Transition elements: general trends

Definition. A transition element is one whose atoms, in the elemental state or in any commonly occurring oxidation state, have an incompletely filled d-subshell. The 3d series runs from Sc (Z = 21) to Zn (Z = 30), with the general configuration

where: $[\text{Ar}]$ is the argon core (18 electrons), the superscripts on $3d$ and $4s$ show how many electrons occupy each subshell across the series, and the order of filling is $4s$ before $3d$ (lower energy first) while the order of removal on ionisation is $4s$ before $3d$ as well — this reversal-on-removal is the single fact behind almost every trend below. Zn ($3d^{10}4s^2$) is a border case: its d-subshell is completely full in every common compound, so it is often excluded from "typical" transition behaviour despite sitting in the d-block. The trends across the series follow from how close in energy the $4s$ and $3d$ electrons are, and from the special stability of half-filled ($d^5$) and fully filled ($d^{10}$) configurations:

• Metallic character. Every 3d element is a hard, high-melting, good-conducting metal, because both $ns$ and $(n-1)d$ electrons take part in metallic bonding — more delocalised electrons than a typical s-block or p-block metal, hence the unusually strong bonding and high melting points.
• Ionisation enthalpy. Rises broadly, but irregularly, across the series as nuclear charge grows faster than the poor shielding from other d-electrons can compensate; small dips and peaks appear near $d^5$ and $d^{10}$ configurations because those arrangements are extra-stable and harder — or, once passed, comparatively easier — to disturb.
• Oxidation states. Because $4s$ and $3d$ electrons are close in energy, a transition metal can lose them in almost any combination, so oxidation states typically differ by one unit (not the two-unit jumps typical of p-block elements). $+2$ and $+3$ are the commonest states; the maximum oxidation state climbs from Sc ($+3$) to Mn ($+7$), then falls back toward Zn ($+2$ only) as later d-electrons become too tightly bound (core-like) to ionise further.
• Ionic radii. Shrink slowly and somewhat irregularly across the series for a fixed oxidation state (say $M^{2+}$ or $M^{3+}$), because d-electrons shield the rising nuclear charge only partially — a real but much gentler contraction than the sharp lanthanoid contraction seen across the f-block.
• Colour of ions. Most 3d ions in solution are coloured (Cu$^{2+}$ blue, Ni$^{2+}$ green, Fe$^{3+}$ yellow-brown) because a partially filled d-subshell allows electrons to absorb visible light and jump between split d-orbitals — a d–d transition. Ions with $d^0$ (Sc$^{3+}$) or $d^{10}$ (Zn$^{2+}$) configurations have no such transition available and are colourless.

The whole family of trends reduces to one structural cause worth stating precisely: across the 3d series, both the $4s$ and $3d$ orbitals sit close enough in energy that electrons in either can be involved in bonding or removed on ionisation, and the resulting d-electron count decides oxidation state, colour, and radius all at once. Unpacking this in stages shows why each observed trend has to look the way it does rather than being a list of memorised facts.

Step 1. Start from $3d^{1-10}4s^{1-2}$. Moving from Sc to Zn, one proton and (mostly) one 3d electron are added at a time, and d-orbitals are diffuse, so each new d-electron shields the rising nuclear charge only partially. Net effect: effective nuclear charge on the outer electrons creeps up across the series — the same cause behind both the overall rise in ionisation enthalpy and the gentle fall in ionic radii for a fixed oxidation state.

Step 2. Because $4s$ and $3d$ are close in energy, ionisation removes $4s$ electrons first, then $3d$ electrons, and either type can be lost depending on how many are stripped away. That freedom to lose electrons in comparable-energy steps is why oxidation states differ by one unit, not the two-unit jumps of the p-block (where a stable $ns^2$ pair is kept whole or lost whole — the inert-pair effect). The ceiling is simply how many $4s + 3d$ electrons are available: Mn ($3d^54s^2$, seven available) reaches $+7$, while later elements have more d-electrons locked into harder-to-ionise configurations, so the maximum falls back toward Zn.

