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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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

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.

Unlock everything free for 14 days