First-row transition elements (Sc to Zn) show systematic trends in their properties — atomic radii, ionisation enthalpies, melting points, and electrode potentials — that NEET questions test as pattern-recognition items.
The trend you need cold: Atomic radii decrease from Sc to Cr (increasing nuclear charge, electrons added to the same 3d subshell), then remain roughly constant from Cr to Cu (electron–electron repulsion in the progressively filled 3d orbitals balances nuclear charge increase), and finally increase at Zn (paired 3d¹⁰ offers no effective d–d shielding advantage, but loss of exchange energy and fully paired configuration weakens effective nuclear pull on 4s electrons). NCERT Class 12 Chemistry Chapter 4, page 96 states this trend explicitly.
Ionisation enthalpy (IE₁): Generally increases left to right but NOT monotonically. Cr and Cu show lower-than-expected IE₁ due to the extra stability of half-filled (3d⁵4s¹) and fully-filled (3d¹⁰4s¹) configurations respectively. Mn shows higher IE₁ than expected because removing an electron disrupts its stable 3d⁵ half-filled set.
Melting points: High across the row (metallic bonding involving unpaired d-electrons). Maximum near the middle (Cr, V) where unpaired electrons are most numerous. Zn has anomalously low melting point — all d-electrons are paired, contributing nothing to metallic bonding.
Standard electrode potential (E°): The trend across the row is irregular. Cu is the only first-row transition metal with a positive E° (Cu²⁺/Cu = +0.34 V), explained by its high atomisation enthalpy, high ionisation enthalpy, and low hydration enthalpy combined.
Watch-out for NEET: Questions often ask "which property does NOT show a regular trend?" The answer is almost always electrode potential — it depends on the combined effect of atomisation, ionisation, and hydration enthalpies, not a single factor.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The atomic radii of first-row transition metals generally decrease from Sc to Cr and then remain nearly constant up to Cu. The near-constancy is primarily due to:
Answer: B. As electrons are added to the same 3d subshell from Cr to Cu, increasing electron–electron repulsion partially counterbalances the increasing nuclear charge, resulting in nearly constant radii (NCERT Class 12 Chemistry Chapter 4, page 96).
Why A is wrong: A is wrong because 3d electrons are poor shielders — the balance is not exact cancellation of nuclear charge by shielding, but rather repulsion effects in the d-orbitals offsetting contraction.
Why C is wrong: C is wrong because the 4s orbital contains 1 or 2 electrons throughout; there is no 'completion of 4s filling after Cr' — Cr itself has 4s¹.
Why D is wrong: D is wrong because lanthanoid contraction applies to 5d elements (post-lanthanoid), not to the 3d first-row series.
Among the first-row transition metals, Zn has an anomalously low melting point compared to its neighbours. This is because:
Answer: A. Metallic bond strength in transition metals correlates with the number of unpaired d-electrons available for bonding. Zn (3d¹⁰4s²) has zero unpaired d-electrons, so its metallic bonding is weaker, giving a low melting point (NCERT Class 12 Chemistry Chapter 4, page 96).
Why B is wrong: B is wrong because Zn does not have the largest atomic radius — its radius increases slightly from Cu but the size difference does not explain the large melting-point drop.
Why C is wrong: C is wrong because the +2 oxidation state alone doesn't determine melting point; metallic bonding in the elemental state depends on unpaired electrons, not ionic oxidation states.
Why D is wrong: D is wrong because Zn's 4s electrons are in fact held MORE tightly than its neighbours' (Zn has the highest first ionisation enthalpy in the series, 906 kJ/mol), yet that is not what lowers its melting point; the absence of unpaired 3d electrons for metallic bonding is.
Which of the following first-row transition metals has the highest first ionisation enthalpy?
Answer: A. Zn has the highest IE₁ among first-row transition metals due to its completely filled 3d¹⁰4s² configuration, which provides maximum exchange energy stabilisation and makes electron removal energetically costly (NCERT Class 12 Chemistry Chapter 4, page 96).
Why B is wrong: B is wrong because although Mn shows a local peak (disrupting 3d⁵ costs energy), its IE₁ is still lower than Zn's fully-filled configuration.
Why C is wrong: C is wrong because Fe (3d⁶4s²) has one extra electron beyond the stable half-filled 3d⁵, making the 3d⁶ less stable than 3d¹⁰ — its IE₁ is lower than Zn's.
