Why is HCl not used to make the medium acidic in oxidation reactions of KMnO₄ in acidic medium? (i) Both HCl a

NCERT Exemplar · Q21

KMnO_4

(i) Both HCl and

KMnO_4

act as oxidising agents.

(ii)

KMnO_4

oxidises HCl into

Cl_2

which is also an oxidising agent.

(iii)

KMnO_4

is a weaker oxidising agent than HCl.

(iv)

KMnO_4

acts as a reducing agent in the presence of HCl.

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Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

Imagine you're holding a ball on a hill. If you place it exactly at the top, it's balanced — but the slightest push sends it rolling down. That's an unstable position. If you place it in a small dip on the hillside, it stays put even if nudged — that's stable. Oxidation states work the same way: some are like the hilltop (easily changed), others like the dip (hard to change).

The Core Intuition

An oxidation state is just a number we assign to an atom to track how many electrons it has gained or lost compared to its neutral state. But atoms don't "want" to stay in arbitrary oxidation states — they want to reach a configuration that minimises their energy.

Stability here means: how reluctant is that oxidation state to change under normal conditions? A stable oxidation state resists being oxidised further or reduced further. An unstable one readily changes into something else.

The Precise Statement

Stability of an oxidation state refers to the tendency of an element to maintain that particular oxidation state under given conditions (temperature, pH, presence of other reagents). A stable oxidation state is one that does not easily undergo redox reactions — it is neither easily oxidised nor easily reduced.

This depends on three key factors:

Electronic configuration – Half-filled and fully-filled d or f subshells confer extra stability (e.g., $\text{Fe}^{3+}$ with $d^5$ is more stable than $\text{Fe}^{2+}$ with $d^6$ in some contexts).
Inert pair effect – Heavier p-block elements (like Tl, Pb, Bi) show lower oxidation states (e.g., +1 for Tl) as more stable than higher ones (+3 for Tl), because the s-electrons become reluctant to participate.
Disproportionation tendency – Some oxidation states are unstable because they spontaneously convert into two other states (e.g., $\text{Cu}^+$ in aqueous solution gives $\text{Cu}^{2+}$ and $\text{Cu}$ ).

Important

Stability is relative — it depends on the environment. $\text{Mn}^{2+}$ is stable in acidic solution but easily oxidised in alkaline medium. Always specify conditions when discussing stability.

Examples That Make It Concrete

Transition metals – $\text{Cr}^{3+}$ ( $d^3$ ) and $\text{Mn}^{2+}$ ( $d^5$ ) are exceptionally stable because half-filled/half-filled-like configurations have low energy. $\text{Cr}^{2+}$ ( $d^4$ ) is easily oxidised to $\text{Cr}^{3+}$ — it's unstable.

p-block elements – $\text{Pb}^{2+}$ is stable, $\text{Pb}^{4+}$ is a strong oxidising agent (unstable). $\text{Sn}^{2+}$ is a reducing agent (easily oxidised to $\text{Sn}^{4+}$ ), so $\text{Sn}^{4+}$ is more stable for tin.

Common pattern – For most elements, the most common oxidation state is the most stable one under standard conditions. But "most common" isn't always "most stable" — e.g., $\text{Fe}^{3+}$ is common but $\text{Fe}^{2+}$ is more stable in acidic solution.

Watch out

Do not confuse "stability" with "occurrence". $\text{Mn}^{7+}$ (as $\text{MnO}_4^-$ ) is common in the lab but is a powerful oxidising agent — it is not stable in the sense of resisting change. It readily accepts electrons.

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

This concept explains why certain oxidation states of an element are more stable than others — and why some states are never observed at all.

The Core Idea: Energy Minimisation

An oxidation state is stable when the total energy of the system is at a minimum. This depends on three competing factors:

Ionisation energy (energy needed to remove electrons)
Lattice energy (for ionic compounds) or bond energy (for covalent compounds)
Electronic configuration (half-filled / fully-filled subshells)

There is no single formula for stability — instead, we use trends and principles that act as "formulae" for prediction.

Key Principle 1: Inert Pair Effect (for p-block elements)

Why it holds:

For heavier elements (e.g., Tl, Pb, Bi), the 6s² electrons are held very tightly due to poor shielding and relativistic effects. They resist removal.

Result: Lower oxidation state (e.g., +1 for Tl, +2 for Pb) becomes more stable than the higher state (+3, +4).
Example: $\ce{Tl^3+}$ is a strong oxidising agent — it readily gains two electrons to become $\ce{Tl^+}$ .

Derivation logic:

The energy cost to remove the 6s² electrons is greater than the energy gained by forming additional bonds or lattice. So the system stays in the lower state.

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Why it holds:

A half-filled ( $d^5$ , $f^7$ ) or fully-filled ( $d^{10}$ , $f^{14}$ ) subshell has extra exchange energy and symmetry — making it unusually stable.

Example: $\ce{Mn^{2+}}$ ( $d^5$ ) is more stable than $\ce{Mn^{3+}}$ ( $d^4$ ). $\ce{Fe^{3+}}$ ( $d^5$ ) is more stable than $\ce{Fe^{2+}}$ ( $d^6$ ).

