarrow 8\,\text{Na}_2\text{CrO}_4 + 2\,\text{Fe}_2\text{O}_3 + 8\,\text{CO}_2$
Acidification ( $\text{H}^+$ ): yellow chromate converts to orange dichromate (Y):

$2\,\text{Na}_2\text{CrO}_4 + 2\,\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\,\text{Na}^+ + \text{H}_2\text{O}$ …
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KCET 2020Set A-11 markMCQ
$Q. Copper is extracted from copper pyrites by (A) Auto reduction (B) Thermal decomposition (C) Reduction by coke (D) Electrometallurgy$
›Reveal solutionSolution
Copper pyrites ( $CuFeS_2$ ) is a sulphide ore, and the extraction of copper from it uses the principle of auto reduction — the ore is partially roasted to produce some oxide, which then reacts with the remaining sulphide in a self-sustaining reaction without an external reducing agent. The correct option is (A).

The key to this question lies in understanding the nature of the ore and the method of extraction that is actually used in industry. Copper pyrites, also known as chalcopyrite, is a mixed sulphide of copper and iron. For sulphide ores, the common route is roasting followed by reduction. But here's the twist: copper extraction from chalcopyrite does not use coke or a separate reducing agent. Instead, it exploits a clever chemical cycle.

The process is called auto reduction (or sometimes "self-reduction"). Here's why it works:

Partial roasting: The concentrated ore is first roasted in a limited supply of air. This converts some of the copper sulphide ( $Cu_2S$ ) into copper oxide ( $Cu_2O$ ), while the iron sulphide ( $FeS$ ) is largely converted to iron oxide ( $FeO$ ), which is then removed as slag with silica.

$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$

The auto reduction step: The remaining $Cu_2S$ (which was not roasted) now reacts with the freshly formed $Cu_2O$ in the absence of air. This is a redox reaction where the sulphide acts as the reducing agent for the oxide:

$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$
Notice: no coke, no external reducing agent. The copper sulphide itself provides the reduction. This is the "auto" part — the ore reduces itself.

Why not the other options?

(B) Thermal decomposition: This works for ores like carbonates (e.g., $ZnCO_3$ ) or hydroxides, which break down on heating. Sulphides like $CuFeS_2$ do not simply decompose to metal; they need a chemical reaction. …
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KCET 2020Set A-11 markMCQ
$BaCl_2$
›Reveal solutionSolution
The salt (A) is a sulfite ( $SO_3^{2-}$ ) that gives a white $BaSO_3$ precipitate with $BaCl_2$ , which dissolves in dilute HCl to release $SO_2$ gas (B). $SO_2$ decolourises acidified $KMnO_4$ . So A = $BaSO_3$ , B = $SO_2$ , which matches option (B).

The key here is to connect each observation to a specific chemical property. The problem gives three clues: a white precipitate with $BaCl_2$ , that precipitate dissolves in dilute HCl to produce a gas, and that gas decolourises acidified $KMnO_4$ . Each clue narrows down the possibilities.

Let’s walk through it step by step.

White precipitate with $BaCl_2$

Barium chloride ( $BaCl_2$ ) is a common test for sulfate ( $SO_4^{2-}$ ) and sulfite ( $SO_3^{2-}$ ) ions. Both form white precipitates:

$Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4$ (white, insoluble in dilute acids)

$Ba^{2+} + SO_3^{2-} \rightarrow BaSO_3$ (white, but soluble in dilute acids) So the precipitate could be either $BaSO_4$ or $BaSO_3$ . We need the next clue to decide.

Precipitate dissolves in dilute HCl to produce a gas

This is the decisive step.

$BaSO_4$ is insoluble even in dilute HCl — it does not dissolve. So if the precipitate were $BaSO_4$ , it would not produce any gas.

$BaSO_3$ , on the other hand, reacts with dilute HCl:

$BaSO_3 + 2HCl \rightarrow BaCl_2 + H_2O + SO_2 \uparrow$
The gas produced is sulfur dioxide ($SO_2$).

