Transition Elements: Question 2

Worked solution

Part (a): The anomalous configuration of chromium

Following the Aufbau (“building-up”) pattern from vanadium ($[\text{Ar}]3d^34s^2$) to manganese ($[\text{Ar}]3d^54s^2$), a naive extension would predict chromium to be $[\text{Ar}]3d^44s^2$. The actual ground-state configuration is: $\text{Cr}:\ [\text{Ar}]3d^54s^1$

Reason: a subshell in which every orbital contains exactly one electron, all with parallel spin, a half-filled subshell, is an unusually stable, low-energy arrangement (it maximises exchange energy between the parallel-spin electrons). By promoting one electron from $4s$ into $3d$, chromium trades the “expected” $3d^44s^2$ arrangement for the more stable $3d^54s^1$ arrangement, in which the $3d$ subshell is exactly half-filled (one electron in each of the five 3d orbitals) and the $4s$ subshell also holds a single electron. This extra stability outweighs the small energy cost of promoting the electron.

(The same effect, for the same reason, gives copper the configuration $[\text{Ar}]3d^{10}4s^1$ rather than $[\text{Ar}]3d^94s^2$. There a completely filled $3d^{10}$ subshell, rather than a half-filled one, provides the extra stability.)

Part (b): Electron configurations of Fe²⁺ and Fe³⁺

The iron atom is $\text{Fe}: [\text{Ar}]3d^64s^2$. Transition-metal atoms always lose their $4s$ electrons before any $3d$ electrons on ionisation, because once the $3d$ subshell is occupied it falls below $4s$ in energy, making the $4s$ electrons the most easily removed.

Removing the two $4s$ electrons gives: $\text{Fe}^{2+}:\ [\text{Ar}]3d^6$

Removing one further electron (now necessarily from the $3d$ subshell, since $4s$ is already empty) gives: $\text{Fe}^{3+}:\ [\text{Ar}]3d^5$

Part (c): Why iron is a transition element

Definition: a transition element is a d-block element that forms at least one stable ion with an incompletely-filled (partly-filled) d subshell. That is, neither a completely empty ($d^0$) nor a completely full ($d^{10}$) subshell.

From part (b), $\text{Fe}^{2+}$ has the configuration $3d^6$ and $\text{Fe}^{3+}$ has the configuration $3d^5$. Both of these are partly-filled subshells (more than zero but fewer than ten electrons), so iron forms stable ions satisfying the definition. Iron is therefore correctly classified as a transition element.

Final answers

• (a) $\text{Cr}: [\text{Ar}]3d^54s^1$. One $4s$ electron is promoted to give the extra stability of a half-filled $3d^5$ subshell, rather than the naively expected $[\text{Ar}]3d^44s^2$.
• (b) $\text{Fe}^{2+}: [\text{Ar}]3d^6$; $\text{Fe}^{3+}: [\text{Ar}]3d^5$ (both lose $4s$ electrons first; $\text{Fe}^{3+}$ additionally loses one $3d$ electron).
• (c) Iron is a transition element because both $\text{Fe}^{2+}$ ($3d^6$) and $\text{Fe}^{3+}$ ($3d^5$) have partly-filled d subshells, satisfying the definition.