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 8 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:
Among the first-row transition metals, Zn has an anomalously low melting point compared to its neighbours. This is because:
Which of the following first-row transition metals has the highest first ionisation enthalpy?
Cu is the only first-row transition metal with a positive standard reduction potential (E° for M²⁺/M). This is attributed to:
The ionisation enthalpy of Mn is higher than that of its immediate neighbours (Cr and Fe). The best explanation is:
Among first-row transition metals, the melting points generally increase up to the middle and then decrease. However, one element shows an anomalously low melting point despite being in the middle. That element is:
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:
Consider the first ionisation enthalpies of Cr (652 kJ/mol), Mn (717 kJ/mol), and Fe (762 kJ/mol). A student claims that IE₁ increases uniformly from Sc to Zn. Which observation contradicts this claim?
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.) ---
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. 8Magnetic moment from n unpaired electrons. 1 unpaired: 1.73 BM; 5: 5.92 BM.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | Bohr magneton |
Predicts paramagnetic moment of d-block ion. n unpaired electrons in d-orbitals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | BM |
Catalogues common stable oxidation states across first-row transition metals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| OS | oxidation state | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
Question gives KMnO4 oxidation in unspecified or specific medium.
Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.
Root cause: concept gap
Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
Root cause: concept gap
Lanthanoid contraction (imperfect 4f shielding) makes 5d elements similar in size to 4d. Zr/Hf, Nb/Ta, Mo/W have nearly identical chemistry.
10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
The correct order of decreasing basic strength of the given amines is:
The pair of lanthanoid ions which are diamagnetic is
Which one of the following statements is correct?
Gadolinium has a low value of third ionisation enthalpy because of
Tritium, a radioactive isotope of hydrogen, emits which of the following particles? (β–)
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
uses acidic formula in basic medium
Assumes Mn²⁺ regardless of medium
predicts large difference due to period shift
Expects 5d to be much larger than 4d
misses d0 d10 stability
Doesn't account for closed-shell stability
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