Transition metals

In a nutshell

The transition metals titanium to copper all have an incomplete d sub-level in their atoms or ions. That one feature gives them four characteristic properties: they form complexes, form coloured ions, show variable oxidation states and have catalytic activity.

This note covers all four: how ligands bond to and swap on a metal ion, the shapes and isomers of complexes, why complexes are coloured (and how colorimetry uses that), redox titrations with manganate(VII), and how transition metals act as catalysts.

Assumed knowledge: Atomic structure (electron configurations), Bonding (co-ordinate bonds), Oxidation, reduction and redox equations, Thermodynamics (ΔG = ΔH − TΔS).

Core content

General properties: the incomplete d sub-level

Transition metal characteristics of the elements Ti to Cu arise from an incomplete d sub-level in their atoms or ions.

The four characteristic properties:

• complex formation
• formation of coloured ions
• variable oxidation state
• catalytic activity

When a transition metal forms an ion, the 4s electrons are lost before the 3d electrons.

Zinc is in the d block but is not a transition metal: its only ion, Zn²⁺, has a full 3d sub-level, and so does the atom. "In the d block" is not the definition; the incomplete d sub-level is.

General properties: ligands, complexes and co-ordination number

A ligand is a molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons.

A co-ordinate bond (dative covalent bond) is a covalent bond in which both electrons in the shared pair come from one atom. In a complex, that atom is on the ligand, never the metal.

A complex is a central metal atom or ion surrounded by ligands.

The co-ordination number is the number of co-ordinate bonds to the central metal atom or ion.

In [Cu(H₂O)₆]²⁺, each water ligand donates a lone pair from its oxygen atom to Cu²⁺, so the co-ordination number is 6.

How to write the formula of a complex ion:

• Put the whole complex in square brackets.
• Work out the overall charge: metal ion charge plus the ligand charges (H₂O and NH₃ are 0, Cl^- is −1, C₂O₄^2- is −2).
• Write that net charge once, outside the bracket. Never show charges on individual ligands inside the bracket.

So Cu²⁺ with four Cl^- ligands is [CuCl₄]^2- (2+ plus 4 × 1− = 2−).

Substitution reactions: water, ammonia and chloride ligands

H₂O, NH₃ and Cl^- are monodentate ligands: each forms one co-ordinate bond to the metal.

NH₃ and H₂O are similar in size and are uncharged, so they swap without a change in co-ordination number (it stays 6):

[Co(H₂O)₆]²⁺(aq) + 6NH₃(aq) → [Co(NH₃)₆]²⁺(aq) + 6H₂O(l)

Pink solution to pale yellow (straw) solution, in excess concentrated ammonia.

Substitution can be incomplete. With copper(II), only four of the six water ligands are replaced:

[Cu(H₂O)₆]²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 4H₂O(l)

Blue solution to deep blue solution in excess ammonia. (A pale blue precipitate of the hydroxide forms first with a little ammonia: that part belongs to Reactions of ions in aqueous solution.)

The Cl^- ligand is larger than NH₃ and H₂O, so only four Cl^- ions fit around the metal. Swapping H₂O for Cl^- therefore changes the co-ordination number from 6 to 4. The reagent is concentrated hydrochloric acid (a high concentration of Cl^-).

[Cu(H₂O)₆]²⁺(aq) + 4Cl^-(aq) → [CuCl₄]^2-(aq) + 6H₂O(l)

[Co(H₂O)₆]²⁺(aq) + 4Cl^-(aq) → [CoCl₄]^2-(aq) + 6H₂O(l)

[Fe(H₂O)₆]³⁺(aq) + 4Cl^-(aq) → [FeCl₄]^-(aq) + 6H₂O(l)

The chloride substitutions are reversible: adding lots of water turns blue [CoCl₄]^2- back to pink [Co(H₂O)₆]²⁺.

Substitution reactions: bidentate and multidentate ligands

A bidentate ligand has two atoms that each donate a lone pair, so it forms two co-ordinate bonds to the metal. Learn two:

• ethane-1,2-diamine, H₂NCH₂CH₂NH₂ (uncharged; donates from each N)
• ethanedioate, C₂O₄^2- (charge 2−; donates from one O on each carbon)

A multidentate ligand forms more than one co-ordinate bond (so bidentate ligands are one kind of multidentate ligand). EDTA^4- donates six electron pairs (from two N and four O atoms), so one EDTA^4- fills all six sites. You do not need to draw EDTA.

