Electron configuration exceptions — Cr, Cu, and why half/fully-filled orbitals are stable

Question

Why do chromium and copper have unexpected electron configurations? Explain the extra stability of half-filled and fully-filled d-orbitals.

Solution — Step by Step

Using the Aufbau principle straightforwardly:

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

In both cases, one electron from the $4s$ orbital “shifts” to the $3d$ orbital. Cr achieves a half-filled $3d^5$ configuration, and Cu achieves a fully-filled $3d^{10}$ configuration.

Two factors contribute:

1. Symmetrical distribution of electrons: In a half-filled configuration ( $d^5$ ), each of the five d-orbitals has exactly one electron. In a fully-filled configuration ( $d^{10}$ ), each has exactly two. This symmetry lowers the overall energy.

2. Maximum exchange energy: Electrons with the same spin in different orbitals can “exchange” positions, releasing energy. The number of exchange pairs is maximised when orbitals are half-filled or fully-filled.

For $d^5$ : exchange pairs = $\binom{5}{2} = 10$ (all 5 electrons have parallel spin)

For $d^4$ : exchange pairs = $\binom{4}{2} = 6$

The jump from 6 to 10 exchange pairs is significant — the extra exchange energy more than compensates for the energy cost of promoting one electron from $4s$ to $3d$ .

In the first transition series, the energy gap between $3d$ and $4s$ is very small. So the energy cost of moving one electron from $4s$ to $3d$ is minimal. The exchange energy gained ( $\approx$ extra stabilisation) outweighs this small promotion energy.

In elements where the $3d$ - $4s$ gap is larger (like V, Mn), the promotion is not worth it, so no exception occurs. The exception only happens at the critical points: $d^4 \to d^5$ (Cr) and $d^9 \to d^{10}$ (Cu).

graph TD
    A[Aufbau filling order] --> B{Near half-filled or fully-filled?}
    B -->|d4 to d5 possible| C["Cr: 3d4 4s2 → 3d5 4s1"]
    B -->|d9 to d10 possible| D["Cu: 3d9 4s2 → 3d10 4s1"]
    B -->|No| E[Normal configuration]
    C --> F[Extra exchange energy gained]
    D --> F
    F --> G[Symmetrical distribution + stability]

Why This Works

Exchange energy is a quantum mechanical effect — electrons with parallel spin repel each other less (due to exchange interaction). More parallel-spin electrons means more exchange pairs, which means lower energy.

The formula for exchange pairs with $n$ parallel electrons is $\frac{n(n-1)}{2}$ . The difference between $d^4$ (6 pairs) and $d^5$ (10 pairs) is 4 extra exchange pairs — this is the largest single-electron jump in exchange energy across the entire d-block.

Similar exceptions occur in the second transition series: Mo ( $[\text{Kr}] 4d^5 5s^1$ ) and Ag ( $[\text{Kr}] 4d^{10} 5s^1$ ). The third series shows Pd ( $[\text{Xe}] 4d^{10} 5s^0$ ) — even more extreme, with 4d taking both electrons.

Alternative Method

For exam questions that ask “write the electronic configuration of Cr/Cu,” just remember the two exceptions and the reason in one line:

Cr and Cu promote one 4s electron to 3d to achieve half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) stability due to extra exchange energy.

Other d-block elements with similar exceptions (Mo, Ag, Pd, Pt, Au) follow the same logic.

Common Mistake

Students often extend this logic and write wrong configurations for other elements. For example, they write Fe (Z=26) as $[\text{Ar}] 3d^7 4s^1$ to “get closer to half-filled.” This is wrong — Fe has the normal $[\text{Ar}] 3d^6 4s^2$ configuration. The promotion only happens when it achieves exactly half-filled or fully-filled d-orbitals. Anything in between does not provide enough exchange energy to justify the promotion.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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