$KMnO_4$ acts as an oxidising agent in acidic medium. The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphide ions in acidic solution is (i) $\frac{2}{5}$ (ii) $\frac{3}{5}$ (iii) $\frac{4}{5}$ (iv) $\frac{1}{5}$

> [!TLDR] > In acidic medium, $MnO_4^-$ is reduced to $Mn^{2+}$ (gain of 5 electrons) while $S^{2-}$ is oxidised to $S$ (loss of 2 electrons). Balancing the electron transfer shows that 2 moles of $MnO_4^-$ react with 5 moles of $S^{2-}$, so 1 mole of $S^{2-}$ requires $\frac{2}{5}$ mole of $KMnO_4$. The correct option is **(i)**. The key to this problem lies in understanding **how oxidation states change** during the reaction. In redox chemistry, the number of moles of oxidising agent needed depends entirely on how many electrons it can accept versus how many the reducing agent can donate. Let’s break it down. 1. **Identify the half-reactions in acidic medium.** Permanganate ion ($MnO_4^-$) is a powerful oxidising agent. In acidic solution, it gets reduced to the $Mn^{2+}$ ion. The manganese in $MnO_4^-$ has an oxidation state of +7 (since each oxygen is -2, and the overall charge is -1: $x + 4(-2) = -1 \Rightarrow x = +7$). In $Mn^{2+}$, the oxidation state is +2. So each $MnO_4^-$ gains **5 electrons** to become $Mn^{2+}$: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$ 2. **Now look at the sulphide ion.** Sulphide ion ($S^{2-}$) is being oxidised. In acidic medium, the most common product is elemental sulphur ($S$). The oxidation state of sulphur in $S^{2-}$ is -2, and in elemental sulphur it is 0. So each $S^{2-}$ loses **2 electrons** to become $S$: $$S^{2-} \rightarrow S + 2e^-$$ 3. **Balance the electrons transferred.** The electrons lost by sulphide must equal the electrons gained by permanganate. The smallest common multiple of 5 and 2 is 10. Multiply the permanganate half-reaction by 2: $$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$ Multiply the sulphide half-reaction by 5: $$5S^{2-} \rightarrow 5S + 10e^-$$ 4. **Combine the half-reactions.** Adding them gives the net ionic equation: $$2MnO_4^- + 5S^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O$$ This tells us the

KMnO₄ acts as an oxidising agent in acidic medium. The number of moles of KMnO₄ that will be needed to react w

NCERT Exemplar · Q11

KMnO_4

(i)

\frac{2}{5}

(ii)

\frac{3}{5}

(iii)

\frac{4}{5}

(iv)

\frac{1}{5}

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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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In acidic medium, $MnO_4^-$ is reduced to $Mn^{2+}$ (gain of 5 electrons) while $S^{2-}$ is oxidised to $S$ (loss of 2 electrons). Balancing the electron transfer shows that 2 moles of $MnO_4^-$ react with 5 moles of $S^{2-}$ , so 1 mole of $S^{2-}$ requires $\frac{2}{5}$ mole of $KMnO_4$ . The correct option is (i).

The key to this problem lies in understanding how oxidation states change during the reaction. In redox chemistry, the number of moles of oxidising agent needed depends entirely on how many electrons it can accept versus how many the reducing agent can donate.

Let’s break it down.

Identify the half-reactions in acidic medium. Permanganate ion ( $MnO_4^-$ ) is a powerful oxidising agent. In acidic solution, it gets reduced to the $Mn^{2+}$ ion. The manganese in $MnO_4^-$ has an oxidation state of +7 (since each oxygen is -2, and the overall charge is -1: $x + 4(-2) = -1 \Rightarrow x = +7$ ). In $Mn^{2+}$ , the oxidation state is +2. So each $MnO_4^-$ gains 5 electrons to become $Mn^{2+}$ :

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

Now look at the sulphide ion. Sulphide ion ( $S^{2-}$ ) is being oxidised. In acidic medium, the most common product is elemental sulphur ( $S$ ). The oxidation state of sulphur in $S^{2-}$ is -2, and in elemental sulphur it is 0. So each $S^{2-}$ loses 2 electrons to become $S$ :

