(a) Complete and balance the following equations : (i) 2KMnO₄ xrightarrow513K (ii) Na₂Cr₂O₇ + 2KCl arrow (b) W

Question

Q. (a) Complete and balance the following equations :

(i)

2KMnO_4 \xrightarrow{513\,K}

(ii)

Na_2Cr_2O_7 + 2KCl \rightarrow

(b) Why is it difficult to separate lanthanoid elements in pure state ?

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Concept understanding — Lanthanoid Contraction

Lanthanoid Contraction – From Intuition to Precision

Imagine you are walking across a row of the periodic table — from lanthanum (atomic number 57) to lutetium (71). Your first instinct might be: as we add more protons and more electrons, the atom should get bigger. But the opposite happens. The atoms actually shrink, steadily and stubbornly, across these 15 elements. That is the lanthanoid contraction.

Why? The answer lies in the 4f orbitals.

The core intuition: a bad shield

Every new electron you add across the lanthanoid series goes into a 4f orbital. These 4f orbitals are shaped like cloverleaves, but they are tucked deep inside the atom — very close to the nucleus. They are also notoriously poor at shielding the outer electrons from the pull of the nucleus.

Here is the key: each step adds one proton to the nucleus. That proton yanks harder on all the electrons. Normally, the new electron you add would partly cancel that pull (shielding). But 4f electrons are so diffuse and so poorly penetrating that they do a terrible job of shielding. So the net effect is that the effective nuclear charge felt by the outer electrons increases steadily across the series. The outer electrons get pulled inward, and the whole atom shrinks.

Note

The 4f orbitals are "inside" the atom — they lie closer to the nucleus than the 5d and 6s orbitals. So adding electrons there does not push the outer shell outward; instead, the increasing nuclear charge dominates.

The precise statement

Lanthanoid contraction is the progressive and regular decrease in atomic and ionic radii of the lanthanoid elements (Ce to Lu) as atomic number increases. The contraction is about 1 pm per element, totalling roughly 15 pm from La to Lu.

This is not a small effect. It is so consistent that the radii of the later lanthanoids are almost identical to those of the 4d transition metals directly above them in the periodic table. For example, zirconium (Zr) and hafnium (Hf) have nearly the same atomic radius — a direct consequence of the lanthanoid contraction.

\text{Atomic radius (pm)} \approx 187 - 0.9 \times (Z - 57) \quad \text{(rough linear fit for trivalent ions)}

Why it matters

The lanthanoid contraction explains several important patterns in chemistry:

Similarity of 4d and 5d transition metals: Elements like Zr and Hf, Nb and Ta, Mo and W are nearly identical in size and chemical behaviour. Without the lanthanoid contraction, the 5d metals would be much larger. This is why separating Hf from Zr is famously difficult — they are chemical twins.
Trend in basicity: Across the lanthanoid series, the ionic radius decreases. This increases the charge density on the ion, making it more polarising. As a result, the basicity of the hydroxides decreases from La(OH)₃ (strong base) to Lu(OH)₃ (weak base). …

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The lanthanide contraction causes nearly identical ionic radii across the lanthanoid series, making chemical separation extremely difficult. For the equations: (i) $2KMnO_4 \xrightarrow{513\,K} K_2MnO_4 + MnO_2 + O_2$ ;

(ii) $Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl$ .

Let’s tackle this in two parts: first the chemical equations, then the conceptual question about lanthanoids.

Part (a): Balancing the equations

1. Equation (i): $2KMnO_4 \xrightarrow{513\,K}$

Potassium permanganate ( $KMnO_4$ ) is a strong oxidising agent. When heated strongly (around 513 K), it decomposes. The key is to recognise that manganese in $KMnO_4$ has an oxidation state of +7. On heating, it disproportionates — meaning the same element both oxidises and reduces itself.

