Q.How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
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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: 1s22s22p63s23p64s23d6
- 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]3d104s2 — the 3d sub-shell is full (10 electrons).
- But Zn commonly forms Zn2+: [Ar]3d10 — still full.
- So zinc is NOT a transition element by the IUPAC definition.
Example: Copper (Cu)
- Neutral Cu: [Ar]3d104s1 — 3d is full.
- But Cu2+: [Ar]3d9 — now the 3d sub-shell is incomplete.
- So copper IS a transition element.
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]3d14s2 — has one d-electron → transition element
- Zinc (Zn): [Ar]3d104s2 — d-subshell is full → not a transition element
2. The "incomplete d-subshell" condition
The definition focuses on incompleteness because:
- A full d-subshell (d10) is exceptionally stable (like a noble gas configuration for d-orbitals)
- Elements with d10 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]3d104s2 — d-subshell is full → not a transition element
- Common ion: Zn2+: [Ar]3d10 — still full → still not a transition element
Now consider Copper (Cu):
- Ground state: [Ar]3d104s1 — d-subshell is full → by atom definition alone, not a transition element
- But Cu2+: [Ar]3d9 — 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:
Transition element⟺(Atom has d1−9)∨(Stable ion has d1−9)
Where:
- d1−9 means incomplete d-subshell (1 to 9 electrons)
- d0 or d10 means complete (empty or full) → not a transition element
Common Exam Exceptions …
The key idea is that first ionisation enthalpy rises only mildly and irregularly across the series (d-electrons shield the 4s electrons somewhat, but not perfectly), while the real, sharp irregularity shows up in the second ionisation enthalpy.
Reasoning:
- From Sc to Zn, nuclear charge increases steadily, so first ionisation enthalpy (ΔiH1) generally rises — but only slightly and with minor bumps, because the added 3d electrons shield the outer 4s electron from the growing nuclear charge almost as effectively as it shields itself, keeping the rise gentle (unlike the steep rise across a normal period of non-transition elements).
- The second ionisation enthalpy (ΔiH2) shows the real irregularity: it is unusually high for Cr and Cu, because their singly-charged ions (Cr+=3d5, Cu+=3d10) already have an extra-stable, half-filled or fully-filled d-subshell — removing a second electron means breaking into that stable subshell, which costs much more energy. …
The first ionisation enthalpy rises only slightly and irregularly across the first transition series, because the added 3d electrons imperfectly shield the 4s electrons from the growing nuclear charge. The sharper irregularity is in the second ionisation enthalpy: it is unusually high for Cr and Cu (whose M+ ions have the extra-stable 3d5/3d10 configurations, so removing a further electron breaks that stability) and comparatively low for Mn and Zn (whose M+ ions still carry one loosely-held 4s electron beyond a stable d5/d10 core).
The Real Data (Table 4.2)
| Element | ΔiH1 | M+ configuration | ΔiH2 |
|---|---|---|---|
| Sc | 631 | 3d14s1 | 1235 |
| Ti | 656 | 3d24s1 | 1309 |
| V | 650 | 3d34s1 | 1414 |
| Cr | 653 | 3d5 | 1592 |
| Mn | 717 | 3d54s1 | 1509 |
| Fe | 762 | 3d64s1 | 1561 |
| Co | 758 | 3d74s1 | 1644 |
| Ni | 736 | 3d84s1 | 1752 |
| Cu | 745 | 3d10 | 1958 |
| Zn | 906 | 3d104s1 | 1734 |
Step-by-Step Reasoning
1. First ionisation enthalpy — a gentle, only mildly irregular rise.
Across Sc→Zn, nuclear charge rises by one unit each step, so ΔiH1 generally increases (631 → 906). But the rise is much gentler than across a normal (non-transition) period, because each new electron is added to an inner 3d orbital rather than the outer shell — a 3d electron shields the 4s electrons from the nucleus almost as effectively as another 4s electron would, so the effective nuclear charge felt by the valence electron increases only slowly. This is why the chapter describes the first-ionisation-enthalpy trend as "irregular... though of little chemical significance," without pinning the irregularity to any one specific element.
2. Second ionisation enthalpy — the real, well-defined break.
ΔiH2 removes an electron from the singly-charged ion M+, and here the electronic configuration of M+ matters directly:
- Chromium: neutral Cr is 3d54s1 (the well-known half-filled-stability exception), so Cr+ is already 3d5 — a stable, half-filled d-subshell. Removing a second electron means breaking into this stable arrangement, so ΔiH2 for Cr (1592) is unusually high.
