The involvement of (n – 1)d electrons in the behaviour of transition elements impart certain distinct characteristics to these elements. Thus, in addition to variable oxidation states, they exhibit paramagnetic behaviour, catalytic properties and tendency for the formation of coloured ions. The transition metals react with a number of non-metals like oxygen, nitrogen and halogens. $KMnO_4$ and $K_2Cr_2O_7$ are common examples. The two series of inner transition elements, lanthanoids and actinoids, constitute the f-block of the periodic table. In the lanthanoids, there is regular decrease in atomic size with increase in atomic number due to the imperfect shielding effect of 4f-orbital electrons which causes contraction. Answer the following questions : (a) Why do transition metals and their compounds act as good catalysts ? (b) What is the cause of contraction in the atomic size of lanthanoids ? (c) Define lanthanoid contraction. How does it affect the atomic radii of the third transition series and the second transition series ? OR (c) In aqueous media, which is a stronger reducing agent – $Cr^{2+}$ or $Fe^{2+}$ and why ?

> [!TLDR] > Part (a): Transition metals catalyse via variable oxidation states + surface adsorption; lanthanoid contraction is caused by poor 4f shielding and is the steady $Ce\to Lu$ size decrease that makes 5d radii ≈ 4d radii. > Part (b): $Cr^{2+}$ is the stronger reducing agent than $Fe^{2+}$ ($E^\circ = -0.41$ V vs $+0.77$ V). --- ## Part (a) ### (a) Why transition metals and their compounds act as good catalysts Two features are responsible: **variable oxidation states** and **vacant $d$-orbitals**. 1. **Variable oxidation states** let the metal change state by one electron at a time, forming temporary bonds with reactants and lowering activation energy (e.g. $V_2O_5$ cycles $V^{5+}\rightleftharpoons V^{4+}$ in the contact process). 2. **Surface adsorption** — in heterogeneous catalysis (Fe in Haber's process) reactants adsorb on the metal surface via vacant $d$-orbitals, weakening their bonds; the product then desorbs. 3. **Intermediate-complex formation** in homogeneous catalysis regenerates the catalyst. ### (b) Cause of contraction in the atomic size of lanthanoids From $La$ ($Z=57$) to $Lu$ ($Z=71$) each added electron enters a deeply buried 4f orbital. 4f electrons shield the outer electrons **imperfectly**, so the increasing nuclear charge (+1 each step) pulls the electron cloud inward, giving a steady decrease in atomic and ionic radii. > [!WARNING] > The contraction is due to *poor shielding* of the 4f electrons, not merely because the 4f subshell is being filled. ### (c) Lanthanoid contraction — definition and effect on transition series **Definition:** the progressive decrease in atomic and ionic radii of the lanthanoids ($Ce\to Lu$) with increasing atomic number, arising from poor 4f shielding. **Effect:** the 5d (third) transition series begins just after the lanthanoids, so its elements are already shrunk. Their radii end up almost identical to the 4d (second) series: | 4d | radius (pm) | 5d | radius (pm) | |----|----|----|----| | Zr

The involvement of (n – 1)d electrons in the behaviour — CBSE Class 12 Chemistry 2024 PYQ

Question

KMnO_4

(a) Why do transition metals and their compounds act as good catalysts ?

(b) What is the cause of contraction in the atomic size of lanthanoids ?

(c) Define lanthanoid contraction. How does it affect the atomic radii of the third transition series and the second transition series ?

(OR)

Cr^{2+}

Fe^{2+}

and why ?

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Part (a)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). …

Part (b)Concept understanding — Standard Reduction Potential

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V. …

Why this formula?

Standard Reduction Potential: Why the Formula Holds

Let's build this from first principles — understanding why before memorising what.

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

A standard reduction potential ( $E^\circ$ ) measures how strongly a species wants to gain electrons (be reduced) under standard conditions (1 M concentration, 1 atm pressure, 25°C).

But why can't we measure this directly? Because every reduction must be paired with an oxidation — you can't have electrons flowing without a complete circuit.

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

Consider a Daniell cell:

Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu

Experimentally, we measure the cell potential as +1.10 V.

Now, we define the standard hydrogen electrode (SHE) as exactly 0.00 V:

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00 \text{ V}

The reasoning step-by-step:

We can only measure differences — like measuring height difference between two points.
If we connect the SHE to the copper half-cell, we measure +0.34 V (Cu²⁺ is reduced).
If we connect the SHE to the zinc half-cell, we measure −0.76 V (Zn²⁺ is reduced less readily than H⁺).

Now, the cell potential is the difference in their tendencies:

E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = (+0.34) - (-0.76) = +1.10 \text{ V}

Key insight: The formula uses subtraction because we're comparing two half-cells against the same reference (SHE). The cathode is where reduction happens (higher $E^\circ$ ), the anode is where oxidation happens (lower $E^\circ$ ).

