3.2.5 Transition Metals

A transition metal forms at least one stable ion with an incomplete d subshell. Check the ion, not just the position of the neutral atom in the d block; Zn is the standard reminder because Zn²⁺ has a full d¹⁰ subshell.

For Ti to Cu, write the electron configuration of the relevant ion if you need to decide whether the d subshell is incomplete.

The characteristic transition-metal chemistry you need is formation of complexes, coloured ions, variable oxidation states and catalytic activity. All four become much easier to organise once you track the d electrons and ligand environment explicitly.

Treat these as separate examinable ideas. Colour comes from d-level splitting and visible-light absorption; variable oxidation state and catalysis are redox ideas; complex formation is coordinate bonding to ligands.

For the definition, I’d learn: a ligand is an ion or molecule that donates a lone pair of electrons to a metal atom or ion to form a coordinate bond.

The donated lone pair is the essential feature. A species being “around the metal” is not enough unless it is actually coordinated through an electron pair.

A complex contains a central metal atom or ion surrounded by ligands joined to it through coordinate bonds. The whole assembly may be neutral or charged.

When you read a complex formula, identify the metal centre, ligand identities, ligand charges and overall charge. Those checks usually expose a missing ligand or an impossible oxidation state.

To work out a coordination number, count coordinate bonds to the central metal atom or ion, not simply ligand molecules. A bidentate or multidentate ligand can contribute more than one bond.

For [Cu(NH₃)₄(H₂O)₂]²⁺ the coordination number is 6. Three bidentate ligands also give coordination number 6 even though only three ligand molecules are present.

H₂O, NH₃ and Cl⁻ are monodentate ligands: each coordinates through one donor atom and forms one coordinate bond to the metal centre.

Remember the ligand charges as well: H₂O and NH₃ are neutral, while Cl⁻ carries −1. That charge bookkeeping matters when you work out the overall charge of a complex.

H₂O and NH₃ are neutral monodentate ligands of similar size. Replacing water ligands by ammonia can therefore leave both the coordination number and overall complex charge unchanged.

That gives you a useful formula check: if a pure H₂O↔NH₃ substitution suddenly changes the charge, inspect what you have written.

Replacing H₂O by similarly sized NH₃ can leave a six-coordinate complex six-coordinate. The ligand identity changes, but each incoming NH₃ still supplies one coordinate bond in place of one H₂O bond.

Use the Co²⁺ and Cu²⁺ examples to practise writing ligand-exchange equations. Keep metal oxidation state and overall charge unchanged when neutral ligands replace neutral ligands.

For Cu²⁺ with excess NH₃, don’t let the word ‘excess’ tempt you into replacing all six water ligands. You form the deep-blue [Cu(NH₃)₄(H₂O)₂]²⁺ complex, so two H₂O ligands remain.

The complex is still six-coordinate: four NH₃ plus two H₂O ligands each make one coordinate bond. That formula is worth recognising on sight.

Compare the common ligands here: Cl⁻ is noticeably larger than H₂O or NH₃. Keep that size difference in mind because it helps explain why fewer chloride ligands fit around some transition-metal ions.

The point of remembering the relative size is to explain later changes in coordination number and geometry during chloride ligand substitution.

When larger Cl⁻ ligands replace smaller H₂O ligands, the coordination number can fall because fewer bulky ligands fit around the metal centre. Co²⁺, Cu²⁺ and Fe³⁺ provide course examples of this behaviour.

Check both charge and coordination number after substitution: Cl⁻ is charged as well as larger, so replacing neutral water can change more than the colour of the complex.

Ethane-1,2-diamine is bidentate because its two nitrogen atoms can each donate a lone pair. Ethanedioate is also bidentate, coordinating through two oxygen donor atoms.

One molecule/ion of either ligand therefore contributes two coordinate bonds. Draw the two attachment points if you find the word “bidentate” too easy to say without actually counting them.

Recognise EDTA⁴⁻ as the classic multidentate ligand in this course. One EDTA can bind through six donor atoms, so a single ligand can occupy several coordination positions that would otherwise need several monodentate ligands.