Two equations you should be able to write:

[Co(H₂O)₆]²⁺(aq) + 3H₂NCH₂CH₂NH₂(aq) → [Co(H₂NCH₂CH₂NH₂)₃]²⁺(aq) + 6H₂O(l)

[Cu(H₂O)₆]²⁺(aq) + EDTA^4-(aq) → [Cu(EDTA)]^2-(aq) + 6H₂O(l)

Partial substitution by ethanedioate also happens. Two ethanedioate ions replace four water ligands, and the co-ordination number is still 6:

[Cu(H₂O)₆]²⁺(aq) + 2C₂O₄^2-(aq) → [Cu(C₂O₄)₂(H₂O)₂]^2-(aq) + 4H₂O(l)

Count co-ordinate bonds, not ligands: 2 ethanedioates × 2 bonds + 2 waters × 1 bond = 6.

Substitution reactions: haem and carbon monoxide

Haem is an iron(II) complex with a multidentate ligand. It is part of haemoglobin.

• Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, which lets haemoglobin transport oxygen in the blood.
• Carbon monoxide is toxic because it replaces the oxygen co-ordinately bonded to Fe(II). CO binds to the Fe(II) more strongly than O₂ does, so the haemoglobin can no longer carry oxygen.

Substitution reactions: the chelate effect

Bidentate and multidentate ligands replace monodentate ligands from complexes. This is called the chelate effect. The complex formed is more stable.

The explanation is a balance between the enthalpy change and the entropy change. Use this example:

[Co(NH₃)₆]²⁺(aq) + 3H₂NCH₂CH₂NH₂(aq) → [Co(H₂NCH₂CH₂NH₂)₃]²⁺(aq) + 6NH₃(aq)

The explanation, link by link:

1. ΔH is approximately zero: the same number of bonds (six) and the same type of bond (Co-N) are broken and made.
2. The number of particles increases: 4 particles (1 + 3) form 7 particles (1 + 6).
3. So disorder increases and ΔS is positive.
4. So ΔG = ΔH − TΔS is negative: the substitution is feasible, and the chelated complex is more stable.

For EDTA^4- replacing six waters, the particle count goes from 2 to 7, so the entropy increase is even larger.

Still don't get it? · the chelate effect

Imagine six people each holding one rope tied to a post. A single person who can hold six ropes at once walks up and takes over all of them. Six people are now free to wander off around the room. The room has become more disordered, and nothing about the ropes themselves changed: same number of ropes, same knots.

Now rebuild it step by step:

• The post is the metal ion. Each rope is one co-ordinate bond. The six people are six monodentate ligands.
• The new arrival is a multidentate ligand. It makes all six co-ordinate bonds by itself.
• Bonds broken and bonds made are the same number and the same type, so almost no energy is taken in or given out. That is why ΔH is close to zero.
• Before: 1 complex + 1 new ligand = 2 particles. After: 1 complex + 6 released ligands = 7 particles. More particles means more ways to arrange them, so entropy increases and ΔS is positive.
• With ΔH about zero and ΔS positive, the −TΔS term makes ΔG negative, so the reaction happens.

The exam version: same number and same type of bonds broken and made, so ΔH ≈ 0; number of particles increases (state the numbers, e.g. 2 → 7); entropy increases, ΔS positive; so ΔG is negative. Saying only "the complex is more stable" scores nothing; the marks are for the entropy argument.

Shapes of complex ions

Note that two different shapes have co-ordination number 4. You have to know which one a given complex takes.

Shapes of complex ions: cis-trans and optical isomerism

Complexes can show two kinds of stereoisomerism.

Cis-trans isomerism is a special case of E-Z isomerism. It happens when a complex has two identical ligands that can sit either side by side or opposite each other:

• cis: the two identical ligands are next to each other (90° apart).
• trans: the two identical ligands are opposite each other (180° apart).

It is shown by octahedral complexes with monodentate ligands, e.g. [Co(NH₃)₄Cl₂]⁺, and by square planar complexes, e.g. [Pt(NH₃)₂Cl₂]. Cisplatin is the cis isomer.

Optical isomerism is shown by octahedral complexes with bidentate ligands, e.g. [Ni(H₂NCH₂CH₂NH₂)₃]²⁺. The two isomers are non-superimposable mirror images of each other.

Formation of coloured ions: why complexes are coloured

Transition metal ions can be identified by their colour.

Colour arises when some wavelengths of visible light are absorbed and the remaining wavelengths are transmitted or reflected. The colour you see is the light that is left, not light that is given out.

