Electronic Configurations of d-Block Elements

Where the d-block sits. The d-block occupies Groups 3 to 12, wedged between the s-block and the p-block. An atom belongs to the d-block when its highest-energy electron, added last, enters an (n−1)d subshell rather than the outer ns or np subshell — even though the ns orbital was filled first, in order of increasing energy.

The general valence configuration for any d-block atom is written as

where: $n$ is the outermost shell's principal quantum number, the exponent on $d$ ranges from 1 to 10 depending on how many d electrons the element carries, and the exponent on $s$ is usually 2 (occasionally 1 or, for the d-block ions, 0). For the first transition series this becomes the concrete, exam-ready form built on the argon core:

running from scandium (Z = 21) to zinc (Z = 30).

The two Aufbau exceptions. A half-filled ($d^5$) or fully-filled ($d^{10}$) d-subshell carries extra stability because it maximises exchange energy — the stabilising effect that arises when several electrons of parallel spin, all in orbitals of identical energy, can swap positions. Two elements in the 3d series adopt this more-stable arrangement instead of the naive Aufbau prediction:

• Chromium (Z = 24): expected [Ar]$3d^4 4s^2$, actual [Ar]$3d^5 4s^1$.
• Copper (Z = 29): expected [Ar]$3d^9 4s^2$, actual [Ar]$3d^{10} 4s^1$.

The figure below lines up the full 3d series so the two anomalies are visible against the regular pattern.

Forming ions. When a transition-metal atom loses electrons to become a cation, the ns electrons are removed first, not the (n−1)d electrons — even though d filled after s during atom-building. Once an atom is ionised, the (n−1)d orbital drops below ns in energy, so the s electrons are the ones lost. For example, Fe ([Ar]$3d^6 4s^2$) loses both 4s electrons to give Fe$^{2+}$ = [Ar]$3d^6$, and loses one further d electron to give Fe$^{3+}$ = [Ar]$3d^5$. Zinc ([Ar]$3d^{10}4s^2$) and its ion Zn$^{2+}$ (= [Ar]$3d^{10}$) always carry a completely filled d-subshell, which is why zinc is usually excluded from "typical" transition-element behaviour.

Explain why chromium (Z = 24) has the configuration [Ar]$3d^5 4s^1$ instead of the Aufbau-predicted [Ar]$3d^4 4s^2$, and give the configuration of Cr$^{3+}$.

Step 1. Apply the Aufbau principle naively: filling 24 electrons in order, following [Ar]$4s^2$, gives [Ar]$3d^4 4s^2$ — 4 electrons among five 3d orbitals plus 2 paired in 4s.

Step 2. Compare stability. A configuration of $3d^5 4s^1$ places one electron in each of the five 3d orbitals (a half-filled d-subshell) with a single 4s electron, instead of leaving one 3d orbital empty (as in $3d^4$) while 4s is fully paired.

Step 3. Apply the stability rule: half-filled ($d^5$) and fully-filled ($d^{10}$) subshells carry extra stability from symmetrical charge distribution and, more importantly, larger exchange energy, since more electron pairs of parallel spin and identical energy can exchange positions. $3d^5 4s^1$ has more such exchangeable pairs among its five unpaired d electrons than $3d^4 4s^2$ has among only four.

Step 4. Conclude the actual ground-state configuration is $\text{Cr} = [\text{Ar}]\,3d^5\,4s^1$, an experimentally observed exception to strict Aufbau ordering.

Step 5. For Cr$^{3+}$, remove three electrons — the single 4s electron first, then two from 3d: $\text{Cr}^{3+} = [\text{Ar}]\,3d^3$. Answer: Cr = [Ar]$3d^5 4s^1$, and Cr$^{3+}$ = [Ar]$3d^3$, with three unpaired electrons in three separate d orbitals, which is why d$^3$ is chromium's most stable and common oxidation state.

Competitive stretch

Write the configuration of Mn$^{2+}$ and state its number of unpaired electrons. Mn (Z = 25) follows the naive count with no exception, giving [Ar]$3d^5 4s^2$. Removing the two 4s electrons first leaves $\text{Mn}^{2+} = [\text{Ar}]\,3d^5$, with all five d electrons unpaired (one per orbital, by Hund's rule). Answer: Mn$^{2+}$ = [Ar]$3d^5$, with 5 unpaired electrons — the same half-filled stability that makes Cr's configuration special also makes Mn$^{2+}$ a particularly stable, common oxidation state.

• Writing Cr as [Ar]$3d^4 4s^2$ or Cu as [Ar]$3d^9 4s^2$ — these are the two standard Aufbau exceptions in the 3d series and must be memorised, not derived from a formula.
• Removing electrons from (n−1)d before ns when forming a cation. Always remove ns electrons first: Fe$^{2+}$ is $3d^6$, never $3d^4 4s^2$.
• Forgetting that Zn, Cd, Hg (fully-filled $d^{10}$ configuration in every common state) are not "typical" transition elements by the strict definition, which requires a partially filled d-subshell in the atom or at least one common ion.
• Confusing the group number with the number of d electrons — Group 6 (Cr) is not simply "6 d electrons"; the configuration must be built from Z, not read off the group number, because of the s/d electron split.

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