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

Balance the electrons transferred. The electrons lost by sulphide must equal the electrons gained by permanganate. The smallest common multiple of 5 and 2 is 10. Multiply the permanganate half-reaction by 2:

2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O

Multiply the sulphide half-reaction by 5:

5S^{2-} \rightarrow 5S + 10e^-

Combine the half-reactions. Adding them gives the net ionic equation:

2MnO_4^- + 5S^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 5S + 8H_2O

This tells us the stoichiometric ratio: 2 moles of $MnO_4^-$ react with 5 moles of $S^{2-}$ .

Find the moles of $KMnO_4$ needed for 1 mole of $S^{2-}$ . From the ratio: $\frac{2 \text{ mol } MnO_4^-}{5 \text{ mol } S^{2-}} = \frac{x \text{ mol } MnO_4^-}{1 \text{ mol } S^{2-}}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Writing the wrong half-reaction for sulphide ( $S^{2-}$ )

The correct half-reaction: $S^{2-}$ is oxidised to elemental sulphur, and balancing charge alone is enough (no $H^+$ or water needed):

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

Charge check: $-2$ on the left, $0 + 2(-1) = -2$ on the right ✓.

The actual trap: Students forget that in acidic medium, $KMnO_4$ is reduced to $Mn^{2+}$ , not $MnO_2$ or $MnO_4^{2-}$ .

Mistake 2: Using the wrong reduction product of $KMnO_4$

The error: Using $MnO_2$ (neutral/alkaline medium) or $MnO_4^{2-}$ (alkaline medium) instead of $Mn^{2+}$ .

How to avoid: Memorise the three media conditions:

Medium	Product	Change in oxidation state
Acidic	$Mn^{2+}$	+7 → +2 (gain of 5 electrons)
Neutral/weakly alkaline	$MnO_2$	+7 → +4 (gain of 3 electrons)
Strongly alkaline	$MnO_4^{2-}$	+7 → +6 (gain of 1 electron)

Key fact: The question explicitly says acidic medium — so $n$ -factor of $KMnO_4$ = 5.

Mistake 3: Forgetting to balance the sulphide oxidation half-reaction

The error: Writing $S^{2-} \rightarrow S + 2e^-$ without checking charge balance in acidic medium.

Check:

Left: charge = -2

Right: charge = 0 + (-2) = -2 ✓ (balanced)

So this half-reaction is actually fine. The real mistake is not balancing the overall redox equation.

Mistake 4: Incorrect mole ratio calculation

The error: Students often do:

\text{Moles of } KMnO_4 = \frac{\text{moles of } S^{2-} \times n\text{-factor of } S^{2-}}{n\text{-factor of } KMnO_4}

but use wrong $n$ -factors.

Correct approach:

Oxidation: $S^{2-} \rightarrow S + 2e^-$

$n$ -factor of $S^{2-}$ = 2
Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

$n$ -factor of $KMnO_4$ = 5
Mole ratio: …

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Q. $KMnO_4$ acts as an oxidising agent in acidic medium. The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphide ions in acidic solution is

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

Concept: Redox Titration — Equivalent Weight & n-Factor Method

Method: n-Factor / Equivalent Method

Common Mistakes & How to Avoid Them

Mistake 1: Writing the wrong half-reaction for sulphide ( $S^{2-}$ )

Mistake 2: Using the wrong reduction product of $KMnO_4$

Mistake 3: Forgetting to balance the sulphide oxidation half-reaction

Mistake 4: Incorrect mole ratio calculation

Q.KMnO4KMnO_4KMnO4​ acts as an oxidising agent in acidic medium. The number of moles of KMnO4KMnO_4KMnO4​ that will be needed to react with one mole of sulphide ions in acidic solution is

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Q. $KMnO_4$ acts as an oxidising agent in acidic medium. The number of moles of $KMnO_4$ that will be needed to react with one mole of sulphide ions in acidic solution is

Unlock everything free for 14 days