Therefore, the precipitate must be $BaSO_3$ , and the gas (B) is $SO_2$ .

Gas (B) decolourises acidified $KMnO_4$ Acidified potassium permanganate ( $KMnO_4$ ) is a strong oxidising agent. $SO_2$ is a reducing agent and readily reduces $MnO_4^-$ (purple) to $Mn^{2+}$ (colourless): …
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KCET 2018Set A-11 markMCQ
$V_2O_5$
›Reveal solutionSolution
Amphoteric oxides react with both acids and bases. Among the given pairs, only $V_2O_5$ and $Cr_2O_3$ are amphoteric — so option (A) is correct.

The key idea: an amphoteric oxide sits in the middle of the acidity scale — it can act as an acid toward a strong base and as a base toward a strong acid. For transition metal oxides, the acidity increases as the oxidation state of the metal increases. A low oxidation state gives a basic oxide (like CrO, where Cr is +2), a very high oxidation state gives an acidic oxide (like Mn $_2$ O $_7$ , where Mn is +7), and intermediate oxidation states give amphoteric behaviour.

Let’s check each oxide in the options.

$V_2O_5$ — Vanadium is in the +5 oxidation state. This is a moderately high state, and indeed $V_2O_5$ is well-known as an amphoteric oxide. It dissolves in strong acids to give vanadyl salts (e.g., $VO_2^+$ ) and in strong bases to give vanadate ions ( $VO_4^{3-}$ ). So it passes the test.

$Cr_2O_3$ — Chromium is in the +3 oxidation state. This is the classic amphoteric oxide of chromium. It reacts with acids to give $Cr^{3+}$ salts and with bases to give chromite ions ( $CrO_2^-$ ). So it also passes.

$Mn_2O_7$ — Manganese is in the +7 oxidation state. This is a very high oxidation state, making the oxide strongly acidic. It reacts with water to give permanganic acid ( $HMnO_4$ ) and does not behave as a base. So it is not amphoteric.

$CrO$ — Chromium is in the +2 oxidation state. This is a low oxidation state, so the oxide is basic. It reacts with acids to give $Cr^{2+}$ salts but does not react with bases. Not amphoteric.

$V_2O_4$ — Vanadium is in the +4 oxidation state. This oxide is actually amphoteric as well (it reacts with both acids and bases), but the question asks for a pair. Let’s see the options.

Now examine each option:

(A) $V_2O_5$ , $Cr_2O_3$ — Both are amphoteric. This looks correct.

(B) $Mn_2O_7$ , $Cr_2O_3$ — $Mn_2O_7$ is acidic, not amphoteric. So wrong.

(C) $CrO$ , $V_2O_5$ — $CrO$ is basic, not amphoteric. So wrong. …
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Q. $KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with KI, iodide ion is oxidised to ____________.

Concept understanding — Inorganic Synthesis

Inorganic Synthesis – What It Really Means

The Precise Statement

A Concrete Example: Making Copper(II) Sulfate Pentahydrate

Why It Matters

A Common Misconception …

Why this formula?

Inorganic Synthesis: Why the Key Formulae Hold

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

Why this holds:

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

Why this holds:

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

Why this holds:

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

Why this holds:

Concept: Oxidation States & Redox Reactions in Alkaline Medium

Method: Half-Reaction Balancing in Alkaline Medium

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Assuming the product is always $I_2$

Mistake 2: Forgetting the oxidation state change of Mn in alkaline medium

Mistake 3: Not balancing the half-reactions to find the correct product

Step 1 — The concept: oxidant is reduced, reductant is oxidised

Step 2 — What must $\ce{MnO4-}$ do?

Step 3 — What must $\ce{H2S}$ do?

Step 4 — Balance the overall reaction

Q.KMnO4KMnO_4KMnO4​ acts as an oxidising agent in alkaline medium. When alkaline KMnO4KMnO_4KMnO4​ is treated with KI, iodide ion is oxidised to ____________.

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Q. $KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with KI, iodide ion is oxidised to ____________.

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