(a) What is lanthanoid contraction ? Actinoid contraction is greater from element to element than lanthanoid c

Question

Q. (a) What is lanthanoid contraction ? Actinoid contraction is greater from element to element than lanthanoid contraction. Why ?

(OR)

(b) Why do transition metals have high enthalpy of atomization ? Which element of 3d-series has lowest enthalpy of atomization ?

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Part (a)Concept understanding — Actinoid Contraction Greater

The Intuition First

Imagine you're holding a rope that runs through a series of rings. As you pull the rope tighter, the rings get squeezed closer together. Something similar happens inside the atoms of the actinoid elements — but instead of a rope, it's the pull of the nucleus, and instead of rings, it's the electron cloud.

The actinoid series runs from thorium ( $Z=90$ ) to lawrencium ( $Z=103$ ). As you go from one element to the next, you add one proton to the nucleus and one electron to the 5f subshell. That extra proton yanks harder on all the electrons. But here's the key: the 5f orbitals are shaped like little dumbbells that poke inside the atom, very close to the nucleus. They don't shield the outer electrons well from the growing nuclear charge.

So the outer electrons feel a stronger and stronger pull as you move right across the series. The entire electron cloud shrinks. The atomic radius decreases — steadily, noticeably. That's the actinoid contraction.

The Precise Statement

Actinoid contraction is the steady decrease in atomic (and ionic) radii across the actinoid series (from Th to Lr), caused by the poor shielding ability of 5f electrons.

The contraction is greater than what you see in the lanthanoid series (the 4f elements). Why? Two reasons:

The 5f orbitals are more diffuse and penetrate less than 4f orbitals. They sit further from the nucleus on average, so they shield even more poorly. Each added 5f electron does almost nothing to cancel the pull of the new proton.
Relativistic effects become significant for heavy nuclei (Z > 90). The 5f and 6d electrons move at speeds close to the speed of light, which contracts their orbitals further. This is a subtle but real effect that amplifies the contraction.

The result: the total decrease in ionic radius across the actinoid series is about 0.03–0.04 Å — roughly 10–15% larger than the lanthanoid contraction.

Why It Matters …

Why this formula?

Actinoid Contraction: Why It Is Greater Than Lanthanoid Contraction

Let’s build this from the ground up — understanding the why before the what.

1. What Is Actinoid Contraction?

Actinoid contraction is the steady decrease in atomic and ionic radii across the actinoid series (from Th to Lr, atomic numbers 90–103).

It is greater in magnitude than the analogous lanthanoid contraction (across the lanthanoids, Ce–Lu).

Key result: The contraction per element is larger in actinoids (~2–3 pm per element) than in lanthanoids (~1–2 pm per element).

2. The Core Reason: Poorer Shielding by 5f Electrons

The Shielding Effect

Electrons in inner shells shield the outer electrons from the full nuclear charge.
The effective nuclear charge ( $Z_{\text{eff}}$ ) felt by an electron is:

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

where $Z$ = atomic number, $S$ = shielding constant.

Why 5f Shielding Is Weaker Than 4f

4f orbitals (lanthanoids) are more penetrating — they have a small but significant probability near the nucleus. This gives them better shielding ability.
5f orbitals (actinoids) are more diffuse and less penetrating. They are spread farther from the nucleus, so they shield the outer electrons less effectively.

Result: For the same increase in $Z$ , the $Z_{\text{eff}}$ increases more in actinoids than in lanthanoids.

A larger $Z_{\text{eff}}$ pulls the electron cloud inward more strongly → greater contraction.

3. Mathematical Expression of the Trend

The contraction is described by the change in radius per added proton:

For lanthanoids:

\Delta r_{\text{4f}} \approx -1 \text{ to } -2 \text{ pm per element}

For actinoids:

\Delta r_{\text{5f}} \approx -2 \text{ to } -3 \text{ pm per element}

The ratio of contraction is roughly:

\frac{\Delta r_{\text{5f}}}{\Delta r_{\text{4f}}} \approx 1.5 \text{ to } 2

4. Why the Difference in Shielding? — Orbital Shape Matters

Property	4f orbital (lanthanoids)	5f orbital (actinoids)
Principal quantum number ( $n$ )	4	5
Radial extent	More compact	More diffuse
Penetration near nucleus	Higher (has a small lobe near nucleus)	Lower (less probability near nucleus)
Shielding efficiency	Better	Poorer

The radial distribution function shows that 4f electrons have a secondary maximum close to the nucleus, while 5f electrons lack this — they are more "spread out."