Derivation logic:

The exchange energy (Hund's rule) is maximum for half-filled configurations. Removing an electron from a half-filled shell costs extra energy — so the half-filled state is favoured.

Key Principle 3: Lattice Energy / Hydration Energy Compensation

For transition metals, stability of a particular oxidation state in aqueous solution depends on:

\Delta G^\circ = \Delta H^\circ_{\text{hydration}} - \Delta H^\circ_{\text{ionisation}}

Why it holds:

Higher oxidation states have higher ionisation energy (harder to remove electrons).
But they also have higher hydration energy (smaller, more charged ions attract water more strongly).
The balance determines which state is stable.

Example:

$\ce{Cu^+}$ is unstable in water because its hydration energy is too low to compensate for the loss of the second electron.
$\ce{Cu^{2+}}$ is stable in water.

Key Principle 4: Disproportionation

Some oxidation states are unstable and spontaneously convert to two other states:

\ce{2Cu^+ -> Cu + Cu^{2+}}

Why it holds: …

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The key idea is that $KMnO_4$ is a strong enough oxidising agent to oxidise chloride ions ( $Cl^-$ ) from HCl to chlorine gas ( $Cl_2$ ), which interferes with the intended oxidation reaction. Therefore, HCl is not used to acidify the medium. The correct option is (ii).

Why This Matters: Stability of Oxidation States

When you use $KMnO_4$ as an oxidising agent in acidic medium, the reaction is:

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

The permanganate ion ( $Mn$ in +7 state) is a very powerful oxidising agent. But the medium itself must not react with $KMnO_4$ — otherwise, you lose the oxidising power to a side reaction instead of using it on your intended target.

HCl is a common strong acid, so you might think it's a natural choice to provide $H^+$ ions. But here's the catch: the chloride ion ( $Cl^-$ ) in HCl is itself easily oxidised. $KMnO_4$ doesn't discriminate — it will oxidise whatever it can, including the very acid you added.

Step-by-Step Reasoning

1. Identify the nature of $KMnO_4$ in acidic medium

In acidic solution, $KMnO_4$ is one of the strongest common oxidising agents. Its standard reduction potential is high:

E^\circ (MnO_4^-/Mn^{2+}) = +1.51\ \text{V}

This means it can oxidise many substances, including halide ions.

2. Check what happens to chloride ions

Chloride ions ( $Cl^-$ ) can be oxidised to chlorine gas ( $Cl_2$ ). The half-reaction is:

2Cl^- \rightarrow Cl_2 + 2e^-

with a standard potential $E^\circ = -1.36\ \text{V}$ (when written as reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$ , $E^\circ = +1.36\ \text{V}$ ).

Since $1.51\ \text{V} > 1.36\ \text{V}$ , the permanganate ion can easily oxidise $Cl^-$ to $Cl_2$ .

3. Write the actual side reaction

When $KMnO_4$ is added to HCl, the following redox reaction occurs:

2MnO_4^- + 10Cl^- + 16H^+ \rightarrow 2Mn^{2+} + 5Cl_2 + 8H_2O

This consumes both $KMnO_4$ and the acid, producing chlorine gas — a coloured, toxic gas that itself is an oxidising agent.

Watch out

A common mistake is to think that HCl is avoided simply because it's "reactive" or because $KMnO_4$ is "weaker" — but option (iii) is exactly backwards. $KMnO_4$ is actually the stronger oxidising agent here, which is precisely why it attacks HCl.

4. Evaluate each option

(i) Both HCl and $KMnO_4$ act as oxidising agents. …

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Q.Why is HCl not used to make the medium acidic in oxidation reactions of $KMnO_4$ in acidic medium?

Concept understanding — Stability of Oxidation States

Stability of Oxidation States – From Intuition to Precision

The Core Intuition

The Precise Statement

Examples That Make It Concrete

How to Think About It in Exams …

Why this formula?

Stability of Oxidation States: Why It Works

The Core Idea: Energy Minimisation

Key Principle 1: Inert Pair Effect (for p-block elements)

Key Principle 2: Half-Filled / Fully-Filled Subshell Stability

Key Principle 3: Lattice Energy / Hydration Energy Compensation

Key Principle 4: Disproportionation

Why This Matters: Stability of Oxidation States

Step-by-Step Reasoning

Method: Redox Reactivity Analysis

Steps:

Common Mistakes Students Make on This Question

Mistake 1: Choosing (i) — "Both HCl and KMnO₄ act as oxidising agents"

Mistake 2: Choosing (iii) — "KMnO₄ is a weaker oxidising agent than HCl"

Mistake 3: Choosing (iv) — "KMnO₄ acts as a reducing agent in the presence of HCl"

Mistake 4: Not recognising the actual reaction

Q.Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4KMnO_4KMnO4​ in acidic medium?

Why This Matters: Stability of Oxidation States

Step-by-Step Reasoning

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Q.Why is HCl not used to make the medium acidic in oxidation reactions of $KMnO_4$ in acidic medium?

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