How do you count for the following observations ? (a) Though alkaline potassium permanganate and acidic potass

Exercises · 7.12

Q. How do you count for the following observations ?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

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Concept understanding — Oxidation State Stability

Oxidation State Stability: The First Meeting

Imagine you have a pile of Lego bricks. Some colours click together easily; others keep falling apart. An atom's oxidation state is like the colour of a Lego brick — it tells you how many electrons the atom has lost (positive state) or gained (negative state) compared to its neutral form. But not all oxidation states are equally happy. Some are rock-solid; others are fragile and want to change.

That's oxidation state stability: why some oxidation states of an element are comfortable and long-lasting, while others are restless and reactive.

The Intuition: Why would an atom "prefer" one state over another?

An atom wants to be stable. In chemistry, stability usually means one of two things:

A full outer shell — like noble gases (8 electrons in the outermost shell, or 2 for helium). Atoms will gain, lose, or share electrons to get there.
A half-filled or fully-filled subshell — this is a special bonus stability. For example, chromium loves having a $3d^5$ configuration (half-filled d-subshell) even if it means sacrificing a 4s electron.

So when an atom takes on an oxidation state, it's essentially deciding how many electrons to give away or accept. The "best" oxidation states are those that leave the atom with a stable electronic configuration.

Note

This is why sodium (Na) almost always shows +1 — losing one electron gives it the same electron configuration as neon (a noble gas). It's not trying to be fancy; it's trying to be comfortable.

The Precise Statement

Oxidation state stability refers to the tendency of an element to exist in a particular oxidation state under given conditions (temperature, pH, presence of other substances). An oxidation state is stable if:

The atom's electronic configuration after gaining/losing electrons is noble-gas-like (octet) or has a half-filled/full-filled d or f subshell.
The energy required to reach that state is low compared to other possible states.
The state does not spontaneously change into another state (e.g., by reacting with air, water, or itself).

Stability order (general trend): For main-group elements: oxidation states that give a noble gas configuration are most stable. For transition metals: +2 and +3 are common, but stability varies with the element and the environment.

Examples to cement the idea

Element	Common stable states	Why?
Sodium (Na)	+1	Losing 1 electron → Ne configuration (2,8)
Chlorine (Cl)	-1	Gaining 1 electron → Ar configuration (2,8,8)
Iron (Fe)	+2, +3	Both leave d-subshell partially filled; +3 is more stable in acidic conditions
Manganese (Mn)	+2, +7	+2 is stable (half-filled d⁵); +7 is stable in permanganate ion (MnO₄⁻) due to strong bonding with oxygen
Carbon (C)	+4, -4	Both give noble gas configuration (He for +4? No — careful: carbon's +4 means losing 4 electrons, leaving 1s², which is He-like. -4 means gaining 4 electrons, giving Ne-like 2,8)

Watch out

Don't confuse "common" with "stable". Carbon's +4 is common but not always stable — CO₂ is stable, but CCl₄ is not very stable in water. Stability depends on the compound, not just the element.

The deeper reason: Why do some states "disproportionate"?

Some oxidation states are so unstable that the atom reacts with itself. This is called disproportionation — one atom in an intermediate state splits into two atoms: one higher, one lower.

Example: Copper(I) in water. Cu⁺ (oxidation state +1) is unstable in aqueous solution. It spontaneously does this:

2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}

One Cu⁺ gets oxidised to Cu²⁺ (more stable in water), the other gets reduced to Cu metal (even more stable). The +1 state was disproportionation-unstable.

Tip

A quick check: If an element has three consecutive oxidation states (say +1, +2, +3), the middle one (+2) is often prone to disproportionation if the +1 and +3 states are both stable. This is common for elements like copper, mercury, and thallium.

What determines stability in practice? …

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The key idea is that the choice of oxidising agent depends on the reaction medium and the desired product. For (a), alcoholic KMnO₄ is used to avoid over-oxidation of toluene to CO₂, giving benzoic acid selectively. For (b), concentrated H₂SO₄ oxidises Br⁻ to Br₂ (red vapour) but not Cl⁻ to Cl₂ because Br⁻ is a stronger reducing agent than Cl⁻.

(a) Why alcoholic KMnO₄ for benzoic acid from toluene?

Concept & Intuition:

Potassium permanganate (KMnO₄) is a powerful oxidising agent. Its oxidising strength depends heavily on the pH of the medium. In acidic medium, MnO₄⁻ is reduced to Mn²⁺ (strongest oxidising power). In alkaline medium, it reduces to MnO₂ (moderate oxidising power). In neutral or alcoholic medium, the reaction is milder and more controlled.