Its −4 charge also contributes to the overall complex charge, so do the charge balance independently of the coordination-number count.

Haem contains Fe²⁺ held within a multidentate porphyrin-type ligand. Several donor atoms bind the iron, while additional coordination can occur above/below the ligand plane.

For this course, the useful connection is that the Fe²⁺ centre can then bind small molecules such as O₂ reversibly; you do not need to turn this line into a full biochemistry detour.

Think reversible ligand binding for oxygen transport in haemoglobin. O₂ coordinates to the Fe²⁺ centre where its partial pressure is high and must be able to come off again where tissues need it.

Frame this as ligand binding to a metal complex. The iron is not being used up in a stoichiometric reaction each time an oxygen molecule is transported.

For carbon-monoxide toxicity, focus on competition for the same Fe²⁺ binding site. CO binds much more strongly than O₂, so once it occupies a site that site is no longer available for normal reversible oxygen transport.

The key comparison is relative binding strength at the same coordination site. A vague statement that CO is poisonous does not explain the transition-metal chemistry.

A bi- or multidentate ligand can displace several monodentate ligands because it forms more than one coordinate bond to the same metal centre. The product contains one or more chelate rings.

Count the coordinate bonds before and after substitution. A single incoming ligand molecule can replace several outgoing ligand molecules while leaving the metal coordination number unchanged.

The chelate effect is the greater stability/favourability often observed when bi- or multidentate ligands replace comparable monodentate ligands in a complex.

Use the term for that substitution advantage. “Multidentate” describes the ligand; the chelate effect describes the favourable substitution/stability effect. The next step is to explain why the overall enthalpy/entropy balance can favour the chelated complex.

When you explain the chelate effect, don’t reduce it to ‘chelates are stronger’. In many substitutions the metal–ligand bond enthalpy changes are fairly small, while binding one multidentate ligand can release several monodentate ligand molecules into solution.

That increase in the number of free particles can give a favourable entropy change. Count species before and after the substitution, then consider ΔH and ΔS together. A blanket claim that “chelates are stronger” misses the thermodynamic argument.

Six coordinate bonds around a transition-metal centre commonly give an octahedral arrangement with 90° bond angles. Small ligands such as H₂O and NH₃ readily form these six-coordinate complexes.

Link geometry to coordination number: six-coordinate is your strong octahedral cue in the course examples.

When you look for stereoisomerism in an octahedral complex, first check the ligand arrangement. Suitable complexes can give cis/trans pairs, and bidentate ligands can also produce non-superimposable mirror images and optical isomerism.

When deciding whether two drawings are isomers, first confirm identical formula, charge and connectivity. Then ask whether one three-dimensional arrangement can be rotated onto the other.

With a four-coordinate complex containing bulky ligands such as Cl⁻, tetrahedral geometry is the common AQA pattern; [CoCl₄]²⁻ is the useful example to recognise. Keep that separate from square-planar four-coordinate complexes.

The geometry links back to ligand size: four bulky chloride ligands can fit around the metal more readily than six. Keep this separate from square-planar examples, which are also four-coordinate.

A square-planar complex has four coordinate bonds in one plane, typically at 90°. With two pairs of suitable ligands, the identical ligands can be adjacent (cis) or opposite (trans).

Draw the square first and place the ligands explicitly. That makes cis/trans assignment much safer than trying to infer it from the formula alone.

For cisplatin, identify the geometry first: it is square planar, and the two Cl ligands sit next to one another. That adjacent arrangement is what makes it the cis isomer.

Draw the square plane if the formula alone is making the name hard to see. The two Cl ligands must occupy adjacent positions for cisplatin; putting them opposite one another gives the trans isomer.

[Ag(NH₃)₂]⁺ is a linear Ag(I) complex: two ammonia ligands give coordination number 2 and a 180° arrangement. It is the silver-containing complex present in Tollens’ reagent.