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

This is the thermodynamic derivation — the real "why."

From Gibbs free energy:

\Delta G = \Delta G^\circ + RT \ln Q

For an electrochemical cell:

\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ

Substituting:

-nFE = -nFE^\circ + RT \ln Q

Rearranging:

E = E^\circ - \frac{RT}{nF} \ln Q

Why this makes physical sense:

$RT \ln Q$ represents the entropy penalty of non-standard concentrations
$nF$ converts charge to energy (Faraday's constant × number of electrons)
The minus sign means: as products accumulate ( $Q$ increases), the cell potential drops — the reaction is approaching equilibrium

At equilibrium ( $Q = K$ ), $E = 0$ — the battery is "dead."

--- …

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Part (a): Transition metals catalyse via variable oxidation states + surface adsorption; lanthanoid contraction is caused by poor 4f shielding and is the steady $Ce\to Lu$ size decrease that makes 5d radii ≈ 4d radii.

Part (b): $Cr^{2+}$ is the stronger reducing agent than $Fe^{2+}$ ( $E^\circ = -0.41$ V vs $+0.77$ V).

Part (a)

(a) Why transition metals and their compounds act as good catalysts

Two features are responsible: variable oxidation states and vacant $d$ -orbitals.

Variable oxidation states let the metal change state by one electron at a time, forming temporary bonds with reactants and lowering activation energy (e.g. $V_2O_5$ cycles $V^{5+}\rightleftharpoons V^{4+}$ in the contact process).
Surface adsorption — in heterogeneous catalysis (Fe in Haber's process) reactants adsorb on the metal surface via vacant $d$ -orbitals, weakening their bonds; the product then desorbs.
Intermediate-complex formation in homogeneous catalysis regenerates the catalyst.

(b) Cause of contraction in the atomic size of lanthanoids

From $La$ ( $Z=57$ ) to $Lu$ ( $Z=71$ ) each added electron enters a deeply buried 4f orbital. 4f electrons shield the outer electrons imperfectly, so the increasing nuclear charge (+1 each step) pulls the electron cloud inward, giving a steady decrease in atomic and ionic radii.

Watch out

The contraction is due to poor shielding of the 4f electrons, not merely because the 4f subshell is being filled.

(c) Lanthanoid contraction — definition and effect on transition series

Definition: the progressive decrease in atomic and ionic radii of the lanthanoids ( $Ce\to Lu$ ) with increasing atomic number, arising from poor 4f shielding.

Effect: the 5d (third) transition series begins just after the lanthanoids, so its elements are already shrunk. Their radii end up almost identical to the 4d (second) series:

4d	radius (pm)	5d	radius (pm)
Zr	160	Hf	159
Nb	146	Ta	146
Mo	139	W	139

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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 markMCQ
$Q. A transition element X is placed in the first series of transition elements. It has the following characteristics:$
(i) It is a coinage metal.
(ii) It has positive reduction potential.
(iii) It does not react with HCl but reacts with nitric acid. The element 'X' is likely to be
(a) Zn
(b) Cu
(c) Au
(d) Ag
›Reveal solutionSolution
Among the first-series transition metals, copper uniquely fits all three clues: it is a coinage metal, has a positive standard reduction potential, and dissolves only in oxidising acids like HNO3, not in HCl.

Coinage metals are Cu, Ag, and Au - but the question specifies the element is in the FIRST series of transition elements (3d series), which rules out Ag (4d series) and Au (5d series), leaving only Cu.

Cu has E°(Cu2+/Cu) = +0.34 V, a positive reduction potential, meaning it is less reactive than hydrogen and will not displace H2 from dilute HCl (a non-oxidising acid). …
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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 2025Set ANNUAL1 markMCQ
$KMnO_4$
(a) reduces permanganate to $Mn^{2+}$
(b) oxidises oxalic acid to $CO_2$ and $H_2O$
(c) gets oxidised by oxalic acid to chlorine
(d) furnishes $H^+$ ions in addition to those from oxalic acid.
›Reveal solutionSolution

KMnO₄ is a strong enough oxidiser to attack the chloride ion of HCl itself, converting it to chlorine gas and introducing a side reaction that spoils the titration — this rules out using HCl as the acidifying agent.

$KMnO_4$ is a powerful oxidising agent. If dilute HCl were used to acidify the oxalic acid solution for this titration, the chloride ion ( $Cl^-$ ) itself would be oxidised by $KMnO_4$ to chlorine gas, as an unwanted side reaction:

$2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 5Cl_2\uparrow + 8H_2O$

…

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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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Part (a)

(a) Why transition metals and their compounds act as good catalysts

(b) Cause of contraction in the atomic size of lanthanoids

(c) Lanthanoid contraction — definition and effect on transition series

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