5. The Chain of Reasoning (Step-by-Step)

Add a proton → nuclear charge $Z$ increases by 1.
Add a 5f electron → it shields poorly because of its diffuse shape.
Net effect: $Z_{\text{eff}}$ increases more than it would for a 4f electron. …

Part (b)Concept understanding — Transition Element Definition

Transition Elements: From Intuition to Definition

Imagine you're building a house with bricks. Most bricks are identical — you stack them in neat rows. But some bricks are special: they have extra slots on their sides where you can attach hooks, magnets, or other bricks. These special bricks can change the shape of the wall, conduct electricity, or even change colour when you heat them.

In the periodic table, transition elements are those special bricks. They are the metals that sit in the middle block — groups 3 to 12 — and they have a unique ability: they can use their inner electrons (not just the outermost ones) to form bonds, change oxidation states, and create colourful compounds.

The Intuition: Why "Transition"?

The word "transition" comes from the idea that these elements form a bridge between the highly reactive metals on the left (like sodium, magnesium) and the less reactive metals / non-metals on the right (like aluminium, silicon). Their properties are not extreme — they are in-between.

But the real reason they are special lies in their electron configuration.

The Precise Definition (IUPAC)

A transition element is an element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

Let's unpack that.

1. The "d" sub-shell

Electrons are arranged in shells (K, L, M, N...) and sub-shells (s, p, d, f). The d sub-shell can hold a maximum of 10 electrons. In transition elements, the d sub-shell is being filled — but not completely.

For example, consider Iron (Fe):

Atomic number 26
Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$
The 3d sub-shell has 6 electrons — it is incomplete (it can hold 10).

So iron is a transition element.

2. The "or" part — cations matter

Some elements have a complete d sub-shell in their neutral atom, but when they lose electrons to form positive ions (cations), the d sub-shell becomes incomplete.

Example: Zinc (Zn)

Neutral Zn: $[Ar] 3d^{10} 4s^2$ — the 3d sub-shell is full (10 electrons).
But Zn commonly forms $Zn^{2+}$ : $[Ar] 3d^{10}$ — still full.
So zinc is NOT a transition element by the IUPAC definition.

Example: Copper (Cu)

Neutral Cu: $[Ar] 3d^{10} 4s^1$ — 3d is full.
But $Cu^{2+}$ : $[Ar] 3d^9$ — now the 3d sub-shell is incomplete.
So copper IS a transition element.

Watch out

A common mistake: thinking that all elements in the d-block (groups 3–12) are transition elements. They are not. Zinc, cadmium, and mercury are d-block elements but NOT transition elements because their common cations have a full d sub-shell.

The "d-block" vs "Transition Elements"

d-block elements	Transition elements
Groups 3 to 12	Groups 3 to 11 (excluding Zn, Cd, Hg)
All have d electrons	Must have incomplete d sub-shell in atom or common cation

Why this formula?

Transition Element Definition: The "Why" Behind the Definition

The Core Definition

A transition element (IUPAC definition) is an element whose atom has an incomplete d-subshell in its ground state or can form stable ions with an incomplete d-subshell.

Key exam point: This definition covers both the neutral atom and its common ions.

Why This Definition? The Reasoning

1. The d-orbital filling pattern

In the periodic table, transition elements belong to the d-block (Groups 3–12). As we move across a period, electrons fill the $(n-1)d$ orbitals after the $ns$ orbital.

For example, in Period 4:

Scandium (Sc): $[Ar]\,3d^1\,4s^2$ — has one d-electron → transition element
Zinc (Zn): $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element

2. The "incomplete d-subshell" condition

The definition focuses on incompleteness because:

A full d-subshell ( $d^{10}$ ) is exceptionally stable (like a noble gas configuration for d-orbitals)
Elements with $d^{10}$ configurations do not show the characteristic properties of transition metals (variable oxidation states, coloured compounds, catalytic activity, paramagnetism)

3. Why include ions?

Consider Zinc (Zn):

Ground state: $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element
Common ion: $\text{Zn}^{2+}$ : $[Ar]\,3d^{10}$ — still full → still not a transition element

Now consider Copper (Cu):

Ground state: $[Ar]\,3d^{10}\,4s^1$ — d-subshell is full → by atom definition alone, not a transition element
But $\text{Cu}^{2+}$ : $[Ar]\,3d^9$ — incomplete d-subshell → is a transition element

Therefore: The definition must include ions to correctly classify elements like Cu, which form stable ions with incomplete d-subshells.

