Part 3: Hydroxides, ammonia and ligand exchange

Separate acid–base precipitation, amphoteric dissolution and ligand exchange, then attach each equation to a visible observation and its conditions.

Bases remove protons from coordinated water

Aqueous metal ions are surrounded by water ligands. The metal ion attracts electron density from oxygen, polarising and weakening the O–H bonds so a coordinated water can release H⁺. Small, highly charged 3+ ions tend to polarise water more strongly than comparable 2+ ions, which explains why many 3+ aqua ions make more acidic solutions.

Adding OH⁻ removes H⁺ from coordinated water. With enough base, neutral hydrated metal hydroxides precipitate. A small amount of NH₃ can also accept protons, becoming NH₄⁺; it therefore initially gives the same hydroxide precipitates as NaOH. At this stage ammonia acts as a Brønsted–Lowry base, not as the ligand in an ammine complex.

We use hydrated precipitate formulae M(OH)₂(H₂O)₄ and M(OH)₃(H₂O)₃ to make proton transfer visible. The simpler M(OH)₂ and M(OH)₃ formulae are useful summaries, but do not mix a simplified formula with an equation whose water balance assumes the full hydrated one. Precipitation does not change the metal oxidation state.

Record limited addition and excess separately

Add reagent dropwise to a small fresh sample, record colour and physical state, then add excess and record whether the solid dissolves and the new solution colour. ‘Blue’ alone cannot distinguish a precipitate from a solution. Concentration, ageing and oxygen exposure matter; the immediate observation and the final observation may differ.

Chromium(III) substitution is kinetically slow. The Edexcel reaction model includes the violet hexaammine complex in excess ammonia, but its formation should not be presented as the same rapid cold-test response as copper. State the conditions and time supplied in an experiment and use its observations. June 2025 9CH0/01 Q6(d) explicitly classifies the chromium hydroxide/excess-ammonia reaction as ligand exchange.

Chromium hydroxide dissolves in acid and excess hydroxide

An amphoteric hydroxide reacts with both acids and bases. Acid protonates its OH groups and regenerates the aqua ion. Excess OH⁻ instead produces a soluble negatively charged hydroxo complex. The dissolved product has an overall −3 charge, so it is attracted to water molecules and no longer forms the neutral hydroxide solid.

In the hydrated representation, excess OH⁻ deprotonates the remaining water ligands in Cr(OH)₃(H₂O)₃ to give [Cr(OH)₆]³⁻. Chromium stays +3. Compare this acid–base behaviour with ammonia substitution: in the latter, NH₃ donates nitrogen lone pairs and replaces oxygen donor ligands. The change is not a redox reaction merely because a green precipitate disappears.

Worked distinction: Cr(OH)₃(H₂O)₃ + 3OH⁻ → [Cr(OH)₆]³⁻ + 3H₂O has charge −3 on both sides. In acid, adding three H⁺ gives the +3 hexaaqua ion. Showing both equations supports the term amphoteric; showing only dissolution in alkali does not establish both behaviours.

Excess ammonia can switch from base to ligand

The copper sequence makes ammonia’s two roles clear. Limited NH₃ removes protons and forms pale-blue Cu(OH)₂(H₂O)₄. Excess NH₃ then replaces four water-derived oxygen donor positions, dissolving the precipitate and producing the deep-blue tetraammine diaqua ion. Four ammonia and two water ligands still provide six coordinate bonds, so the course geometry remains octahedral and Cu remains +2.

Ammonia exchange in cobalt can give [Co(NH₃)₆]²⁺. This complex is readily oxidised by oxygen in air to cobalt(III) ammine species, causing a later darker brown colour. Keep initial substitution and subsequent oxidation separate. Fe²⁺ and Fe³⁺ hydroxides do not dissolve in excess ammonia under these test conditions.

For chromium, the formal excess-ammonia equation below represents formation of [Cr(NH₃)₆]³⁺ when the exchange proceeds; the kinetic limitation described above still applies. In all three equations, check the unchanged metal oxidation state and the OH⁻ released rather than mistaking the overall complex charge for oxidation number.

Concentrated chloride changes ligand and coordination number

A high chloride concentration, supplied by an appropriate concentrated chloride solution, replaces six waters around Cu²⁺ or Co²⁺ with four chlorides. The larger ligands favour four-coordinate complexes, treated as tetrahedral in this course. Both the ligand environment and coordination number change; the metal oxidation number remains +2.

Pink [Co(H₂O)₆]²⁺ gives blue [CoCl₄]²⁻. Blue aqueous copper(II) gives a yellow chloride-complex contribution; mixed solutions often appear yellow-green or green because both copper complexes are present. Use ‘yellow/green’ for the chloride-rich copper system with this explanation, rather than claiming every green solution contains only pure [CuCl₄]²⁻.

Adding water lowers chloride concentration and favours the aqua side, restoring pink cobalt or blue copper solution. Dissolving a metal chloride salt in plenty of water does not guarantee a tetrachloro complex: solution concentration and equilibrium determine coordination. A concentrated HCl demonstration also introduces acid, so follow laboratory controls and identify chloride as the ligand causing this exchange.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources