Why Cr is [Ar]3d⁵4s¹ and Cu is [Ar]3d¹⁰4s¹ — exceptions to Aufbau

Question

According to the Aufbau principle, the expected electronic configurations of Cr (Z=24) and Cu (Z=29) are $[\text{Ar}]3d^4 4s^2$ and $[\text{Ar}]3d^9 4s^2$ respectively. But the actual configurations are $[\text{Ar}]3d^5 4s^1$ and $[\text{Ar}]3d^{10} 4s^1$ . Explain why.

(NCERT Class 12, d- and f-Block Elements)

Solution — Step by Step

Element	Expected (Aufbau)	Actual
Cr (Z=24)	$[\text{Ar}]3d^4 4s^2$	$[\text{Ar}]3d^5 4s^1$
Cu (Z=29)	$[\text{Ar}]3d^9 4s^2$	$[\text{Ar}]3d^{10} 4s^1$

In both cases, one electron “shifts” from the 4s orbital to the 3d orbital.

The key reason is the extra stability associated with:

Half-filled d subshell ( $d^5$ ): all 5 d orbitals singly occupied — maximum exchange energy
Fully-filled d subshell ( $d^{10}$ ): all 5 d orbitals doubly occupied — symmetric electron distribution

This extra stabilisation energy exceeds the small energy cost of moving one electron from 4s to 3d.

When electrons with parallel spins occupy different orbitals of the same subshell, each pair contributes an exchange energy that stabilises the atom. For $n$ electrons with parallel spins, the number of exchange pairs is $\binom{n}{2} = n(n-1)/2$ .

$3d^4$ : 4 parallel-spin electrons → $\binom{4}{2} = 6$ exchange pairs
$3d^5$ : 5 parallel-spin electrons → $\binom{5}{2} = 10$ exchange pairs

That’s 4 extra exchange pairs — a significant stabilisation that makes $3d^5 4s^1$ more stable than $3d^4 4s^2$ .

$3d^9 4s^2$ : 5 spin-up + 4 spin-down in 3d → exchange pairs from each spin set
$3d^{10} 4s^1$ : Completely filled d subshell with perfect spherical symmetry

The fully-filled $d^{10}$ configuration has maximum exchange energy for both spin-up and spin-down sets ( $\binom{5}{2} = 10$ pairs each = 20 total), plus the additional stability from symmetrical charge distribution. This outweighs having a paired 4s.

Why This Works

The Aufbau principle is based on the approximate order of orbital energies for hydrogen-like atoms. In multi-electron atoms, inter-electronic repulsion and exchange interactions modify this order. The 3d and 4s orbitals are very close in energy — the energy gap is small enough that the exchange energy gain from half-filling or fully-filling the d subshell can tip the balance.

Think of it this way: the Aufbau principle gives you the default filling order, but nature picks whatever configuration has the lowest total energy. For most elements, Aufbau works perfectly. For Cr and Cu (and a few others like Mo, Ag), the exchange energy correction matters.

Alternative Method

A simpler way to remember: the d subshell is most stable when it’s either half-full ( $d^5$ ) or completely full ( $d^{10}$ ). If an element is just one electron away from these configurations AND the electron can come from the nearby 4s orbital, it will.

For NEET and CBSE, you need to remember only two exceptions in the 3d series: Cr and Cu. Both lose one 4s electron to achieve $d^5$ or $d^{10}$ . For JEE, also know that similar exceptions occur in 4d (Mo, Ag) and 5d series.

Common Mistake

Students often extend this logic incorrectly and predict that all elements near half-filling show anomalies. For example, they might expect V (Z=23) to be $3d^4 4s^1$ instead of $3d^3 4s^2$ . But vanadium follows Aufbau normally — the exchange energy gain from going $d^3 \rightarrow d^4$ (6 → 6 pairs for one spin set) is not enough to overcome the 4s pairing advantage. The exception only works when you’re exactly one electron away from $d^5$ or $d^{10}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next