The "Formula" — A Decision Tree

The definition can be expressed as a logical condition:

\text{Transition element} \iff \left( \text{Atom has } d^{1-9} \right) \lor \left( \text{Stable ion has } d^{1-9} \right)

Where:

$d^{1-9}$ means incomplete d-subshell (1 to 9 electrons)
$d^{0}$ or $d^{10}$ means complete (empty or full) → not a transition element

Common Exam Exceptions …

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Part (a): lanthanoid contraction = steady size decrease across the 4f series from poor 4f shielding; actinoid contraction is greater because 5f electrons shield even more poorly. Part (b): transition metals atomize with high enthalpy because unpaired $d$ -electrons give strong metallic bonding; Zn ( $3d^{10}$ ) has the lowest in the 3d series.

Part (a)

Across the lanthanoids, electrons enter the deeply buried, diffuse 4f subshell. Because f-orbitals penetrate poorly toward the nucleus, they shield the outer 5d/6s electrons only weakly. As each proton is added, the effective nuclear charge felt by the outer electrons therefore increases, pulling the electron cloud inward. The result is a small, steady decrease in atomic and ionic radii from cerium to lutetium - the lanthanoid contraction (e.g. $r(La^{3+})\approx106$ pm falls to $r(Lu^{3+})\approx85$ pm). …

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

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set 56/1/11 markMCQ
$Q. Which of the following is not a transition metal ? (A) Sc (B) Ag (C) Hg (D) Cu$
›Reveal solutionSolution
Transition metals are defined by having partially filled d-orbitals in their common oxidation states. Mercury (Hg) has a full d¹⁰ configuration in both its elemental and common +2 state, so it is not a transition metal. The correct answer is (C).

Why This Definition Matters

The classification of transition metals isn't about being a metal or having d-electrons — it's about partially filled d-orbitals in the atom or in any common ion. This is the IUPAC definition. If an element's d-subshell is completely full (d¹⁰) in all its common forms, it doesn't qualify, even if it sits in the d-block of the periodic table.

Let's check each option against this rule.

Scandium (Sc, Z=21)

Electronic configuration: $[\text{Ar}]\,3d^1 4s^2$ .

Its common ion is Sc³⁺: $[\text{Ar}]$ — the 3d orbital is empty.

But the atom has a partially filled d-orbital (3d¹). The definition says "in the atom or in any common oxidation state." Since the atom itself has an incomplete d-subshell, Sc is a transition metal.

Silver (Ag, Z=47)

Configuration: $[\text{Kr}]\,4d^{10} 5s^1$ .

Common ion: Ag⁺ → $[\text{Kr}]\,4d^{10}$ — full d-subshell.

However, silver also forms Ag²⁺ (e.g., in AgF₂), which has $4d^9$ — a partially filled d-orbital. Because at least one common oxidation state (Ag²⁺) has an incomplete d-subshell, Ag qualifies as a transition metal.

Mercury (Hg, Z=80)

Configuration: $[\text{Xe}]\,4f^{14} 5d^{10} 6s^2$ .

Common ions: Hg²⁺ → $[\text{Xe}]\,4f^{14} 5d^{10}$ (full d¹⁰); Hg₂²⁺ (dimeric) also has each Hg with a full d¹⁰ core.

Mercury does not form any stable ion with a partially filled d-orbital. Its d-subshell is always full. Therefore, Hg is not a transition metal.

Copper (Cu, Z=29)

Configuration: $[\text{Ar}]\,3d^{10} 4s^1$ . …
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CBSE 2026Set 56/2/11 markMCQ
$Q. Assertion (A) : Zinc, cadmium and mercury are not considered as transition elements. Reason (R) : These elements have completely filled orbitals in their ground state as well as in their common oxidation states.$
›Reveal solutionSolution
The key idea is that transition elements are defined by having partially filled d-orbitals in their ground state or common oxidation states. Since Zn, Cd, and Hg have completely filled d¹⁰ configurations in both, they fail this criterion. Thus, both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).

