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What gives transition metals their colour, variable oxidation states and catalysis?

The characteristic properties of transition metals, variable oxidation states, the formation and shapes of complex ions, ligand substitution, the origin of colour from d-orbital splitting, and the catalytic behaviour of transition metals.

A CCEA A-Level Chemistry answer on transition metals, covering their characteristic properties, variable oxidation states, the formation and shapes of complex ions, ligand substitution, the origin of colour from d-orbital splitting, and their catalytic behaviour.

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  1. What this dot point is asking
  2. Characteristic properties
  3. Complex ions and ligand substitution
  4. The origin of colour
  5. Catalysis
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA wants you to describe the characteristic properties of transition metals, explain variable oxidation states, describe the formation and shapes of complex ions and ligand substitution, explain the origin of colour from d-orbital splitting, and describe the catalytic behaviour of transition metals.

Characteristic properties

The variable oxidation states arise because the 4s4s and 3d3d sub-shells are close in energy, so successive electrons can be lost without a large jump in ionisation energy, letting an element such as manganese exist in states from +2+2 to +7+7. The higher oxidation states are usually found combined with oxygen or fluorine (as in MnO4\text{MnO}_4^- or VO2+\text{VO}_2^+), and the relative stability of the states underlies the redox chemistry exploited in titrations.

Complex ions and ligand substitution

The number of dative bonds the metal forms is its coordination number: six is the most common (octahedral, as in [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+}), but bulky ligands such as chloride give four (tetrahedral, as in [CuCl4]2[\text{CuCl}_4]^{2-}). The overall charge on the complex is the metal-ion charge plus the charges of any ionic ligands. Ligand substitution can be partial or complete and often produces a striking colour change: adding concentrated ammonia to pale blue [Cu(H2O)6]2+[\text{Cu}(\text{H}_2\text{O})_6]^{2+} gives the deep royal-blue [Cu(NH3)4(H2O)2]2+[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}, and adding concentrated hydrochloric acid gives the yellow-green tetrahedral [CuCl4]2[\text{CuCl}_4]^{2-}. These reactions are used to test for and identify transition-metal ions.

The origin of colour

Because the colour depends on the size of ΔE\Delta E, changing the ligand changes the colour: a stronger-field ligand widens the gap, so higher-energy (shorter-wavelength) light is absorbed and the transmitted colour shifts. This is why Sc3+\text{Sc}^{3+} (empty d sub-shell) and Zn2+\text{Zn}^{2+} (full d sub-shell) are colourless, there being no partly filled d level for an electron to be promoted within. The intensity of the colour is the basis of colorimetry, used to find the concentration of a coloured ion from a calibration graph.

Catalysis

Transition metals and their compounds are good catalysts because their variable oxidation states let them accept and donate electrons readily, providing an alternative reaction pathway of lower activation energy. In heterogeneous catalysis the reactants adsorb onto the metal surface, react, and the products leave (iron in the Haber process, V2O5\text{V}_2\text{O}_5 in the Contact process, nickel in hydrogenation). In homogeneous catalysis the catalyst is in the same phase and works by changing oxidation state during the cycle, as Fe2+/Fe3+\text{Fe}^{2+}/\text{Fe}^{3+} does in catalysing the reaction between S2O82\text{S}_2\text{O}_8^{2-} and I\text{I}^-.

Examples in context

Transition-metal chemistry is everywhere in industry and biology. The intense colours of complexes give pigments and gemstones their hue (the green of emerald and the red of ruby both come from chromium ions in a crystal where ΔE\Delta E differs). Haemoglobin is an iron(II) complex in which oxygen binds reversibly as a ligand, and carbon monoxide is toxic because it binds the same iron site far more strongly, an irreversible ligand substitution. Industrially, the catalytic power of iron, vanadium and nickel makes the Haber, Contact and hydrogenation processes economic, while colorimetric analysis of Cu2+\text{Cu}^{2+} or MnO4\text{MnO}_4^- relies on the d-orbital colour CCEA asks you to explain.

Try this

Q1. State what is meant by a ligand. [2 marks]

  • Cue. A molecule or ion that donates a lone pair to a metal ion to form a dative covalent bond.

Q2. Explain why transition metal complexes are coloured. [2 marks]

  • Cue. The d-orbitals split, and electrons absorb visible light to move between the levels.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20194 marksExplain, in terms of d-orbitals, why aqueous copper(II) ions are blue, and predict the effect on the colour of replacing the water ligands with ammonia.
Show worked answer →

In a complex the five d-orbitals are split by the ligands into two energy levels separated by an energy gap ΔE\Delta E. The copper(II) ion has a partly filled d sub-shell, so an electron can absorb a photon of visible light whose energy equals ΔE\Delta E and jump from the lower to the higher level.

The light absorbed by [Cu(H2O)6]2+[\text{Cu}(\text{H}_2\text{O})_6]^{2+} is in the orange-red region, so the transmitted (complementary) colour is blue.

Replacing water with ammonia (a stronger-field ligand) increases ΔE\Delta E, so light of higher energy (shorter wavelength) is absorbed; the colour shifts to a deeper royal blue ([Cu(NH3)4(H2O)2]2+[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}).

Markers reward (1) d-orbitals split into two levels by the ligands, (2) an electron absorbs visible light equal to ΔE\Delta E, (3) the complementary colour is transmitted, (4) ammonia changes ΔE\Delta E and hence the colour.

CCEA 20225 marksA series of copper(II) standards is used to construct a colorimetric calibration graph that is a straight line through the origin with absorbance =2.50×[Cu2+]= 2.50 \times [\text{Cu}^{2+}] (concentration in mol dm3\text{mol dm}^{-3}). An unknown copper(II) solution gives an absorbance of 0.450.45. Calculate its concentration and outline how the calibration graph was obtained.
Show worked answer →

The calibration graph is a straight line through the origin, absorbance =2.50×[Cu2+]= 2.50 \times [\text{Cu}^{2+}] (this is the Beer-Lambert relationship at low concentration).

Rearranging for the unknown: [Cu2+]=absorbance2.50=0.452.50=0.18 mol dm3[\text{Cu}^{2+}] = \dfrac{\text{absorbance}}{2.50} = \dfrac{0.45}{2.50} = 0.18\ \text{mol dm}^{-3}.

The graph is obtained by measuring the absorbance of several standard copper(II) solutions of known concentration in the colorimeter (using a filter of the complementary colour, here red/orange light), plotting absorbance against concentration, and drawing the best-fit straight line through the origin.

Markers reward (1) reading or using the straight-line relationship, (2) correct rearrangement, (3) the value 0.18 mol dm30.18\ \text{mol dm}^{-3}, (4) measuring known standards, (5) plotting absorbance against concentration and reading the unknown from the line.

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