Why D is wrong: D is wrong because Cr (3d⁵4s¹) has a relatively lower IE₁ — the half-filled stability helps but the single 4s electron is comparatively easy to remove.
Cu is the only first-row transition metal with a positive standard reduction potential (E° for M²⁺/M). This is attributed to:
Answer: B. E° depends on the sum of sublimation (atomisation), ionisation, and hydration enthalpy. For Cu, the high energy cost of atomisation and ionisation is not sufficiently compensated by hydration enthalpy, making the overall process thermodynamically unfavourable — hence positive E° (NCERT Class 12 Chemistry Chapter 4, page 96).
Why A is wrong: A is wrong because Cu actually has HIGH ionisation enthalpy (not low) — if IE were low, E° would be negative (metal would dissolve readily).
Why C is wrong: C is wrong because Cu²⁺ is d⁹, not d¹⁰. Cu⁺ is d¹⁰, but the E° for M²⁺/M refers to Cu²⁺, which has no special closed-shell stability.
Why D is wrong: D is wrong because alloy formation is a bulk metallurgical property irrelevant to the thermodynamic E° value, which is defined for the pure element.
The third ionisation enthalpy of Mn is higher than that of its immediate neighbours (Cr and Fe). The best explanation is:
Answer: C. Mn is 3d⁵4s², so Mn²⁺ is 3d⁵. The third ionisation must remove an electron from this half-filled subshell, which has extra exchange-energy stabilisation, so Mn's third ionisation enthalpy (3260 kJ/mol) is higher than Cr's (2990) and Fe's (2962) (values from Table 4.2; NCERT Class 12 Chemistry Chapter 4, page 96, notes that the third ionisation enthalpy of Fe is lower than that of Mn). Mn's first ionisation enthalpy (717) is not higher than Fe's (762), so the question is about the third.
Why A is wrong: A is wrong because nuclear charge increases continuously (Cr=24, Mn=25, Fe=26) — if nuclear charge alone determined it, Fe would have the highest third ionisation enthalpy of the three, but Fe's (2962 kJ/mol) is lower than Mn's (3260).
Why B is wrong: B is wrong because both 4s electrons are gone before the third ionisation (Mn²⁺ is 3d⁵), so 4s occupancy cannot explain it; Fe also has 4s², yet its third ionisation enthalpy is lower than Mn's. The explanation lies in 3d⁵ stability.
Why D is wrong: D is wrong because the highest oxidation state in compounds doesn't directly determine atomic ionisation enthalpy — the third ionisation enthalpy concerns removing one electron from Mn²⁺, not reaching +7.
Among the first-row transition metals, melting points generally rise towards the middle of the series and then fall. One element in the middle has an anomalously LOW melting point. That element is:
Answer: C. C is correct. Manganese sits in the middle of the series, where melting points should be near their maximum, yet it melts well below both of its neighbours, chromium and iron. Its 3d⁵ 4s² configuration has an exactly half-filled d subshell, which is especially stable, so its d-electrons take less part in metallic bonding than the trend would suggest (NCERT Class 12 Chemistry Chapter 4, page 92).
Why A is wrong: A is wrong because vanadium fits the rising part of the trend: with three unpaired d-electrons taking part in metallic bonding, its melting point is high, as expected for its position.
Why B is wrong: B is wrong because chromium has one of the HIGHEST melting points in the row. Its 3d⁵ 4s¹ configuration gives six unpaired electrons, the most available for metallic bonding.
Why D is wrong: D is wrong because iron sits on the falling side of the trend, after the middle, and its melting point fits that position. It is lower than chromium's but not anomalously so.
The standard electrode potentials (E° for M²⁺/M) of first-row transition metals do NOT show a regular trend across the series. The primary reason is that E° depends on:
Answer: D. E° is determined by the thermodynamic cycle involving sublimation (atomisation) of solid metal, ionisation of gaseous atom, and hydration of the gaseous ion. Since these three quantities vary irregularly and independently across the series, their sum (which determines E°) shows no simple trend (NCERT Class 12 Chemistry Chapter 4, page 96).
Why A is wrong: A is wrong because ionisation enthalpy is only one of three contributing factors — atomisation and hydration enthalpies also vary across the series and affect E° significantly.
Why B is wrong: B is wrong because hydration enthalpy alone cannot determine E°; the energy required to first get the ion into the gas phase (atomisation + ionisation) is equally important.
