Transition elements (d-block) are elements whose atoms (or common ions) have a partially filled d sub-shell. They occupy groups 3–12, spanning four series (3d, 4d, 5d, 6d) corresponding to periods 4, 5, 6, and 7.
Group 12 elements (Zn, Cd, Hg) have a fully-filled $(n-1)d^{10}$ configuration in both their atomic and common ionic states, so they don't qualify as true transition elements by the strict "partially filled d-orbital" definition — though they're studied alongside the d-block for convenience.
Atomic radii decrease initially across a transition series (increasing nuclear charge), but the decrease levels off and radii become roughly similar towards the middle-to-end of the series — because the added d electrons (in an inner sub-shell) shield the outer 4s electrons from the increasing nuclear charge fairly effectively, partly cancelling the contraction.
Ionization enthalpy generally increases across a transition series, though with irregularities, due to the interplay between increasing nuclear charge and the extra stability of half-filled/fully-filled d configurations.
Explain why the atomic radii of transition metals don't decrease as sharply across a period as they do for main-group (s and p block) elements.
Across a transition series, electrons are added to the inner (n−1)d sub-shell rather than the outermost shell. These d electrons shield the outer ns electrons from the increasing nuclear charge quite effectively, so the pull on the outermost electrons doesn't increase as sharply as it does when electrons are added directly to the valence shell (as in s/p-block elements) — leading to a much more gradual decrease in radius.
Transition elements characteristically show multiple oxidation states, differing usually by 1 (unlike main-group elements, which often differ by 2), because both the $ns$ and $(n-1)d$ electrons are close in energy and can be involved in bonding.
Manganese shows the widest range of oxidation states in the 3d series (from +2 to +7), since it has a half-filled d⁵ configuration that allows access to many stable states. The most common/stable oxidation state across most of the 3d series is +2, corresponding to loss of the two 4s electrons.
All transition elements are metals — hard, with high melting/boiling points (due to strong metallic bonding involving both ns and (n−1)d electrons), and good conductors of heat and electricity.
Why do transition metals generally show a larger number of oxidation states than typical main-group metals?
In transition metals, both the outer ns electrons and the inner (n−1)d electrons are close enough in energy to be involved in bonding/ionisation, allowing several different combinations of electron loss — resulting in multiple accessible oxidation states, unlike main-group metals where usually only the outer ns (and sometimes np) electrons are involved.
Most transition metal ions are coloured due to d-d electronic transitions — when surrounded by ligands (in an aqueous solution or complex), the five d-orbitals split into two sets of slightly different energies. An electron can absorb a specific wavelength of visible light to jump from the lower to the higher energy set; the colour observed is the complementary colour of the wavelength absorbed.
Ions with either an empty ($d^0$) or completely filled ($d^{10}$) d sub-shell are colourless, since there's no d-d transition possible — no electron can be promoted within the d-orbitals if they're all empty or all full. E.g., Sc³⁺ (d⁰) and Zn²⁺ (d¹⁰) are both colourless.
Transition metal ions with unpaired d electrons are paramagnetic (attracted by a magnetic field); those with all d electrons paired are diamagnetic. Magnetic moment increases with the number of unpaired electrons.
where $n$ = number of unpaired electrons, and BM = Bohr Magneton (unit of magnetic moment).
Calculate the spin-only magnetic moment of Fe³⁺ (3d⁵ configuration, all electrons unpaired).
$n = 5$ unpaired electrons. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\ \text{BM}$
Transition metals and their compounds are widely used as catalysts (e.g., Fe in the Haber process, V₂O₅ in the Contact process, Ni in hydrogenation of oils), largely because of their ability to show variable oxidation states and to form intermediate complexes with reactants, providing an alternate lower-energy reaction pathway.
Transition metals readily form alloys with each other, since their atomic radii are quite similar, allowing atoms of one metal to easily replace atoms of another in the crystal lattice without significant distortion.
Small non-metal atoms (H, C, N, B) can fit into the empty spaces (interstitial voids) within a transition metal's crystal lattice, forming interstitial compounds. These compounds are typically very hard, retain metallic conductivity, and have very high melting points (e.g., tungsten carbide, WC).
Prepared from chromite ore, ultimately via sodium chromate, which is converted to sodium dichromate and then to potassium dichromate. In acidic medium, it is a strong oxidising agent, with Cr going from +6 to +3: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$.
Dichromate ions ($\text{Cr}_2\text{O}_7^{2-}$, orange) and chromate ions ($\text{CrO}_4^{2-}$, yellow) exist in a pH-dependent equilibrium: $\text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \rightleftharpoons 2\text{CrO}_4^{2-} + 2\text{H}^+$. Adding acid shifts this towards the orange dichromate form; adding base shifts it towards the yellow chromate form.
Prepared from pyrolusite ore (MnO₂), via potassium manganate as an intermediate, which is later oxidised and acidified to give permanganate. A strong oxidising agent in acidic medium, with Mn going from +7 to +2: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$.
Explain why the colour of a potassium dichromate solution changes from orange to yellow on adding a base.
Dichromate ($\text{Cr}_2\text{O}_7^{2-}$, orange) exists in equilibrium with chromate ($\text{CrO}_4^{2-}$, yellow): $\text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O} \rightleftharpoons 2\text{CrO}_4^{2-} + 2\text{H}^+$. Adding a base removes H⁺ ions from the system, which (by Le Chatelier's principle) shifts this equilibrium forward, converting more dichromate into chromate — turning the solution yellow.
The 14 elements following lanthanum (atomic numbers 58–71), where the 4f sub-shell is progressively filled. All show a dominant +3 oxidation state, with occasional +2 or +4 states for a few elements with special stability (e.g., Ce⁴⁺, Eu²⁺).
A steady, cumulative decrease in atomic/ionic radii across the lanthanoid series, caused by the poor shielding ability of 4f electrons — as nuclear charge increases across the series, the poorly-shielding 4f electrons allow the effective nuclear charge felt by outer electrons to increase steadily, contracting the atom at each step.
Explain why zirconium (Zr) and hafnium (Hf), though in the same group and separated by an entire lanthanoid series, have almost identical atomic radii.
Normally, adding an entire new shell (going from period 5 to period 6) would be expected to increase atomic radius noticeably. However, the intervening lanthanoid contraction (across the 14 elements between La and Hf) causes a substantial cumulative decrease in radius that almost exactly cancels out this expected increase — leaving Zr and Hf with nearly identical atomic radii and very similar chemical properties.
The 14 elements following actinium (atomic numbers 90–103), where the 5f sub-shell is progressively filled. All actinoids are radioactive, and only the first few (up to uranium) occur naturally in significant amounts — the rest are synthetic, man-made elements.
| Feature | Lanthanoids | Actinoids |
|---|---|---|
| Sub-shell filled | 4f | 5f |
| Radioactivity | Non-radioactive (except promethium) | All radioactive |
| Oxidation states | Mostly +3, limited variability | Show a much wider range (+3 to +7), since 5f, 6d, 7s are close in energy |
| Natural occurrence | All occur naturally | Only up to uranium occurs naturally in useful amounts; rest synthetic |