That gives you a useful connection between coordination chemistry and the later redox test: recognise the complex formula first, then treat its reduction to Ag separately.

Characteristic solution colours can help identify transition-metal ions, but colour depends on oxidation state and ligand environment as well as the metal itself. Use the conditions and observations supplied with the question.

I would treat colour as supporting chemical evidence, not a universal name-tag for an element: changing ligand or oxidation state can give a different colour for the same metal.

Ligands split the metal ion’s d orbitals into different energy levels. If visible light of the right energy is absorbed, a d electron is promoted across the gap; the wavelengths not absorbed make up the colour you observe.

The observed colour comes from the wavelengths that remain after selective absorption. The absorbed and observed light are complementary parts of the explanation.

Use ΔE = hν = hc/λ for the absorbed photon. A larger d-level energy gap requires higher-frequency, shorter-wavelength light; a smaller gap corresponds to lower frequency and longer wavelength.

Keep SI units consistent if you calculate: wavelength normally needs converting to metres before using c/λ.

If a transition-metal complex changes colour, ask what changed around the metal. A different oxidation state, ligand or coordination number changes the d-orbital splitting, so a different ΔE and therefore a different set of visible wavelengths is absorbed.

That is why a ligand-substitution or redox reaction can produce a strong colour change even though the central element is unchanged.

When you use a colorimeter or spectrophotometer, choose a wavelength the coloured species absorbs strongly. That way a concentration change gives you a clear change in absorbance.

The measurement turns the qualitative idea of selective absorption into quantitative data. Keep the same wavelength and measurement conditions when comparing standards with an unknown.

For a calibration experiment, prepare standards of known concentration, measure all their absorbances under the same conditions and plot absorbance against concentration. You can then interpolate the unknown from the calibration line.

The unknown should fall within the calibrated range. If it is too concentrated, dilute it by a known factor, remeasure it, then correct the concentration for that dilution.

Transition metals often have several accessible oxidation states because the 3d and 4s electrons are close enough in energy for different numbers of electrons to be removed or involved in bonding.

Show the oxidation state explicitly when you follow a redox sequence. The element name alone does not tell you which transition-metal species is present.

In acid, zinc can reduce vanadium stepwise from V(V) to V(IV), V(III) and V(II). Track the oxidation state at each stage and use the associated colour observations supplied/learnt for the sequence.

The important chemistry is progressive electron gain by vanadium while Zn is oxidised. If you write half-equations, make each oxidation-state change visible. One memorised overall equation hides the useful electron bookkeeping.

Never treat an electrode potential as a property of a bare metal ion in isolation. If the pH or ligand environment changes, one oxidation state may be stabilised relative to another, so the redox equilibrium and measured potential can shift.

So do not treat an E value as an immutable property of the bare metal. Check the species, ligands and conditions associated with the couple.

In the Tollens test, keep the redox pair explicit: [Ag(NH₃)₂]⁺ is reduced to Ag(s) while the aldehyde is oxidised. A silver mirror is the positive observation for an aldehyde; an ordinary ketone gives no corresponding reaction under the test conditions.

This is both complex-ion and redox chemistry. State which species is reduced and connect the observation to the aldehyde/ketone distinction.

In acidified permanganate titrations, MnO₄⁻ is reduced to Mn²⁺. Fe²⁺ can be oxidised to Fe³⁺, and ethanedioate can also act as the reducing species in the appropriate reaction.

Write and combine the relevant half-equations to obtain the mole ratio before doing any titration arithmetic. The purple permanganate itself provides the end-point colour in these reactions.

For one of these redox-titration calculations, get the redox stoichiometry right before touching cV. Convert the titre to moles of known reagent, use the electron-balanced mole ratio, then work on to the requested amount or concentration.

Do the redox balancing before the volume calculation. A perfectly executed cV calculation with the wrong MnO₄⁻:Fe²⁺ or MnO₄⁻:ethanedioate ratio will still be wrong.

