General Properties of Transition Elements

Definition and metallic character. IUPAC defines a transition element as one whose atom, or a common ion, has a partially filled d-subshell. The 3d series runs Sc([Ar]$3d^14s^2$) to Zn([Ar]$3d^{10}4s^2$). Strong metallic bonding — both $ns$ and $(n-1)d$ electrons delocalised into the electron sea — makes every member hard, high-melting, and a good conductor.

Ionisation enthalpy. It rises gradually across the series, with irregular jumps at Cr and Cu, because increasing nuclear charge is poorly screened by d-electrons. Values sit well above those of s-block metals, so transition metals are far less reactive than alkali or alkaline-earth metals.

Oxidation states. Because $ns$ and $(n-1)d$ electrons have comparable energies, both can be lost on oxidation, giving each element several accessible states rather than one fixed value. The lowest common state is usually +2 (loss of the two $4s$ electrons); the highest often equals the total ns + (n-1)d electron count, topping out at +7 for Mn ([Ar]$3d^54s^2$); the spread is widest near the middle of the series (Mn, Fe).

Ionic radii. M$^{2+}$/M$^{3+}$ radii shrink gradually across the series — poor d-electron shielding against rising nuclear charge, the same mechanism (on a smaller scale) as lanthanide contraction — with a slight dip-and-rise superimposed on the trend.

Colour. A partially filled d-subshell lets an electron absorb visible light and jump to a higher d orbital — a d–d transition, treated qualitatively here and developed fully under crystal-field theory later. That is why most transition-metal ions are coloured, while $d^0$ ions (Sc$^{3+}$, Ti$^{4+}$) and $d^{10}$ ions (Zn$^{2+}$, Cu$^+$) are colourless — there is no d-electron to promote.

Magnetism. Unpaired d-electrons make a species paramagnetic (drawn into a field); an all-paired configuration is diamagnetic. Ignoring orbital contribution, the spin-only magnetic moment is

where: $n$ is the number of unpaired electrons in the ion and BM (Bohr Magneton) is the unit of magnetic moment. $\mu$ rises with $n$ and peaks near the middle of the series, where half-filled $d^5$ configurations maximise unpairing.

Catalysis and interstitial compounds. Variable oxidation states let transition metals offer low-activation-energy pathways through unstable intermediates (Fe in the Haber process, V$_2$O$_5$ in the Contact process, Ni in hydrogenation), and their surfaces adsorb and weaken reactant bonds. Because 3d atoms are similar in size they readily form substitutional alloys with each other, while small atoms (H, C, N, B) squeeze into interstitial voids of the lattice to give hard, high-melting interstitial compounds such as TiC and Fe$_3$N.

Calculate the spin-only magnetic moment of Fe$^{2+}$, and separately rank Sc$^{3+}$, Ti$^{3+}$, Mn$^{2+}$ and Zn$^{2+}$ by unpaired electrons, stating which are coloured.

Step 1. Fe (Z = 26) is [Ar]$3d^64s^2$; removing the two $4s$ electrons gives Fe$^{2+}$ = [Ar]$3d^6$. By Hund's rule, four of the five 3d orbitals hold one electron each and the fifth holds a paired second electron, so $n = 4$.

Step 2. Apply the spin-only formula: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\ \text{BM}$. Since $n \neq 0$, Fe$^{2+}$ is paramagnetic.

Step 3. Find the d-electron count of each remaining ion after removing $4s$ (then 3d, if needed) electrons: Sc$^{3+}$ = [Ar]$3d^0$; Ti$^{3+}$ = [Ar]$3d^1$; Mn$^{2+}$ = [Ar]$3d^5$; Zn$^{2+}$ = [Ar]$3d^{10}$.

Step 4. Count unpaired electrons by Hund's rule: Sc$^{3+}$ ($d^0$) has 0; Ti$^{3+}$ ($d^1$) has 1; Mn$^{2+}$ ($d^5$, one electron in each of the five orbitals) has 5; Zn$^{2+}$ ($d^{10}$, all orbitals doubly filled) has 0. Order: Sc$^{3+}$ = Zn$^{2+}$ (0) less than Ti$^{3+}$ (1) less than Mn$^{2+}$ (5).

Step 5. Colour needs a partially filled d-subshell so a d–d transition is possible. Sc$^{3+}$ ($d^0$) and Zn$^{2+}$ ($d^{10}$) are colourless; Ti$^{3+}$ ($d^1$) and Mn$^{2+}$ ($d^5$) are coloured in principle, though Mn$^{2+}$'s colour is unusually faint because its particular d–d transitions are spin-forbidden. Answer: Fe$^{2+}$ has $\mu \approx 4.90$ BM (paramagnetic); unpaired-electron order is Sc$^{3+}$ = Zn$^{2+}$ (0) less than Ti$^{3+}$ (1) less than Mn$^{2+}$ (5), with only Ti$^{3+}$ and Mn$^{2+}$ coloured.

Competitive stretch

Quick check: what is the spin-only moment of Cr$^{3+}$? Cr (Z = 24) is the exception [Ar]$3d^54s^1$; removing three electrons (the one $4s$, then two $3d$) gives Cr$^{3+}$ = [Ar]$3d^3$, with all three electrons unpaired by Hund's rule ($n = 3$). So $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\ \text{BM}$ — a reminder that Cr's own atomic configuration (one 4s electron, not two) does not change the removal order once ionised.

• Removing d-electrons before s-electrons when forming a cation — always strip the $ns$ electrons first, then $(n-1)d$ if more are needed (Cr$^{3+}$ above needs one $4s$ and two $3d$ removed, in that order).
• Assuming more unpaired electrons always means a more intense colour — colour intensity depends on whether the d–d transition is spin- and symmetry-allowed, not simply on the unpaired-electron count (Mn$^{2+}$'s 5 unpaired electrons give only a very faint colour).
• Forgetting that $d^0$ (Sc$^{3+}$) and $d^{10}$ (Zn$^{2+}$) ions are colourless — a common assumption is that all transition-metal ions are coloured.
• Treating the ionisation-enthalpy trend as perfectly smooth — Cr and Cu (and neighbours) show irregular values from the extra stability of half-filled and fully-filled configurations.

Ask the tutor: Want to work through the spin-only moment for a few more ions, like Cr$^{2+}$ or Co$^{2+}$, to get faster at the removal-order and Hund's-rule steps?