- Copper: neutral Cu is 3d104s1, so Cu+ is 3d10 — a stable, fully-filled subshell. Breaking into it likewise makes Cu's ΔiH2 (1958) the highest in the row.
- Manganese: neutral Mn is 3d54s2, so Mn+ is 3d54s1 — the stable 3d5 core is already intact, with one "spare" 4s electron still to remove. Taking that easy 4s electron gives Mn a comparatively low ΔiH2 (1509) — a dip just below Cr's spike, not a peak. …
Method: Configuration-and-Data Analysis (Table 4.2)
This method explains the irregular variation by pairing each element's electronic configurations (atom and M+ ion) with the actual printed data, instead of relying on remembered "dip" rules.
Steps
Step 1: Write the configurations that matter
ΔiH1 removes an electron from the neutral atom M; ΔiH2 removes one from the M+ ion. So write both configurations for each element (Table 4.2's M and M+ rows).
Step 2: Check the first ionisation enthalpy against the data
Table 4.2 (kJ/mol): Sc 631, Ti 656, V 650, Cr 653, Mn 717, Fe 762, Co 758, Ni 736, Cu 745, Zn 906. The rise is gentle and only mildly irregular — each added 3d electron shields the 4s electrons from the growing nuclear charge, so the effective nuclear charge climbs slowly. Note there is no clean dip at Cr or Cu: Cr (653) is slightly above V (650), and Cu (745) is slightly above Ni (736).
Step 3: Explain the second ionisation enthalpy — where the sharp irregularity lives
- Cr⁺ is 3d5 (stable, half-filled) — removing another electron breaks into it, so ΔiH2 is unusually high (1592).
- Cu⁺ is 3d10 (stable, fully filled) — breaking into it makes Cu's ΔiH2 the highest in the row (1958).
- Mn⁺ is 3d54s1 — the second electron comes from the spare 4s, leaving the stable 3d5 intact, so ΔiH2 is comparatively low (1509).
- Zn⁺ is 3d104s1 — same pattern, so Zn's ΔiH2 (1734) sits below Cu's.
Step 4: Summarise (all values from Table 4.2, kJ/mol)
| Element | ΔiH1 | ΔiH2 | Reason |
|---|---|---|---|
| Cr | 653 — no dip (just above V's 650) | 1592 — unusually high | Cr⁺ is stable 3d5; a second removal breaks it |
| Mn | 717 — above Cr | 1509 — comparatively low | Mn⁺ is 3d54s1; the spare 4s electron goes easily |
Here is a breakdown of the common mistakes students make when explaining the irregular variation of ionisation enthalpies in the first transition series, along with how to avoid each.
The Core Concept (The "Why")
The irregular variation is attributed to the varying degrees of stability of the different 3d configurations (e.g., d0, d5, d10 are exceptionally stable). Two things follow from the real data (Table 4.2):
- First ionisation enthalpy (ΔiH1): rises gently and only mildly irregularly across Sc → Zn (631 → 906 kJ/mol), because each added 3d electron partially shields the 4s electrons from the growing nuclear charge. There is no clean dip at Cr or Cu.
- Second ionisation enthalpy (ΔiH2): shows the sharp, well-defined irregularity — unusually high for Cr (1592) and Cu (1958), whose M+ ions are the stable 3d5 and 3d10; comparatively low for Mn (1509) and Zn (1734), whose M+ ions still carry one spare, easily-removed 4s electron.
Common Mistake #1: Claiming the First Ionisation Enthalpy "Dips" at Cr and Cu
The Mistake: Students reason "Cr and Cu each have only one 4s electron, and removing it gives a stable d5/d10 ion — so their ΔiH1 must dip below their neighbours'."
Why it's wrong: Table 4.2's own numbers refute it: Cr's ΔiH1 (653) is slightly higher than V's (650), and Cu's (745) is slightly higher than Ni's (736). The stable-configuration effect shows up cleanly in ΔiH2, not as first-ionisation dips.
How to Avoid: Quote the data, not the remembered rule. If asked about ΔiH1, say the rise is gentle and only mildly irregular; save the d5/d10 stability argument for ΔiH2.
Common Mistake #2: Telling the Same Story for ΔiH1 and ΔiH2
The Mistake: Students say "Cr has low ΔiH1 and also low ΔiH2" (or high for both), applying one blanket rule to both quantities.
The Correct Logic: The two quantities remove electrons from different species:
- ΔiH1 (M → M⁺) depends on the neutral atom.
