Complexes, ligands and colour

A transition element forms at least one stable ion with a partially filled d sub-shell. (Sc and Zn are d-block but not transition elements: Sc³⁺ is d⁰ and Zn²⁺ is d¹⁰ — which is also why their compounds are colourless.)

A ligand donates a lone pair into a vacant orbital on the metal ion, forming a dative covalent (coordinate) bond. The coordination number is the number of coordinate bonds.

• Monodentate — donates one lone pair: H₂O, NH₃, Cl⁻, CN⁻.
• Bidentate — donates two, from two different atoms: ethane-1,2-diamine, ethanedioate.

Why they are coloured: ligands split the five d orbitals into two energy levels. An electron absorbs a photon of visible light and jumps the gap — a d–d transition, with ΔE = hν. The colour you see is the complement of the light absorbed. Change the ligand, the coordination number or the oxidation state and the size of the gap changes — so the colour changes.

Ligand substitution to know cold: [Cu(H₂O)₆]²⁺ (pale blue) + excess concentrated HCl → [CuCl₄]²⁻ (yellow-green). Note the coordination number falls from 6 to 4, because Cl⁻ is much bulkier than H₂O. With excess NH₃ you get the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺.

Ligand or not?

Tap a species, then tap what kind of ligand it is (if any).

Think it through

Catalysis and redox titrations

Transition metals catalyse because their variable oxidation states let them accept and donate electrons — they can be reduced by one reagent and re-oxidised by another, returning to their original state.

Redox titration with manganate(VII) is self-indicating: the end-point is the first permanent pale pink from the first trace of excess MnO₄⁻.

Your turn — calculation 1

Your turn — calculation 2

Electrode potentials and cells

A standard electrode potential, E°, is measured against the standard hydrogen electrode (defined as 0.00 V) under standard conditions: 298 K, 100 kPa, all solutions 1.00 mol dm⁻³.

• A reaction is feasible if E°_cell is positive.
• The more positive the E°, the stronger the oxidising agent on the left of the half-equation.

The two limitations that carry marks: (1) E° applies only under standard conditions; change the concentrations and the potential changes. (2) A positive E°_cell shows the reaction is thermodynamically feasible but says nothing about the rate — a large activation energy can make a feasible reaction immeasurably slow.

Hydrogen fuel cell: H₂ + ½O₂ → H₂O. Chemical energy is converted directly into electrical energy (so it is not limited by the efficiency of a heat engine), the only product is water, and it never needs recharging while fuel is supplied. Against that: hydrogen is hard to store, and producing it usually consumes energy from elsewhere.

Your turn — calculation 3

Think it through

Electrolysis and Faraday

In electrolysis, an external supply drives a non-spontaneous redox reaction. Reduction happens at the cathode (negative); oxidation at the anode (positive).

Then use the half-equation to convert moles of electrons into moles of product. For copper: Cu²⁺ + 2e⁻ → Cu, so 2 mol of electrons deposit 1 mol of Cu.

Your turn — calculation 4

Amines, amides, nitriles and amino acids

Amines are bases — the lone pair on nitrogen accepts a proton. Base strength depends on how available that lone pair is:

• Ethylamine > ammonia — the alkyl group releases electron density onto N (positive inductive effect), making the lone pair more available.
• Phenylamine < ammonia — the nitrogen lone pair is delocalised into the benzene ring, so it is far less available. Phenylamine is a very weak base.

Nitriles (R–C≡N) are made from a haloalkane + ethanolic KCN — this adds a carbon. They can be reduced (LiAlH₄ or H₂/Ni) to a primary amine, or hydrolysed (dilute acid, reflux) to a carboxylic acid.

Amides (R–CONH₂) come from an acyl chloride + ammonia. Despite the NH₂, an amide is not basic — the nitrogen lone pair is delocalised onto the adjacent C=O.

Amino acids contain both –NH₂ and –COOH, so they are amphoteric. At intermediate pH they exist as a zwitterion, ⁺H₃N–CHR–COO⁻ — which explains their unexpectedly high melting points and their solubility in water. The pH at which the zwitterion dominates is the isoelectric point.

Zwitterion behaviour: in acid, the COO⁻ is protonated → the ion is positively charged overall. In alkali, the ⁺NH₃ loses a proton → the ion is negatively charged. This is why amino acids buffer, and how electrophoresis separates them.

Think it through

Think it through

Chromatography, IR, mass spectrometry and NMR

Chromatography separates components between a stationary and a mobile phase. The more strongly a component is adsorbed onto the stationary phase, the more slowly it travels.

IR identifies functional groups: O–H (acid) very broad 2500–3300 cm⁻¹ · C=O strong and sharp near 1700 cm⁻¹ · O–H (alcohol) broad 3200–3600 cm⁻¹ · N–H 3300–3500 cm⁻¹.

Mass spectrometry: the peak at the highest m/z is the molecular ion, M⁺, giving the M_r directly. Fragments reveal the structure (loss of 15 = CH₃; loss of 29 = C₂H₅ or CHO).

NMR: TMS is the reference, δ = 0.

• ¹³C NMR — the number of peaks = the number of different carbon environments.
• ¹H NMR — shift gives the environment; integration gives the ratio of hydrogens; splitting follows the n + 1 rule, where n = hydrogens on the adjacent carbon.
• O–H and N–H protons appear as broad singlets and disappear on shaking with D₂O.

Your turn — calculation 5

Your turn — calculation 6

Match the technique to what it tells you

Tap a technique on the left, then what it reveals on the right.

Think it through

The big ideas to know

That is Unit A2 2 — how a chemist proves what they have actually made. Press Finish to see your score.