On addition of small amount of $KMnO_4$ to concentrated $H_2SO_4$, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following. (i) $Mn_2O_7$ (ii) $MnO_2$ (iii) $MnSO_4$ (iv) $Mn_2O_3$

> [!TLDR] > Concentrated $H_2SO_4$ dehydrates $KMnO_4$ to form the highly explosive, green oily $Mn_2O_7$. The correct option is **(i)**. The key here is recognising what concentrated sulphuric acid does to permanganate. It’s not a simple redox — it’s a dehydration. $H_2SO_4$ is a powerful dehydrating agent, and $KMnO_4$ contains water of constitution (the oxygen atoms bonded to manganese). When you add a little solid $KMnO_4$ to conc. $H_2SO_4$, the acid rips away the potassium and water, leaving behind the anhydrous oxide of manganese in its highest oxidation state. That green oily liquid is $Mn_2O_7$ — manganese heptoxide. It’s famously unstable and explodes on contact with organic matter or even on heating. The reaction is: $$2KMnO_4 + 2H_2SO_4 \rightarrow Mn_2O_7 + 2KHSO_4 + H_2O$$ Now let’s walk through the reasoning step by step. 1. **Oxidation state of manganese in $KMnO_4$** Permanganate ion is $MnO_4^-$. Oxygen is $-2$ each, so total from four oxygens is $-8$. The ion has a $-1$ charge, so Mn must be $+7$ to balance: $+7 - 8 = -1$. So Mn is in its highest common oxidation state. 2. **What concentrated $H_2SO_4$ does** Conc. $H_2SO_4$ is a strong dehydrating agent. It doesn’t just remove water molecules — it can remove the elements of water from compounds, including oxygen and hydrogen from oxyacids and oxysalts. Here, it removes $K^+$ and $O^{2-}$ in the form of $K_2O$ (which reacts with acid to form $KHSO_4$), leaving behind $Mn_2O_7$. 3. **Why $Mn_2O_7$ is green and oily** $Mn_2O_7$ is a covalent molecule (not ionic), with Mn in $+7$ state. Covalent oxides of transition metals in high oxidation states are often volatile, oily liquids. The green colour is characteristic — it’s the same deep green you see in the permanganate ion in strongly acidic conditions, but here it’s the pure oxide. 4. **Why it’s explosive** $Mn_2O_7$ is a very strong oxidising agent. The Mn–O bonds are strained and the molecule is thermodynamically unstab

On addition of small amount of KMnO₄ to concentrated H₂SO₄, a green oily compound is obtained which is highly

NCERT Exemplar · Q5

KMnO_4

(i)

Mn_2O_7

(ii)

MnO_2

(iii)

MnSO_4

(iv)

Mn_2O_3

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Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

Imagine you are walking through a dense forest. With every step forward, you push through thick undergrowth. The deeper you go, the more tired you become — each step feels a little harder, and you find yourself hunching forward, your shoulders pulling inward. That inward pull is exactly what happens inside the lanthanide atoms.

The lanthanides are the 14 elements from cerium ( $Ce$ , atomic number 58) to lutetium ( $Lu$ , atomic number 71). As you move from one element to the next, you add one proton to the nucleus and one electron to the atom. The new electron goes into a 4f orbital — a set of orbitals that are shaped like clover leaves and sit deep inside the atom, close to the nucleus.

Here is the key: 4f orbitals are poorly shielded. They do not spread out far from the nucleus, and they do not block the nuclear charge from pulling on the outer electrons. So when you add a proton, the nucleus gets stronger, and the 4f electrons do almost nothing to stop that extra pull. The result? The entire electron cloud — especially the outermost electrons — gets pulled inward. The atom shrinks.

Note

Shielding is the ability of inner electrons to "block" the outer electrons from feeling the full positive charge of the nucleus. Electrons in s and p orbitals shield well; 4f electrons shield very poorly.

The Precise Statement

Lanthanide contraction is the steady and significant decrease in the atomic and ionic radii of the lanthanide elements as atomic number increases from 58 (Ce) to 71 (Lu).

\text{Atomic radius} \propto \frac{1}{Z_{\text{eff}}}

The total contraction across the entire series is about 15–20 picometers — roughly 10–15% of the initial radius. That is a substantial shrinkage for a single row of the periodic table.