- ΔiH2 (M⁺ → M²⁺) depends on the M+ ion:
- Cr⁺ (3d5) → Cr²⁺ (3d4): destroys stable d5 → high ΔiH2.
- Cu⁺ (3d10) → Cu²⁺ (3d9): destroys stable d10 → high ΔiH2.
- Mn⁺ (3d54s1) → Mn²⁺ (3d5): removes the spare 4s electron, leaving stable d5 → low ΔiH2.
- Zn⁺ (3d104s1) → Zn²⁺ (3d10): same pattern → ΔiH2 below Cu's.
How to Avoid: Write the configurations of M, M⁺, and M²⁺ for each element. Ask: "Is the electron being removed a spare 4s electron (easy), or does it break a stable d5/d10 core (hard)?"
Common Mistake #3: Writing Wrong Ground-State Configurations
The Mistake: Students write Cr as 3d44s2 or Cu as 3d94s2, and then cannot explain the irregularities.
The Correct Logic: The actual ground-state configurations are the two exceptions:
- Cr: [Ar]3d54s1 (not 3d44s2)
- Cu: [Ar]3d104s1 (not 3d94s2)
How to Avoid: Memorise the two exceptions (Cr and Cu). For every other 3d element, the general 3dn4s2 rule holds.
Common Mistake #4: Blaming Only "Nuclear Charge" or "Shielding"
The Mistake: Students say "ionisation enthalpy increases because nuclear charge increases" and stop there. That explains the general rise, not the irregularities. …
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.Identify the incorrect statement regarding the interstitial compounds (A) They have high melting points (B) They lose electrical conductivity during the formation from metal (C) They are chemically inert (D) They are very hard.
›Reveal solutionSolution
This tests properties of interstitial compounds (metal lattices with small non-metal atoms in the voids). The answer is (B): they retain, not lose, metallic conductivity.
Concept and Intuition
Interstitial compounds (e.g., TiC, TiN, Fe3C, VH0.56) form when small atoms such as H, C, N, or B fit into the interstitial (empty) spaces of a metal's crystal lattice without drastically disrupting the metallic bonding framework. Because the delocalised electron sea of the metal lattice is largely preserved, these compounds keep several metal-like characteristics: high melting point, hardness, and — crucially — metallic electrical conductivity.
Step-by-Step Solution
- (A) High melting points: interstitial compounds are known for even higher melting points than the parent metal (interstitial atoms strengthen the lattice). TRUE.
- (B) Loses electrical conductivity: since the metallic bonding/electron sea is retained, these compounds actually conduct electricity like the parent metal — they do NOT lose conductivity. FALSE — this is the incorrect statement.
- (C) Chemically inert: interstitial compounds are indeed chemically quite inert/unreactive. TRUE. …
- AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQQ.The transition metal with highest melting point is (A) Re (B) Cr (C) Mo (D) W
›Reveal solutionSolution
Tungsten (W) has the highest melting point of all the transition metals (and of all metals), around 3422°C.
Concept and Intuition
Melting points of the d-block transition metals rise toward the middle of each series (peaking around Group 6) because of strong metallic bonding reinforced by (n−1)d electron participation, then fall off toward both ends. Among all transition metals, tungsten holds the record for the highest melting point.
Step-by-Step Solution
- Compare typical high melting points: W ≈ 3422°C, Re ≈ 3186°C, Mo ≈ 2623°C, Cr ≈ 1907°C.
- Tungsten's melting point is the highest among these (and the highest of any metal), due to very strong metallic/covalent-like bonding involving its d-electrons.
Common Mistakes …
- AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQQ.Identify the correctly matched pairs i. TiO – pigment industry ii. MnO2 – dry battery cells iii. Cu/Ni alloy – UK 'copper' coins (A) i, ii, iii (B) ii, iii only (C) i, ii only (D) i, iii only
›Reveal solutionSolution
TiO2-pigment and MnO2-dry cell are standard correct facts; the Cu/Ni-"copper coins" pairing is a mismatch (Cu/Ni is used for the UK's "silver" coins, not its "copper" ones), so only i and ii are correct.
Concept and Intuition
This is a fact-recall matching question about industrially important compounds/alloys and their real-world uses.
Step-by-Step Solution
- i. TiO2 – pigment industry: True. Titanium dioxide is the most widely used white pigment (titanium white) in paints, plastics, and paper.