Step 3. Configurations at $d^5$ (half-filled) and $d^{10}$ (fully filled) are unusually stable — a half-filled set spreads one electron into each of five d-orbitals with parallel spins, maximising exchange energy, and a full set is maximally symmetric. These show up as dips or bumps in the ionisation-enthalpy curve, and explain why ions such as Mn$^{2+}$ ($d^5$) or Zn$^{2+}$ ($d^{10}$) resist further oxidation or reduction.

Step 4. Colour follows the same d-electron count. Surrounding ligands or water split the five d-orbitals into two energy groups; if there is at least one d-electron and one vacancy, visible light can promote an electron between groups, and the ion appears the complementary colour. $d^0$ ions (no electron to promote) and $d^{10}$ ions (no vacancy) cannot do this and are colourless — the same partially-filled-d-subshell condition that defines a transition element also decides whether its ions are coloured.

Problem: For the 3d ions Sc$^{3+}$, Ti$^{3+}$ and Mn$^{2+}$, (a) give the d-electron configuration of each, (b) state which are coloured in aqueous solution, and (c) explain why Mn$^{2+}$ is unusually stable compared with, say, Fe$^{2+}$.

Step 1. Remove electrons in the order $4s$ then $3d$ from the parent atom. Sc ($[\text{Ar}]3d^14s^2$) loses all three outer electrons to form Sc$^{3+}$: $[\text{Ar}]3d^0$. Ti ($[\text{Ar}]3d^24s^2$) loses three electrons to form Ti$^{3+}$: $[\text{Ar}]3d^1$. Mn ($[\text{Ar}]3d^54s^2$) loses its two $4s$ electrons to form Mn$^{2+}$: $[\text{Ar}]3d^5$.

Step 2. Apply the colour rule: colour needs a partially filled d-subshell, $d^1$ to $d^9$. Sc$^{3+}$ is $d^0$ — no electron to promote, colourless. Ti$^{3+}$ is $d^1$ — the electron promotes across the ligand-split d-orbitals on absorbing visible light, giving Ti$^{3+}$(aq) its violet colour. Mn$^{2+}$ is $d^5$ — formally partially filled, but the transition is spin-forbidden, so Mn$^{2+}$(aq) is only very faintly coloured in practice.

Step 3. Compare Mn$^{2+}$ ($d^5$) with Fe$^{2+}$ ($d^6$, from $[\text{Ar}]3d^64s^2$ losing two $4s$ electrons). Mn$^{2+}$'s $d^5$ places one electron in each of the five d-orbitals with parallel spin, maximising exchange energy — a symmetric, extra-stable arrangement. Fe$^{2+}$'s $d^6$ must pair two electrons in one orbital, which is less stable and more easily oxidised further, to Fe$^{3+}$ ($d^5$).

Final answer: Sc$^{3+}$ = $d^0$ (colourless), Ti$^{3+}$ = $d^1$ (coloured, violet), Mn$^{2+}$ = $d^5$ (formally coloured but very faint); Mn$^{2+}$'s stability comes from its symmetric half-filled $d^5$ configuration, which resists further oxidation more than the neighbouring $d^6$ configuration of Fe$^{2+}$.

• Assuming oxidation states always jump by two units, as in the p-block — 3d transition metals lose $4s$ and $3d$ electrons in comparable-energy steps, so states typically differ by one unit.
• Assuming every d-block ion is coloured — $d^0$ ions (Sc$^{3+}$, Ti$^{4+}$) and $d^{10}$ ions (Zn$^{2+}$, Cu$^{+}$) are colourless because no d–d transition is possible.
• Treating the ionic-radius decrease across the 3d series as smooth and steep like the lanthanoid contraction — it is real but small and irregular, since d-electrons shield the nucleus only partially.
• Forgetting that Sc and Zn are borderline members: Sc$^{3+}$ is $d^0$ and Zn$^{2+}$ is $d^{10}$, so neither ion shows the partially-filled-d-subshell behaviour that defines "typical" transition character, even though both elements sit in the d-block.

Ask the tutor: if you would like to work through predicting the d-electron count and colour for any other 3d ion, or trace the full ionisation-enthalpy graph across the series, ask the tutor.