Why It Matters

This contraction has two enormous consequences in chemistry:

Similarity of post-lanthanide elements: After lutetium, the next elements are hafnium ( $Hf$ , 72), tantalum ( $Ta$ , 73), and tungsten ( $W$ , 74). Because the lanthanide contraction has made the atoms so small, these elements have almost identical atomic and ionic radii to their counterparts directly above them in the periodic table — zirconium ( $Zr$ ), niobium ( $Nb$ ), and molybdenum ( $Mo$ ). This is why zirconium and hafnium are chemically almost inseparable — they are the same size.
Difficulty in separating lanthanides: All lanthanide ions ( $Ln^{3+}$ ) have nearly identical chemical properties because their radii change so gradually. Separating them requires hundreds of repeated steps (ion-exchange chromatography, solvent extraction) — a painstaking process that was a major challenge in early nuclear chemistry.

Watch out

A common mistake is to think lanthanide contraction means the atoms get smaller because the 4f orbitals are "full" or because of some repulsion effect. It is purely due to poor shielding of the 4f electrons, which lets the nuclear charge pull everything inward.

The Numbers (for reference)

Element	Atomic Number	Ionic Radius ( $Ln^{3+}$ , pm)
Ce	58	103.4
Pr	59	101.3
Nd	60	99.5

Why this formula?

Lanthanide Contraction: Why It Happens

The Lanthanide Contraction is the steady decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) as atomic number increases. The key observation: the radii shrink by about 1–2 pm per element, despite adding electrons to the 4f subshell.

The Core Question

Why does adding electrons not increase the size, but instead decrease it?

The Formula That Governs It

The effective nuclear charge ( $Z_{\text{eff}}$ ) experienced by an electron is:

Z_{\text{eff}} = Z - S

Where:

$Z$ = atomic number (protons in nucleus)
$S$ = shielding constant (screening by inner electrons)

The key formula for the trend in ionic radii ( $r$ ) across the lanthanides is:

r \propto \frac{n^2}{Z_{\text{eff}}}

Where $n$ is the principal quantum number of the outermost electron (here, $n=6$ for the 6s orbital).

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

Each lanthanide adds:

+1 proton to the nucleus (increases $Z$ by 1)
+1 electron to the 4f subshell

2. The 4f orbital is "penetrating" but poorly shielding

The 4f orbital has a radial distribution that peaks close to the nucleus (inside the 5s and 5p shells).
However, 4f electrons are very poor at shielding the outer 6s electrons from the nuclear charge.

Why?

The 4f orbital is diffuse and deeply buried — it does not effectively screen the outer electrons because:

Its shape (complex, multi-lobed) means it doesn't occupy the space between the nucleus and the 6s electrons efficiently.
The 4f electrons are inside the 5s/5p shells, so they don't block the nuclear pull on the 6s electrons.

3. The net effect on $Z_{\text{eff}}$

When you add one proton ( $\Delta Z = +1$ ) and one 4f electron ( $\Delta S \approx 0.85$ to 0.95), the change in effective nuclear charge is:

\Delta Z_{\text{eff}} \approx +1 - 0.85 = +0.15 \text{ to } +0.05

Result: $Z_{\text{eff}}$ increases slightly with each element.

4. How this shrinks the radius

From the formula $r \propto \frac{n^2}{Z_{\text{eff}}}$ :

$n$ (the principal quantum number of the 6s orbital) stays constant at 6.
$Z_{\text{eff}}$ increases.
Therefore, $r$ decreases. …

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Concentrated $H_2SO_4$ dehydrates $KMnO_4$ to form the highly explosive, green oily $Mn_2O_7$ . The correct option is (i).

The key here is recognising what concentrated sulphuric acid does to permanganate. It’s not a simple redox — it’s a dehydration. $H_2SO_4$ is a powerful dehydrating agent, and $KMnO_4$ contains water of constitution (the oxygen atoms bonded to manganese). When you add a little solid $KMnO_4$ to conc. $H_2SO_4$ , the acid rips away the potassium and water, leaving behind the anhydrous oxide of manganese in its highest oxidation state.

That green oily liquid is $Mn_2O_7$ — manganese heptoxide. It’s famously unstable and explodes on contact with organic matter or even on heating. The reaction is:

2KMnO_4 + 2H_2SO_4 \rightarrow Mn_2O_7 + 2KHSO_4 + H_2O

Now let’s walk through the reasoning step by step.