- ii. MnO2 – dry battery cells: True. In the Leclanché dry cell, MnO2 acts as a depolarizer, oxidizing the hydrogen gas produced at the cathode. …
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.Among V, Cr, Zn, Fe, the metal having lowest enthalpy of atomization is (A) V (B) Cr (C) Zn (D) Fe
›Reveal solutionSolution
This tests why enthalpy of atomization varies across the 3d transition series. Answer: Zn has the lowest enthalpy of atomization.
Concept and Intuition
Enthalpy of atomization reflects the strength of metallic bonding, which comes largely from the overlap of unpaired d-orbital electrons between neighbouring metal atoms (in addition to the delocalized s-electrons). Metals with more unpaired d-electrons form stronger, more extensive metallic bonds and so have higher atomization enthalpies. Zinc has the electronic configuration [Ar]3d104s2 — its d-subshell is completely filled, leaving no unpaired d-electrons to participate in interatomic bonding, so its metallic bonding is comparatively weak.
Step-by-Step Solution
- Write electron configurations: V = [Ar]3d34s2 (3 unpaired d-electrons), Cr = [Ar]3d54s1 (6 unpaired electrons total incl. 4s, exceptionally high atomization enthalpy), Fe = [Ar]3d64s2 (4 unpaired d-electrons), Zn = [Ar]3d104s2 (0 unpaired d-electrons). …
- AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQQ.Which of the following are correct? i. V2+ liberates hydrogen from a dilute acid ii. The earlier members of lanthanide series behave more like aluminium iii. The 'silver' UK coins are made of Cu/Ni alloy iv. The maximum oxidation state exhibited by Neptunium is +7 (A) i, iii only (B) ii, iv only (C) i, iii, iv only (D) i, ii, iii only
›Reveal solutionSolution
This tests recall of d- and f-block facts from NCERT: reducing power of V2+, which metal the early lanthanoids resemble, coinage alloys, and actinoid oxidation states. Three of the four statements (i, iii, iv) are correct.
Concept and Intuition
- Statement (i): A metal ion liberates H2 from a dilute acid when its reduction potential is more negative than that of the H+/H2 couple (taken as 0V). For vanadium, E∘(V3+/V2+)=−0.26V. Since this is negative, the reverse reaction (V2+→V3++e−) coupled with 2H++2e−→H2 is spontaneous — so V2+ is a strong enough reducing agent to liberate hydrogen gas from dilute acid.
- Statement (ii): Lanthanoid contraction means ionic radii shrink steadily across the series. The early members (La, Ce, Pr…) have relatively large Ln3+ radii, close in size to Ca2+ — this is exactly why rare-earth minerals substitute for calcium in nature. They do not behave like aluminium (aluminium chemistry — small, highly charge-dense Al3+ — is a different comparison used elsewhere, e.g. for beryllium/diagonal relationships). So (ii) is false as stated.
- Statement (iii): Historically 'silver' coins in the UK were sterling silver, but since 1947 they have been struck in cupro-nickel (75% Cu, 25% Ni) — a genuine transition-metal alloy fact.
- Statement (iv): Actinoids show a wider range of oxidation states than lanthanoids because 5f, 6d and 7s levels are close in energy. Np, Pu, and Am can all reach +7 (e.g. as NpO53−) under strongly oxidising alkaline conditions, though +5/+6 are more common. So Np's maximum oxidation state of +7 is correct.
Step-by-Step Solution …
- AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQQ.Assertion (A): Transition metals and their complexes show catalytic activity. Reason (R): The activation energy of a reaction is lowered by the catalyst. (A) Both (A) and (R) are correct and (R) is the correct explanation of (A). (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (C) (A) Is correct but (R) is incorrect. (D) (A) Is incorrect but (R) is correct.
›Reveal solutionSolution
The key idea is that while both statements are factually correct, the Reason (R) is a general definition of a catalyst and does not specifically explain why transition metals and their complexes are particularly good at catalysis. The correct option is (B).
Concept and Intuition (Transition Element Definition)
Transition metals (like Fe, Ni, Pt, Pd) and their complexes are famous for their catalytic activity. This is not just because they lower activation energy — all catalysts do that. The special reason lies in their unique electronic structure: they have partially filled d-orbitals, which allow them to:
- adopt multiple oxidation states,
- form temporary bonds with reactants,
- provide a surface or coordination site where reactants can come together in the right orientation.
The Reason (R) simply states the universal property of any catalyst. It is true, but it does not explain why transition metals in particular are so effective. So (R) is not the correct explanation of (A).