Why this approach works

The definition of a transition element is not arbitrary — it stems from the unique chemistry of d-block elements. The IUPAC defines a transition element as an element whose atom has a partially filled d-subshell, or which can give rise to cations with an incomplete d-subshell. This means we must check two things: the ground state electron configuration of the neutral atom, and the configurations of its common oxidation states. If either has a partially filled d-orbital, the element qualifies. If both are completely filled, it does not.

Zinc, cadmium, and mercury sit at the end of their respective d-block series (Group 12). Their neutral atoms have the configuration $(n-1)d^{10} ns^2$ — the d-subshell is full. Their common oxidation state is +2, formed by losing the two s-electrons, leaving $(n-1)d^{10}$ — still full. No partially filled d-orbital appears anywhere. Hence, they are not transition elements.

Let’s walk through the reasoning step by step.

Recall the defining criterion for transition elements.

A transition element must have an atom or a common ion with an incomplete d-subshell. This is the official IUPAC definition. Elements that have completely filled d-orbitals in both the ground state and all common oxidation states are excluded.

Examine the ground state configurations of Zn, Cd, and Hg.

Zinc ( $Z = 30$ ): $[Ar] 3d^{10} 4s^2$

Cadmium ( $Z = 48$ ): $[Kr] 4d^{10} 5s^2$

Mercury ( $Z = 80$ ): $[Xe] 4f^{14} 5d^{10} 6s^2$ In each case, the d-subshell is completely filled ( $d^{10}$ ). No partially filled d-orbital exists in the neutral atom.

Check their common oxidation states.

The most stable and common oxidation state for all three is +2. For example:

Zn loses its two 4s electrons to form $Zn^{2+}$ : $[Ar] 3d^{10}$

Cd loses its two 5s electrons to form $Cd^{2+}$ : $[Kr] 4d^{10}$

Hg loses its two 6s electrons to form $Hg^{2+}$ : $[Xe] 4f^{14} 5d^{10}$ In every case, the d-subshell remains completely filled. No partially filled d-orbital appears in any common oxidation state. …
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CBSE 2026Set DZ1 markMCQ
$Q. Which of the following is not a transition metal?$
(a) Cu
(b) Cr
(c) Fe
(d) Na
›Reveal solutionSolution
A transition metal has a partially filled d-subshell (in the atom or a common ion). Cu, Cr and Fe qualify; Na does not — so option (d).

A transition element is defined as one whose atom or one of its stable ions has an incompletely filled d-orbital.

Cu ( $[Ar]3d^{10}4s^1$ ): forms $Cu^{2+}$ ( $3d^9$ ) — partially filled d → transition metal.

Cr ( $[Ar]3d^{5}4s^1$ ) — partially filled d → transition metal. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Transition element among the following is$
(a) Zinc
(b) Cadmium
(c) Cerium
(d) Rutherfordium
›Reveal solutionSolution
A transition element is defined as one having a partially (incompletely) filled d-subshell in the ground state or in any of its common oxidation states.

Zinc (3d10 4s2) and Cadmium (4d10 5s2) have a completely filled d-subshell in the atom AND in their only common +2 ion, so they are NOT classified as transition elements (they are 'post-transition'/group 12 metals).

Cerium is an f-block (lanthanide) element, not d-block. …
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CBSE 2026Set ANNUAL1 mark
$Q. Why is Zn not considered as transition element? (Z = 30)$
›Reveal solutionSolution

Transition elements must have partly filled d-orbitals in the elemental or common ionic state; zinc's d-subshell is always completely full, so it does not qualify.

Zinc (Z = 30) has the electronic configuration [Ar] 3d10 4s2. When it forms its common oxidation state, Zn2+, it loses the two 4s electrons, giving [Ar] 3d10 - the d-subshell remains completely filled (all 10 electrons present) in both the metal and its ion.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Chromium belongs to which block?$
(a) s-block
(b) p-block
(c) d-block
(d) f-block
›Reveal solutionSolution

Chromium (Z=24) has the configuration [Ar] 3d⁵ 4s¹, placing it in the d-block.