Manganese goes from +7 to two different states: +6 (in manganate) and +4 (in manganese dioxide). Oxygen is released as a gas. The balanced equation is:

2KMnO_4 \xrightarrow{513\,K} K_2MnO_4 + MnO_2 + O_2

Check: Left side has 2 K, 2 Mn, 8 O. Right side: $K_2MnO_4$ gives 2 K, 1 Mn, 4 O; $MnO_2$ gives 1 Mn, 2 O; $O_2$ gives 2 O. Total: 2 K, 2 Mn, 8 O. Balanced.

Watch out

A common mistake is to write $KMnO_2$ or forget the oxygen gas. Remember: heating permanganate always produces oxygen — that’s why it’s used in some old-style oxygen generators.

2. Equation (ii): $Na_2Cr_2O_7 + 2KCl \rightarrow$

This is a double displacement reaction. Sodium dichromate reacts with potassium chloride. The products swap cations: potassium dichromate and sodium chloride. Both dichromates are soluble, but potassium dichromate is less soluble than sodium dichromate in cold water — this reaction is used to prepare potassium dichromate from the cheaper sodium salt.

The balanced equation is:

Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl

No change in oxidation states here — it’s purely a metathesis (exchange) reaction.

Tip

In the lab, this reaction is done by mixing concentrated solutions and cooling. $K_2Cr_2O_7$ crystallises out because its solubility drops sharply with temperature, while $NaCl$ stays in solution.

Part (b): Why is it difficult to separate lanthanoid elements in pure state?

3. The core concept: Lanthanide contraction

As you move across the lanthanoid series (from La to Lu, atomic numbers 57 to 71), the 4f orbitals are being filled. These f-orbitals are deeply buried inside the atom — they have poor shielding ability. Each added proton pulls the outer electrons inward, but the f-electrons don’t shield each other well. So the atomic and ionic radii decrease steadily but very slightly across the series.

This steady decrease is called the lanthanide contraction.

4. Why this makes separation hard

All lanthanoid ions (typically $Ln^{3+}$ ) have almost identical:

Ionic radii (differ by only ~1 pm per element)
Charge (+3)
Chemical behaviour (they form similar complexes, salts, and have similar solubility) …

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

CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Due to ............... Zirconium (Zr) and Hafnium (Hf) have identical radii.$
›Reveal solutionSolution

Zirconium (period 5) and hafnium (period 6) have almost identical atomic/ionic radii because of lanthanide contraction, which offsets the expected increase in size on descending a group.

Normally, atomic/ionic radius increases on going down a group as a new shell is added. But between Zr (Z=40) and Hf (Z=72), the 4f orbitals are filled across the lanthanides (Z=58–71). The 4f electrons shield the nuclear charge poorly, so the effective nuclear charge felt by the outer electrons increases steadily across the lanthanide series, causing a steady contraction in atomic/ionic size — the lanthanide contraction.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The reason of lanthanoid contraction is$
(a) negligible screening effect of f-orbital
(b) increasing nuclear charge
(c) decreasing nuclear charge
(d) decreasing screening effect
›Reveal solutionSolution

The lanthanoid contraction is caused by the poor ability of the diffuse, deeply-buried 4f orbitals to shield one another from the increasing nuclear charge, so effective nuclear charge rises steadily and atomic/ionic radii shrink across the series.

Mechanism: Going from $Ce$ to $Lu$ , the atomic number (and hence nuclear charge $Z$ ) increases by one unit at each step as an electron is added to the inner $4f$ subshell. The $4f$ orbitals are radially compact and have poor overlap/penetration properties, so one $4f$ electron shields another $4f$ electron (or the outer $5d^16s^2$ electrons) from the nucleus very inefficiently — far less efficiently than, say, $s$ or $p$ electrons shield each other. Because this poor screening does not keep pace with the steadily rising nuclear charge, the effective nuclear charge ( $Z_{eff}=Z-S$ ) experienced by the outer electrons increases continuously across the series, pulling the electron cloud in and steadily shrinking the atomic and ionic radii. The cumulative shrinkage across all 14 lanthanoids is the lanthanoid contraction, and it is large enough that the third-row transition elements following the lanthanoids (e.g. $Zr$ and $Hf$ , $Nb$ and $Ta$ ) end up with almost identical atomic radii to their second-row counterparts.