Step-by-step reasoning:
-
Check Assertion (A):
Transition metals and their complexes are indeed widely used as catalysts — e.g., iron in the Haber process, platinum in catalytic converters, nickel in hydrogenation. This is a well-known fact.
→ So (A) is correct.
-
Check Reason (R):
A catalyst, by definition, lowers the activation energy of a reaction, thereby increasing the rate without being consumed. This is a fundamental principle of catalysis.
→ So (R) is also correct.
-
Determine if (R) explains (A): …
- AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQQ.Which of the following elements are not regarded as transition elements? (A) Zn, Cd, Hg (B) Cu, Zn, Hg (C) Ag, Zn, Hg (D) Ag, Cd, Hg
›Reveal solutionSolution
Group 12 elements (Zn, Cd, Hg) have a fully filled d10 configuration and so fail the IUPAC definition of a transition element.
Concept and Intuition
IUPAC defines a transition element as one whose atom (in the ground state) or common ion has an incompletely filled d-subshell. Zinc, cadmium and mercury all have the configuration (n−1)d10ns2 and lose only the ns2 electrons to form M2+, which is still d10 — no partially filled d-orbital ever appears, so they are excluded from the transition series even though they sit in the d-block.
Step-by-Step Solution
- Write electron configurations: Zn = [Ar]3d104s2; Cd = [Kr]4d105s2; Hg = [Xe]4f145d106s2.
- In each case the d-subshell is completely filled (d10), both in the atom and in the common M2+ ion. …
- AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQQ.Assertion (A): Transition elements have higher enthalpies of atomization. Reason (R): Large number of unpaired electrons present in transition elements facilitate strong interatomic interaction and strong bonding between atoms. (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct and (R) is not the correct explanation of (A). (C) (A) Is correct and (R) is incorrect. (D) (A) Is incorrect and (R) is correct.
›Reveal solutionSolution
Both statements are true, and the reason genuinely explains the assertion: transition metals have high atomization enthalpies precisely because their unpaired d-electrons enable extra interatomic (covalent-like) bonding on top of ordinary metallic bonding. Answer: (A).
Concept and Intuition
Enthalpy of atomization measures the energy needed to convert one mole of metal atoms in the solid state into gaseous atoms — essentially, the strength of the metallic bonding holding the solid lattice together. Transition metals show unusually high atomization enthalpies compared to their neighbouring s- and p-block metals. NCERT explains this by noting that in transition metals, in addition to the delocalized valence-electron ('electron sea') metallic bonding common to all metals, the partially filled (n-1)d orbitals allow additional localized, covalent-like overlap between neighbouring atoms' d-orbitals. The greater the number of unpaired d-electrons available for this extra overlap, the stronger the overall interatomic bonding — which is exactly why atomization enthalpies of transition metals peak somewhere in the middle of each series (where the number of unpaired d-electrons is often highest) and are generally much larger than for s-/p-block metals.
Step-by-Step Solution
- Check Assertion (A): transition elements have higher enthalpies of atomization — this is a well-established, textbook-supported fact (compare, e.g., atomization enthalpies of 3d transition metals to those of Ca, K, or Ga/Ge). True.
- Check Reason (R): a large number of unpaired electrons facilitate strong interatomic interaction and strong bonding between atoms — also a textbook-supported mechanistic explanation. True. …
- AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQQ.The general trend of enthalpies of atomisation of d-block elements is ______ (A) Series-1 > Series-2 > Series-3 (B) Series-1 > Series-3 > Series-2 (C) Series-3 > Series-2 > Series-1 (D) Series-2 > Series-1 > Series-2
›Reveal solutionSolution
This tests the periodic trend in enthalpies of atomisation across the three transition series; the answer is Series-3 (5d) > Series-2 (4d) > Series-1 (3d).
Concept and Intuition
Enthalpy of atomisation measures the energy needed to convert one mole of metal atoms in the solid (metallic) state into gaseous atoms — essentially a measure of the strength of metallic bonding. In transition metals, metallic bonding strength depends on the number of unpaired d electrons and how well the d-orbitals overlap between neighbouring atoms.
Step-by-Step Solution
- Across a transition series, atomisation enthalpy is influenced by the number of unpaired electrons — it rises to a maximum near the middle of the series (where the number of unpaired electrons is highest) and falls off toward both ends.
- Comparing the same group across the three transition series (3d, 4d, 5d), the outer d-orbitals become progressively larger and more diffuse — 5d orbitals overlap more effectively with neighbouring atoms' orbitals than 4d, which in turn overlap better than 3d. …
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