When toluene (methylbenzene) is oxidised, the methyl group (-CH₃) is first converted to -CH₂OH, then -CHO, and finally -COOH (benzoic acid). If the oxidising agent is too strong (like acidic KMnO₄), the benzene ring itself may get attacked, leading to ring cleavage and CO₂ formation — over-oxidation. Alcoholic KMnO₄ provides a mild, controlled oxidation that stops at the carboxylic acid stage without destroying the aromatic ring. Just as importantly, alcohol dissolves BOTH reactants — toluene (an organic liquid, immiscible with water) and KMnO₄ — so the two meet in a single phase, with far better contact than a water-only medium could give.

Watch out

A common mistake is to think that "alkaline" and "alcoholic" KMnO₄ are the same. They are not. Alkaline KMnO₄ (aqueous KOH + KMnO₄) is still quite strong and can over-oxidise. Alcoholic KMnO₄ (KMnO₄ used in an alcohol medium) is milder, and — crucially — the alcohol acts as a common solvent that brings the water-loving oxidant and the water-repelling hydrocarbon into one phase.

Balanced redox equation:

The oxidation of toluene to benzoic acid by KMnO₄ in alcoholic medium:

\text{C}_6\text{H}_5\text{CH}_3 + 2\text{KMnO}_4 \rightarrow \text{C}_6\text{H}_5\text{COOK} + 2\text{MnO}_2 + \text{KOH} + \text{H}_2\text{O}

Then acidify to get benzoic acid:

\text{C}_6\text{H}_5\text{COOK} + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{COOH} + \text{KCl}

The half-reactions are:

Oxidation: $\text{C}_6\text{H}_5\text{CH}_3 + 2\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{COOH} + 6\text{H}^+ + 6e^-$
Reduction: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

Multiplying reduction by 2 to balance electrons (6e⁻ each side) gives the net ionic equation.

Tip

In exam problems, you don't need to write the full ionic equation every time. The molecular equation with KMnO₄ → MnO₂ + benzoate salt is sufficient for most board-level questions.

(b) Why HCl gas from chloride but Br₂ vapour from bromide with conc. H₂SO₄?

Concept & Intuition:

Concentrated sulphuric acid (H₂SO₄) is both a strong acid and an oxidising agent. When added to halide salts, it first displaces the hydrogen halide (HX). But for Br⁻ and I⁻, the H₂SO₄ further oxidises the HX to the elemental halogen.

The key lies in the reducing power of the halide ions:

\text{Reducing power: } \text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-

Chloride (Cl⁻): Cl⁻ is a weak reducing agent. Concentrated H₂SO₄ cannot oxidise Cl⁻ to Cl₂. So only the acid displacement occurs:

\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \uparrow

HCl is a colourless, pungent gas. …

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

GUJCET 2025Set 031 markMCQ
$\mathrm{MnF_{4}}$
›Reveal solutionSolution

Multiple-bond formation by oxygen lets Mn reach a higher oxidation state in the oxide than in the fluoride.

Concept — why highest oxide > highest fluoride. Although fluorine is more electronegative, it forms only single bonds. Oxygen can form multiple bonds (including $p\pi$ – $d\pi$ ), so it can stabilise a metal in a higher oxidation state. …
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GUJCET 2025Set 031 markMCQ
$\mathrm{MnO_{4}^{2-}}$
›Reveal solutionSolution
[!TLDR]

In acid, $MnO_4^{2-}$ disproportionates to $MnO_4^-$ (Mn +7) and $MnO_2$ (Mn +4).

Concept

Manganese in $MnO_4^{2-}$ is in the +6 state, which is unstable in acidic medium. It undergoes disproportionation, being simultaneously oxidised to $MnO_4^-$ (+7) and reduced to $MnO_2$ (+4).

Solution

The balanced reaction is:

$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ …
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GUJCET 2024Set 131 markMCQ
$Q. Name the member of lanthanide series which is well known to exhibit +4 oxidation state? (A) Cerium (B) Thulium (C) Gadolinium (D) Samarium$
›Reveal solutionSolution

Cerium attains the noble-gas-like $4f^0$ configuration in $Ce^{4+}$ , making $+4$ a well-known, stable oxidation state.

Concept: Lanthanides are mostly $+3$ , but a few show $+4$ or $+2$ when it gives an empty, half-filled or filled $4f$ subshell. $Ce^{4+}$ ( $[Xe]4f^0$ ) is especially stable and is a common oxidising agent (ceric …
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Q.How do you count for the following observations ?

(a) Why alcoholic KMnO₄ for benzoic acid from toluene?

(b) Why HCl gas from chloride but Br₂ vapour from bromide with conc. H₂SO₄?

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