Why the other options are wrong/incomplete: …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Lanthanoid contraction is due to increase in$
(a) atomic number
(b) effective nuclear charge
(c) atomic radius
(d) valence electrons
›Reveal solutionSolution

Lanthanoid contraction is the steady decrease in atomic/ionic radii across the lanthanoid series, caused by an increase in effective nuclear charge that is imperfectly screened by the diffuse 4f electrons.

Across the lanthanoid series (Ce to Lu), each successive element adds one proton to the nucleus and one electron to the inner 4f subshell. The 4f orbitals have poor shielding ability (diffuse, non-directional shapes), so they do not effectively screen the outer electrons from the increasing nuclear charge.

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Which of the following pair of ions have same size due to lanthanoid contraction?$
(a) Fe2+, Ni2+
(b) Zr4+, Ti4+
(c) Zr4+, Hf4+
(d) Zn2+, Hf4+
›Reveal solutionSolution

Zr4+ (4d series) and Hf4+ (5d series) end up nearly identical in size because lanthanoid contraction cancels out the size increase normally expected on adding a new shell.

Going down a group, ionic radius normally increases as a new principal shell is added. But between the 4d series (e.g. Zr, Z = 40) and the 5d series (e.g. Hf, Z = 72), the 14 lanthanoid elements (57-71) intervene, each adding a poorly-shielding 4f electron. The resulting steady contraction in radius across the lanthanoids ('lanthanoid contraction') almost exactly cancels the expected size increase from Zr to Hf. The net result is that Zr4+ and Hf4+ have almost identical ionic radii (~72 pm) …

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CBSE 2022Set ANNUAL1 mark
$Q. What is lanthanoid contraction?$
›Reveal solutionSolution

Across the lanthanoid series, each added electron goes into the deeply buried, diffuse 4f subshell, which shields the increasing nuclear charge poorly — so the effective nuclear charge felt by outer electrons rises steadily, causing atomic/ionic size to shrink steadily from La to Lu.

Definition

Lanthanoid contraction refers to the steady and cumulative decrease in the atomic and ionic (especially $Ln^{3+}$ ) radii of the lanthanoid elements as one moves from lanthanum ( $Z=57$ ) to lutetium ( $Z=71$ ) across the 4f series.

Cause

As atomic number increases across the lanthanoids, each successive electron is added to an inner $4f$ orbital. The $4f$ orbitals are radially very diffuse and penetrate poorly, so electrons within the same $4f$ subshell shield each other from the nuclear charge very inefficiently (poor mutual shielding). Consequently, as protons are added to the nucleus, the effective nuclear charge ( $Z_{eff}$ ) experienced by the outer ( $5s$ , $5p$ , $6s$ ) electrons increases steadily across the series (since the extra positive charge is not being cancelled out by the added, poorly-shielding $4f$ electron). This progressively stronger pull draws the outer electron cloud inward, so the atomic/ionic radius contracts steadily with each additional element.

Consequences

…

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CBSE 2021Set A1 markMCQ
$Q. In which of the following elements 4f energy level is gradually filled?$
(a) Lanthanides
(b) Actinides
(c) Transition metals
(d) Coinage metals
›Reveal solutionSolution
The 4f orbitals fill across the lanthanides (Ce → Lu).

The lanthanides (the first inner-transition series, elements 58 Ce to 71 Lu) are characterised by the gradual filling of the 4f energy level.

Actinides fill the 5f orbitals. …
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Q.(a) Complete and balance the following equations :

Part (a): Balancing the equations

Part (b): Why is it difficult to separate lanthanoid elements in pure state?

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