Why C is wrong: C is wrong because the number of accessible oxidation states is a consequence of electronic structure but does not directly determine the M²⁺/M electrode potential, which is a specific thermodynamic quantity.
The first ionisation enthalpies of the 3d series, in kJ mol⁻¹, are Sc 631, Ti 656, V 650, Cr 653, Mn 717, Fe 762, Co 758, Ni 736, Cu 745, Zn 906. A student claims that IE₁ increases uniformly from Sc to Zn. Which observation contradicts this claim?
Answer: D. D is correct. Read the row and look for any step that goes DOWN. There are three: Ti 656 → V 650, Fe 762 → Co 758, and Co 758 → Ni 736. Each is a decrease where the claim demands an increase, so each one alone refutes "increases uniformly", and both are offered. The overall trend across the series is upward — as nuclear charge grows the 4s electron is held more tightly — but it is not monotonic, because the energy needed also depends on the particular 3d/4s configuration being ionised (NCERT Class 12 Chemistry Chapter 4, Table 4.2, page 94).
Why A is wrong: A is true but incomplete. Ti 656 → V 650 is a genuine decrease and does contradict the claim, but it is not the only one in the data given, so D is the better answer.
Why B is wrong: B is true but incomplete. Co 758 → Ni 736 is a genuine decrease and does contradict the claim, but Ti → V is another, so D is the better answer.
Why C is wrong: C is wrong because Zn having the highest value is exactly what a uniform increase would predict. A statement that AGREES with the claim cannot contradict it — the contradiction has to be a step that goes down, and the steps that go down are Ti → V, Fe → Co and Co → Ni.
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Given
First-row transition elements Fe (Z=26), Co (Z=27), Ni (Z=28), Cu (Z=29), Zn (Z=30). Approximate atomic radii (pm): Fe ≈ 126, Co ≈ 125, Ni ≈ 124, Cu ≈ 128, Zn ≈ 138.
Required
Arrange in increasing order of atomic radii and explain the anomalies at Cu and Zn.
Concept
Across the first-row transition series, increasing nuclear charge contracts the atom, but poor shielding by 3d electrons and increasing electron–electron repulsion partially offset this. At Zn (3d¹⁰4s²), the fully-filled d-orbitals and loss of exchange stabilisation lead to an increase in size.
Formula/Principle
No quantitative formula — this is a qualitative trend question based on effective nuclear charge and d-electron shielding arguments.
Substitution/Application
• Ni (Z=28, 3d⁸): smallest in this set — high Z_eff, incomplete d-shell.• Co (Z=27, 3d⁷): slightly larger than Ni.• Fe (Z=26, 3d⁶): slightly larger than Co.• Cu (Z=29, 3d¹⁰4s¹): increases from Ni — the filled 3d¹⁰ shell provides better inter-electron repulsion; also 4s¹ configuration shifts electron density outward.• Zn (Z=30, 3d¹⁰4s²): largest — fully filled d-shell, 4s² fills the outermost orbital.
Calculation
Order of increasing atomic radii: Ni < Co < Fe < Cu < Zn (124 < 125 < 126 < 128 < 138 pm).
Final answer
Ni < Co < Fe < Cu < Zn. The expected monotonic decrease breaks at Cu and Zn because Cu achieves the stable 3d¹⁰ configuration (reducing effective nuclear pull on 4s) and Zn's fully filled 3d¹⁰4s² has no d-electron participation in size contraction.
Common trap
Students often expect Zn to be similar in size to Cu (both have 3d¹⁰). The key difference: Cu is 3d¹⁰4s¹ while Zn is 3d¹⁰4s². The extra 4s electron in Zn adds significant shielding and size.
Similar NEET-style question
"Among Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺, and Zn²⁺, which ion has the largest ionic radius?" (Answer: Zn²⁺ does NOT follow the same pattern as atoms — for ions, the d-electron count and charge matter. This is a follow-up that tests whether you can transfer the atomic trend reasoning to ions.)
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Atomic radii: decrease then increase (lanthanide-like contraction). Density: increases. Melting point: high (Cr, Mn anomalies due to half-filled d⁵). Ionisation enthalpy: gradual increase across series.
-- NCERT Class 12 Chemistry, Ch. 4, p. 94More in The d and f Block Elements: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
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