Oxidation state of manganese in $KMnO_4$

Permanganate ion is $MnO_4^-$ . Oxygen is $-2$ each, so total from four oxygens is $-8$ . The ion has a $-1$ charge, so Mn must be $+7$ to balance: $+7 - 8 = -1$ . So Mn is in its highest common oxidation state.
What concentrated $H_2SO_4$ does

Conc. $H_2SO_4$ is a strong dehydrating agent. It doesn’t just remove water molecules — it can remove the elements of water from compounds, including oxygen and hydrogen from oxyacids and oxysalts. Here, it removes $K^+$ and $O^{2-}$ in the form of $K_2O$ (which reacts with acid to form $KHSO_4$ ), leaving behind $Mn_2O_7$ .
Why $Mn_2O_7$ is green and oily

$Mn_2O_7$ is a covalent molecule (not ionic), with Mn in $+7$ state. Covalent oxides of transition metals in high oxidation states are often volatile, oily liquids. The green colour is characteristic — it’s the same deep green you see in the permanganate ion in strongly acidic conditions, but here it’s the pure oxide.
Why it’s explosive

$Mn_2O_7$ is a very strong oxidising agent. The Mn–O bonds are strained and the molecule is thermodynamically unstable. A slight shock, heat, or contact with a reducing agent (like organic solvents, paper, or even dust) triggers a violent decomposition to $MnO_2$ and $O_2$ :

2Mn_2O_7 \rightarrow 4MnO_2 + 3O_2

This releases a lot of energy very fast — hence explosive. …

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Let’s break this down step-by-step — first the concept, then the common mistakes and how to avoid them.

🧪 The Concept

When a small amount of $KMnO_4$ is added to concentrated $H_2SO_4$ , the reaction produces:

2KMnO_4 + 2H_2SO_4 \rightarrow Mn_2O_7 + 2KHSO_4 + H_2O

$Mn_2O_7$ is a green oily liquid that is highly explosive — it decomposes violently on heating or contact with organic matter.

✓ Correct answer: (i) $Mn_2O_7$

✗ Common Mistakes & How to Avoid Them

1. ✗ Choosing $MnO_2$ (Option (ii))

Why students pick it:

They recall that $KMnO_4$ decomposes to $MnO_2$ on heating, or that $MnO_2$ is a common product in redox reactions of permanganate.

How to avoid:

Remember: $MnO_2$ is a black solid, not a green oily liquid.
The green oily description is a dead giveaway for $Mn_2O_7$ .
$MnO_2$ forms in neutral/alkaline medium, not in conc. $H_2SO_4$ .

2. ✗ Choosing $MnSO_4$ (Option (iii))

Why students pick it:

They think $KMnO_4$ in acidic medium always reduces to $Mn^{2+}$ (as $MnSO_4$ ).

How to avoid:

That’s true for dilute $H_2SO_4$ — but here it’s concentrated $H_2SO_4$ , which acts as a dehydrating agent, not just an acid.
Conc. $H_2SO_4$ removes water, forcing the formation of the anhydride $Mn_2O_7$ .

3. ✗ Choosing $Mn_2O_3$ (Option (iv))

Why students pick it:

They confuse the oxidation states or think $Mn_2O_3$ is also an explosive oxide.

How to avoid:

$Mn_2O_3$ is a brown-black solid, not green or oily.
It is not explosive like $Mn_2O_7$ .
The explosive nature is unique to $Mn_2O_7$ among common manganese oxides.

4. ✗ Forgetting the “green oily” and “explosive” clues

Why students miss it:

They focus only on the chemical equation and ignore the physical description.

How to avoid: …

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

AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$\mathrm{CrO}$
›Reveal solutionSolution
This tests several standard inorganic-chemistry facts; the false statement is the lanthanide-aluminium reactivity comparison — early lanthanoids are compared to calcium, not aluminium.

Concept and Intuition

This is an elimination-style question testing multiple independent facts. Oxidation-state trends in chromium oxides (CrO basic, $\mathrm{CrO_3}$ acidic, $\mathrm{Cr_2O_3}$ amphoteric — the classic 'basic-to-acidic with increasing oxidation state' trend), the oxidising power of acidified permanganate, and the industrial Wacker process catalyst are all standard textbook facts. The lanthanoids, being highly electropositive with large ionic radii similar to $\mathrm{Ca^{2+}}$ , are conventionally noted to have reactivity resembling calcium — not the much smaller, covalent-leaning, amphoteric aluminium.

Step-by-Step Solution

(A): CrO is basic (low oxidation state, +2), $\mathrm{Cr_2O_3}$ is amphoteric (+3) — this trend of increasing acidity with oxidation state is textbook-correct.

(B): $\mathrm{KMnO_4}$ in acidic medium is a strong oxidiser and readily oxidises nitrite ( $\mathrm{NO_2^-}$ ) to nitrate ( $\mathrm{NO_3^-}$ ) — correct.

