Sc (Z=21) is a transition element but Zinc (Z=30) is not because a) both Sc³⁺ and Zn²⁺ ions are colourless and

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Q. Sc (Z=21) is a transition element but Zinc (Z=30) is not because

a) both Sc³⁺ and Zn²⁺ ions are colourless and form white compounds.

b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

c) last electron as assumed to be added to 4s level in case of zinc

d) both Sc and Zn do not exhibit variable oxidation states

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Concept understanding — Position and Electronic Configuration of Transition Elements

By the IUPAC definition, a transition metal is an element whose atom has an incomplete d sub-shell, or which can give rise to a cation with an incomplete d sub-shell. Transition metals occupy the central block of the periodic table (groups 3-12), forming a bridge between the highly reactive s-block metals and the mostly non-metallic p-block elements -- their properties are genuinely transitional between the two. Except for the group-11 metals (Cu, Ag, Au), transition metals are hard solids with very high melting points, and they are good conductors of heat and electricity.

There are three complete transition series: the 3d series (Sc to Zn, period 4, 10 elements), the 4d series (Y to Cd, period 5, 10 elements), and the 5d series (La, then Hf to Hg, period 6, 10 elements) -- plus a partial, entirely radioactive 7th-period series (Ac, Rf to Cn) whose members (bar actinium) are all synthetic with very short half-lives. Zn, Cd and Hg (group 12) have a completely filled d¹⁰ configuration in both their elemental and normal +2 states, so strictly they fail the IUPAC test -- but they are conventionally still classed as transition metals because their properties are a direct extension of their transition-metal neighbours.

Following the Aufbau principle, 4s fills before 3d, so 3d filling genuinely begins only at scandium, [Ar]3d¹4s², and completes at zinc, [Ar]3d¹⁰4s². Chromium and copper are the two well-known exceptions: Cr is [Ar]3d⁵4s¹ (not the 'expected' 3d⁴4s²) and Cu is [Ar]3d¹⁰4s¹ (not 3d⁹4s²) -- in both, one 4s electron shifts into 3d to reach the extra-stable half-filled (d⁵) or fully-filled (d¹⁰) configuration, a stability rooted in the symmetrical, exchange-energy-favoured distribution of electrons across the five d orbitals. Accounting for these exceptions, the general configuration of the d-block is Noble gasd¹⁻¹⁰ns¹⁻², n = 4 to 7; in periods 6 and 7 (except La/Ac) this extends to include a filled (n-2)f¹⁴ term, foreshadowing the f-block.

Step 1. Recall the IUPAC definition: a transition metal is an element whose atom has an incomplete d sub-shell, or which can give rise to a cation with an incomplete d sub-shell.

Step 2. Write out the configurations: Sc (Z=21) is [Ar]3d¹4s² -- its 3d sub-shell holds only 1 of the 10 possible electrons, so it is genuinely INCOMPLETE. Zn (Z=30) is [Ar]3d¹⁰4s² -- its 3d sub-shell holds all 10 electrons, so it is COMPLETE, both in the neutral atom and in its normal Zn²⁺ cation ([Ar]3d¹⁰).

Step 3. Check the other options against this fact: (a) is wrong because Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) are indeed both colourless, but that is a consequence of their d-configurations, not the reason Sc is transitional and Zn is not. (c) is a true statement about the filling ORDER (4s fills before 3d) but does not itself explain why zinc fails the transition-metal test. (d) is false -- Sc genuinely shows only one oxidation state (+3), so 'both do not exhibit variable oxidation states' is not the distinguishing reason either.

Step 4. The only statement that correctly identifies the actual IUPAC criterion being applied is (b): Sc's 3d sub-shell is partially filled, Zn's is completely filled.

✓Final answer

(b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled

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

CBSE 2025Set ANNUAL1 markMCQ
$Q. In Ni(CO)4 electronic configuration and oxidation state of Ni is$
(a) [Ar]3d⁸ 4s², 0
(b) [Ar]3d⁸, +2
(c) [Ar]3d¹⁰, 0
(d) [Ar]3d⁸ 4s², +2
›Reveal solutionSolution

$Ni(CO)_4$ is a zero-oxidation-state complex; Ni's $4s^2$ electrons rearrange into the 3d subshell, giving $[Ar]3d^{10}4s^0$ .

CO is a neutral ligand (it donates a lone pair from carbon but carries no charge), so it does not oxidise the metal. Since all four ligands in $Ni(CO)_4$ are neutral CO molecules and the overall complex is neutral, the oxidation state of Ni must be 0.

A free, ground-state Ni atom has the configuration $[Ar]3d^8 4s^2$ . In the zero-oxidation-state complex, the 4s electrons are promoted/rearranged into the 3d subshell to give a fully-filled, more symmetric $3d^{10}4s^0$ configuration, which then undergoes $sp^3$ hybridisation (using the empty 4s and three 4p orbitals) to bond tetrahedrally with the four CO ligands.

✓Final answer
$[Ar]3d^{10}$
CBSE 2018Set ANNUAL1 markMCQ
$Q. Which of the following ion will give colourless aqueous solution ?$
(a) $Ni^{2+}$
(b) $Sc^{3+}$
(c) $Cu^{2+}$
(d) $Mn^{3+}$
›Reveal solutionSolution

Colour in transition-metal ions arises from d-d electronic transitions, which require partially-filled d-orbitals ( $d^1$ to $d^9$ ). An ion with $d^0$ (or $d^{10}$ ) configuration cannot undergo a d-d transition and is therefore colourless.

Scandium ( $Z=21$ ) has configuration $[Ar]3d^1 4s^2$ ; losing 3 electrons to form $Sc^{3+}$ removes both $4s$ electrons and the single $3d$ electron, leaving $[Ar]3d^0$ — no d-electrons at all, so $Sc^{3+}$ is colourless (this is why it is sometimes excluded from the 'true' transition series). $Ni^{2+}$ is $3d^8$ (green), $Cu^{2+}$ is $3d^9$ (blue), and $Mn^{3+}$ is $3d^4$ (violet/red) — all have partially-filled d-shells and show characteristic colours from d-d transitions.

✓Final answer
$Sc^{3+}$

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Q.Sc (Z=21) is a transition element but Zinc (Z=30) is not because

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