Chromium has atomic number 24 and electronic configuration [Ar] 3d⁵ 4s¹ (an exception to the expected 3d⁴4s², arising from the extra stability of a half-filled d-subshell). Since its last-entering electron occupies a …

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CBSE 2026Set ANNUAL1 mark
$Q.______ block elements are called transition elements.$
›Reveal solutionSolution

d-block elements are called transition elements.

The periodic table is divided into four blocks (s, p, d, f) based on which subshell receives the last (differentiating) electron. Elements in which the last electron enters a d-orbital form the d-block, spanning groups 3 to 12. These are called transition elements because most of them have partially filled d-orbitals in their elemental or common ionic states, giving them properties 'transitional' between the highly reactive s-b …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Transition metals have incomplete$
(a) s-orbital
(b) p-orbital
(c) d-orbital
(d) f-orbital
›Reveal solutionSolution

Transition metals have incomplete d-orbitals.

Transition elements are defined as those whose atoms or stable ions have partially filled (incomplete) d-orbitals. This partly filled d-subshell is responsible for their characteristic properties: var …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Total number of elements are in d-block is ______$
(a) 39
(b) 40
(c) 38
(d) 36
›Reveal solutionSolution
Four transition series of 10 elements each -> 40 d-block elements.

The d-block (transition elements) consists of elements in which the last electron enters a d-subshell. There are four such series:

3d series (Sc to Zn),

4d series (Y to Cd),

5d series (La/Hf to Hg),

6d series (Ac/Rf onward). …
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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Zn, Cd and Hg generally do not considered as a ______ elements.$
›Reveal solutionSolution

Zn, Cd and Hg are not true transition elements because their d-orbitals are completely filled.

A transition element is defined as one that has a partially filled d-subshell in its atomic state or in one of its common oxidation (ionic) states. Zinc, cadmium and mercury have the configuration (n-1)d10 ns2, and even in their common +2 ions they are (n-1)d10 - the d-subshell is completely filled. Since they …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Choose the correct order of atomic and ionic radii of chlorine atom and chloride (Cl-) ion (in pm) is$
(a) a) 136 and 90
(b) b) 167 and 99
(c) c) 99 and 181
(d) d) 186 and 90
›Reveal solutionSolution
[!TLDR]

c) 99 and 181

Why

The covalent/atomic radius of Cl is about 99 pm, while the Cl- ion is larger (extra electron, same n …
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CBSE 2025Set 56/6/11 markMCQ
$Q. Which of the following is the softest metal ? (A) Zn (B) Sc (C) Cu (D) Fe$
›Reveal solutionSolution
Softness in metals is determined by how easily their atoms slide past one another; among the transition elements listed, scandium (Sc) has the weakest metallic bonding due to having only one $d$ -electron available for bonding, making it the softest. The answer is (B).

Understanding Metallic Softness

Hardness in metals arises from the strength of metallic bonding—the electrostatic attraction between the "sea" of delocalized electrons and the positive metal ions. The more electrons available for delocalization (especially from $d$ -orbitals in transition metals), and the smaller the atomic radius, the stronger the bonding and the harder the metal.

Conversely, a soft metal has weaker metallic bonds. Its atoms can slide past one another more easily under stress, making it malleable and easy to deform.

Comparing the Four Metals

Let's examine each candidate by looking at their electronic configurations and the number of electrons contributing to metallic bonding:

Zinc (Zn): Electronic configuration $[\text{Ar}] \, 3d^{10} \, 4s^2$

All ten $d$ -electrons are paired in filled orbitals. While the two $4s$ electrons participate in bonding, the filled $d^{10}$ subshell contributes relatively little to directional bonding. Zinc is moderately soft but not the softest here.

Scandium (Sc): Electronic configuration $[\text{Ar}] \, 3d^1 \, 4s^2$

Only one $d$ -electron is available. This is the first transition metal in the series, with minimal $d$ -orbital participation in bonding. The metallic bond is weak because there are fewer electrons to delocalize and the $d$ -orbitals are only just beginning to fill. Scandium is notably soft and can be cut with a knife.

Copper (Cu): Electronic configuration $[\text{Ar}] \, 3d^{10} \, 4s^1$ …
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Q.(a) What is lanthanoid contraction ? Actinoid contraction is greater from element to element than lanthanoid contraction. Why ?

Part (a)

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