(C): The Wacker process (oxidation of ethylene to acetaldehyde) uses a $\mathrm{PdCl_2}$ / $\mathrm{CuCl_2}$ catalytic system — correct. …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$Q. Chemistry of lanthanides (Ln) is dominated by +3 oxidation state. Which of the following is incorrect? (A) Ln (III) compounds are generally colourless (B) Ionic size of Ln (III) decreases with increasing atomic number (C) Due to large size of Ln (III) ions the bonding in their compounds is ionic (D) Ln (III) hydroxides are basic in character$
›Reveal solutionSolution
Most Ln(III) ions are actually coloured (due to f–f transitions), so the claim that they are "generally colourless" is the incorrect statement among the four.

Concept and Intuition

Lanthanide chemistry is dominated by the +3 oxidation state because that is the most stable configuration once the outer $6s$ and often one $5d$ / $4f$ electron are lost. The 4f orbitals in Ln(III) ions are only partially filled for most of the series (except La³⁺: 4f⁰, Gd³⁺: 4f⁷ half-filled sometimes near-colourless, Lu³⁺: 4f¹⁴ fully filled) which allows f–f transitions that absorb visible light and give many lanthanide ions their characteristic (often pastel) colours — Nd³⁺ is lilac, Pr³⁺ is green, Sm³⁺ is yellow, Er³⁺ is pink, and so on. Only the ions with an empty, fully-filled, or (nearly) exactly half-filled f-subshell are colourless or nearly so.

Separately, the lanthanide contraction (steady decrease in ionic radius across the series due to poor shielding by 4f electrons) is a well-established, correct fact; the large ionic size of Ln³⁺ leading to predominantly ionic bonding is also standard; and Ln(OH)₃ compounds are indeed basic (basicity decreasing across the series as the ion shrinks) — all textbook-correct.

Step-by-Step Solution

Evaluate (A): "Ln(III) compounds are generally colourless" — FALSE, since most have partially-filled 4f orbitals giving rise to f–f transitions and visible colour.

Evaluate (B): "Ionic size of Ln(III) decreases with increasing atomic number" — TRUE, this is the lanthanide contraction. …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. From the list given below, the number of lanthanides which exhibit +4 state in their oxides is Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy (A) 5 (B) 4 (C) 6 (D) 2$
›Reveal solutionSolution
The lanthanide contraction stabilises certain oxidation states; only Pr, Nd, Tb, and Dy form stable +4 oxides among the listed elements, giving a total of 4.

The key to this question lies in the Lanthanide Contraction — the steady decrease in ionic radii across the lanthanide series due to poor shielding by 4f electrons. This contraction makes higher oxidation states (+4) accessible only for elements that can achieve a stable, half-filled (f⁷), empty (f⁰), or fully-filled (f¹⁴) 4f subshell after losing four electrons. Let’s work through the list systematically.

Understand the stability rule for +4 state

A lanthanide in the +4 oxidation state has lost its two 6s electrons and two 4f electrons (or one 4f and one 5d, but effectively the 4f count drops by 2). The +4 state is stable only if the resulting electron configuration is either:

f⁰ (empty subshell) — very stable, like La³⁺ but La itself doesn’t easily give +4.

f⁷ (half-filled) — extra stability from exchange energy.

f¹⁴ (fully filled) — also very stable. For the elements listed, we check which ones, upon losing 4 electrons, reach f⁰, f⁷, or f¹⁴.

Write the ground-state 4f configurations for the neutral atoms

Pr (Z=59): 4f³

Nd (Z=60): 4f⁴

Pm (Z=61): 4f⁵

Sm (Z=62): 4f⁶

Eu (Z=63): 4f⁷

Gd (Z=64): 4f⁷ 5d¹ (but often written 4f⁷ 5d¹; in ions, Gd³⁺ is 4f⁷)

Tb (Z=65): 4f⁹

Dy (Z=66): 4f¹⁰

Determine which give stable +4 configurations

Pr: 4f³ → lose 4e⁻ → 4f⁻¹? That’s impossible. Actually, Pr⁴⁺ has configuration 4f¹ (since Pr³⁺ is 4f²). Wait — careful: Pr loses 3 electrons to become Pr³⁺ (4f²). To get Pr⁴⁺, it loses one more 4f electron → 4f¹. That’s not f⁰, f⁷, or f¹⁴. So why does Pr show +4? Because Pr⁴⁺ has the same configuration as Ce⁴⁺ (4f¹) — but Ce⁴⁺ is stable due to achieving noble gas core? Actually Ce⁴⁺ is 4f⁰ (since Ce is [Xe]4f¹5d¹6s², Ce⁴⁺ is [Xe]). For Pr, Pr⁴⁺ is [Xe]4f¹ — not particularly stable. However, Pr does form PrO₂ (Pr⁴⁺) because the lattice energy of the oxide compensates. So Pr is borderline but does exhibit +4 in oxides (PrO₂, Pr₆O₁₁). So Pr counts.

