Write the general electronic configuration of the first-row (3d) transition elements, and give the actual conf

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Q. Write the general electronic configuration of the first-row (3d) transition elements, and give the actual configurations of chromium and copper, explaining why they deviate from the general pattern.

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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.

Every first-row transition metal atom, built on the $[\text{Ar}]$ core, has the general outer configuration $3d^{1-10}4s^{1-2}$ : scandium is $3d^14s^2$ , titanium $3d^24s^2$ , vanadium $3d^34s^2$ , and so on up to zinc, $3d^{10}4s^2$ .

Two members break from the pattern a simple, orbital-by-orbital filling would predict. A straightforward extension of the pattern would give chromium the configuration $3d^44s^2$ and copper the configuration $3d^94s^2$ . Instead, chromium's actual ground-state configuration is $[\text{Ar}]3d^54s^1$ , and copper's is $[\text{Ar}]3d^{10}4s^1$ : in both cases, one electron that "should" occupy $4s$ is instead found in $3d$ .

The reason is the extra stability associated with a half-filled ( $d^5$ ) or completely filled ( $d^{10}$ ) subshell. When five electrons occupy five different $d$ orbitals with parallel spin (as in $d^5$ ), or when all five orbitals are doubly occupied (as in $d^{10}$ ), the electron-electron repulsion within the subshell is minimized, and the quantum-mechanical exchange stabilization between same-spin electrons in different orbitals is maximized. This stabilization is large enough that the atom "prefers" to promote one electron out of $4s$ and into $3d$ to reach one of these two special configurations, even though doing so leaves the $4s$ orbital singly (rather than doubly) occupied.

✓Final answer

3d^{1-10}4s^{1-2}

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.Write the general electronic configuration of the first-row (3d) transition elements, and give the actual configurations of chromium and copper, explaining why they deviate from the general pattern.

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