To decide whether a catalyst is homogeneous or heterogeneous, start with phase. A heterogeneous catalyst is in a different phase and works at a surface; a homogeneous catalyst is in the same phase and works through catalyst-containing intermediates in the mixture.

For transition-metal examples, identify which model applies before explaining the mechanism. Surface adsorption belongs to heterogeneous catalysis; oxidation-state/intermediate cycles belong to homogeneous catalysis.

For heterogeneous catalysis, picture the surface sequence: reactants adsorb at active sites, bonds are weakened or favourably oriented, reaction occurs and products desorb. That surface route gives a lower activation energy.

Mention the active surface and regeneration of the site. Simply saying “the catalyst speeds up the reaction” does not explain the heterogeneous mechanism.

If you spread a heterogeneous catalyst over a high-surface-area support, more catalyst atoms are exposed as usable active sites. That can give more activity per gram and lets you use less of an expensive transition metal.

The support is useful because of surface area and distribution of the active material; do not describe it as though it were necessarily the catalytic substance itself.

For the Contact process, the catalyst I’d expect you to recognise is V₂O₅. It catalyses the oxidation of SO₂ to SO₃, with vanadium changing oxidation state during the route and then being regenerated.

If the question asks only for the catalyst, V₂O₅ is the key fact. If it asks for the route, then show the separate redox steps that regenerate the vanadium(V) species.

For the Haber process, remember that iron is a heterogeneous catalyst. Picture N₂ and H₂ adsorbing on the iron surface, reacting by a lower-energy route and NH₃ desorbing so the active sites can be used again.

Keep the explanation at the surface level required here: adsorption, lower-energy pathway and desorption. The catalyst changes the rate, not the equilibrium constant.

If an impurity poisons a heterogeneous catalyst, it binds strongly to active sites and blocks reactants from using them. You can therefore lose a large fraction of the effective catalytic surface even with quite a small amount of poison.

The practical consequence is lower efficiency and possible catalyst replacement/regeneration cost. Tie the economic effect to the chemical loss of active sites.

For the word homogeneous, go straight to phase: the catalyst is in the same phase as the reactants. It can then form catalyst-containing molecular or ionic intermediates throughout the mixture without needing a solid surface.

Use phase as the definition first. The presence of an intermediate explains the alternative route, but it is not what the word “homogeneous” itself means.

Homogeneous catalysis proceeds through one or more intermediates containing the catalyst. The catalyst is used in one elementary step and regenerated in another.

Add the catalytic steps as a check: intermediates and catalyst should cancel, leaving the overall reaction. If the supposed catalyst remains in the net equation, your cycle is incomplete.

Accessible oxidation states let a transition-metal ion accept electrons in one step and donate them in another. That can connect two redox partners through lower-barrier electron-transfer steps while regenerating the original catalyst.

Follow the oxidation state through the cycle. A catalyst can change temporarily during the mechanism; the requirement is that it is regenerated overall.

When you show the Contact-process catalytic cycle, make the vanadium oxidation-state changes do the work. Vanadium(V) is reduced while SO₂ is oxidised, then O₂ re-oxidises the vanadium species so the catalyst is regenerated.

When you add the catalytic steps, they must give the overall SO₂ + ½O₂ → SO₃ reaction while the vanadium species cancel. That cancellation shows that the catalyst participates in the route but is regenerated by the end.

The Fe²⁺/Fe³⁺ couple provides an alternative redox route between I⁻ and S₂O₈²⁻. One step oxidises Fe²⁺ to Fe³⁺; the other reduces Fe³⁺ back to Fe²⁺ while oxidising iodide.

Write the two equations so Fe²⁺/Fe³⁺ cancel on addition and the net equation is S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂. That is a very good self-check on the mechanism.

For autocatalysis in the permanganate–ethanedioate reaction, the useful thing to notice is that Mn²⁺ is both a product and a catalyst. As you make more Mn²⁺, the reaction speeds up.

The defining point is that a reaction product acts as the catalyst. Do not confuse this line with the redox-titration stoichiometry line just because the same reacting system is used.