Nd: Nd³⁺ is 4f³; Nd⁴⁺ would be 4f² — not a special stability. Yet Nd also forms NdO₂ under strong oxidising conditions. So Nd does exhibit +4, though less common. It counts.

Pm: Pm³⁺ is 4f⁴; Pm⁴⁺ would be 4f³ — no special stability. Also Pm is radioactive and scarce; no stable +4 oxide known. Does not count.

Sm: Sm³⁺ is 4f⁵; Sm⁴⁺ would be 4f⁴ — no. Sm does not show +4 in oxides (SmO₂ is not stable). Does not count. …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{Ce(OH)_3 < Lu(OH)_3 < Eu(OH)_3}$
›Reveal solutionSolution
This tests how lanthanide contraction affects the basicity of lanthanide hydroxides across the series. Answer: $\mathrm{Lu(OH)_3 < Eu(OH)_3 < Ce(OH)_3}$ .

Concept and Intuition

Across the lanthanide series, as atomic number increases, the ionic radius of the $\mathrm{Ln^{3+}}$ ion steadily contracts (the lanthanide contraction, due to poor shielding by the $4f$ electrons). A smaller, more highly charged-density cation polarizes the O–H bond in its hydroxide more strongly, making the M–OH bond more covalent and the hydroxide less able to dissociate to release $\mathrm{OH^-}$ — i.e. less basic. So basicity of $\mathrm{Ln(OH)_3}$ decreases steadily from the lighter (larger) lanthanides to the heavier (smaller) ones.

Step-by-Step Solution

Identify atomic numbers: Ce (Z=58, early lanthanide, largest ionic radius among the three), Eu (Z=63, mid-series), Lu (Z=71, last lanthanide, smallest ionic radius).

By lanthanide contraction, ionic radius order: $\mathrm{Ce^{3+}} > \mathrm{Eu^{3+}} > \mathrm{Lu^{3+}}$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$_3$
›Reveal solutionSolution
I, II and III are established periodic-trend facts (electron gain enthalpy of P, the Eu/Yb exception to lanthanide contraction, and Ce(OH) $_3$ as the most basic hydroxide among the strict Ce–Lu lanthanoid series); IV has the Na/Na $^+$ radii swapped.

Concept and Intuition

This question strings together several classic periodic-trend exceptions that examiners love: (a) half-filled/fully-filled configurations giving anomalously low reactivity/EGE, (b) exceptions within the otherwise smooth lanthanide contraction, and (c) the direct relationship between ionic size and hydroxide basicity for the lanthanoids.

Step-by-Step Solution

Statement I: Electron gain enthalpy magnitudes (kJ/mol, approx.): F ≈ −328, Cl ≈ −349, S ≈ −200, P ≈ −72. Phosphorus's extra stability from a half-filled $3p^3$ configuration makes it reluctant to accept another electron, giving it the least negative (smallest magnitude) EGE of the four. True.

Statement II: Lanthanide contraction is the steady radius decrease across Ce→Lu due to poor $4f$ shielding. Eu ( $4f^7$ ) and Yb ( $4f^{14}$ ) have special stability in the +2 oxidation state, giving them abnormally larger radii that break the smooth contraction — i.e. the contraction is not observed as a smooth trend at these two elements. True. …
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Q.On addition of small amount of $KMnO_4$ to concentrated $H_2SO_4$ , a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

The Precise Statement

Why It Matters

The Numbers (for reference)

Why this formula?

Lanthanide Contraction: Why It Happens

The Core Question

The Formula That Governs It

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

2. The 4f orbital is "penetrating" but poorly shielding

3. The net effect on $Z_{\text{eff}}$

4. How this shrinks the radius

Concept: Oxidation States of Manganese & Oxidative Properties of $KMnO_4$

Method: Oxidation State Analysis + Known Reaction of $KMnO_4$ with Conc. $H_2SO_4$

🧪 The Concept

✗ Common Mistakes & How to Avoid Them

1. ✗ Choosing $MnO_2$ (Option (ii))

2. ✗ Choosing $MnSO_4$ (Option (iii))

3. ✗ Choosing $Mn_2O_3$ (Option (iv))

4. ✗ Forgetting the “green oily” and “explosive” clues

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.On addition of small amount of KMnO4KMnO_4KMnO4​ to concentrated H2SO4H_2SO_4H2​SO4​, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

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Q.On addition of small amount of $KMnO_4$ to concentrated $H_2SO_4$ , a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

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