What happens inside the ion:

1. In a complex, the d orbitals of the metal ion are at different energies.
2. d electrons move from the ground state to an excited state when light is absorbed.
3. The energy absorbed matches the energy gap, ΔE, between those d-electron levels.
4. Light of that frequency is removed from white light; the rest is transmitted or reflected, and that is the colour seen.

The energy gap is linked to the light absorbed by:

$\Delta E = h\nu = \frac{hc}{\lambda}$

A larger ΔE means higher-frequency, shorter-wavelength light is absorbed.

Hexaaquacopper(II) absorbs mainly at the red and orange end of the spectrum and absorbs little blue light. The blue light is transmitted, so the solution looks blue.

Changes in oxidation state, co-ordination number and ligand alter ΔE, and this changes the colour:

• ligand changes: [Co(H₂O)₆]²⁺ (pink) → [Co(NH₃)₆]²⁺ (pale yellow)
• ligand and co-ordination number change: [Co(H₂O)₆]²⁺ (pink) → [CoCl₄]^2- (blue)
• oxidation state changes: [Co(NH₃)₆]²⁺ (pale yellow) → [Co(NH₃)₆]³⁺ (dark brown)

Ions with no d electrons (Sc³⁺, 3d⁰) or a full d sub-level (Zn²⁺, 3d¹⁰) have no d-electron transitions, so they cannot absorb visible light this way: their solutions are colourless and their solid compounds are white.

Still don't get it? · why a complex is coloured

Shine white light through a sheet of blue cellophane. The cellophane does not make blue light. White light already contains every colour; the cellophane soaks up the reds, oranges and yellows, and only the blue gets through to your eye.

Now rebuild it step by step:

• White light is a mixture of all the visible wavelengths.
• In a complex ion, the d electrons have a lower level they sit in (the ground state) and a higher level they can jump to (an excited state). The gap between them is ΔE.
• Light whose energy exactly matches ΔE can be absorbed: an electron uses it to jump up. Light of other energies cannot be used, so it passes straight through.
• So one band of colours is removed from the white light. What reaches your eye is everything that was not absorbed.
• Change the ligand, the co-ordination number or the oxidation state, and the size of the gap changes. A different band is absorbed, so a different colour is left over.

The exam version: visible light is absorbed; d electrons are excited from the ground state to a higher energy level; the remaining light is transmitted or reflected and that is the colour seen. Never write that the complex "emits" light or that electrons "fall back and give out" the colour: examiners penalise it every time.

Formation of coloured ions: colorimetry

The absorption of visible light is used in spectroscopy, and a simple colorimeter can be used to determine the concentration of coloured ions in solution. The more concentrated the coloured ion, the more light it absorbs.

Method for finding an unknown concentration:

1. If the colour is pale, add a suitable ligand to intensify the colour (for example excess ammonia with copper(II) ions). Treat every solution the same way.
2. Choose a filter of the colour that the solution absorbs most strongly.
3. Zero the colorimeter with a cuvette of water.
4. Make up a series of solutions of known concentration and measure the absorbance of each.
5. Plot a graph of absorbance against concentration (a calibration curve). It should pass through, or very close to, the origin.
6. Measure the absorbance of the unknown and read its concentration from the graph.

Use the same cuvette (same path length) every time, because absorbance also depends on the distance the light travels through the solution.

Colorimetry is also a good way to follow a reaction: it measures the concentration without using up or disturbing the reaction mixture, and gives readings quickly.

Variable oxidation states: vanadium

Transition elements show variable oxidation states. Vanadium shows four in solution, each with its own colour:

Vanadium species in oxidation states IV, III and II are formed by the reduction of vanadate(V) ions by zinc in acidic solution. Add zinc to an acidified solution of ammonium vanadate(V) and the colour changes yellow → (green) → blue → green → violet.

The first green is not a new oxidation state: it is a mixture of yellow vanadium(V) and blue vanadium(IV) while the first step is part-way through.

Each step, with zinc as the reducing agent (Zn → Zn²⁺ + 2e^-):

2VO₂⁺(aq) + 4H⁺(aq) + Zn(s) → 2VO²⁺(aq) + Zn²⁺(aq) + 2H₂O(l)

2VO²⁺(aq) + 4H⁺(aq) + Zn(s) → 2V³⁺(aq) + Zn²⁺(aq) + 2H₂O(l)

2V³⁺(aq) + Zn(s) → 2V²⁺(aq) + Zn²⁺(aq)

Electrode potentials explain why zinc can go all the way to V²⁺. Zinc's E^⦵ is more negative than every vanadium couple below:

Variable oxidation states: pH, ligands and Tollens' reagent

The redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH and by the ligand.

pH. Many reduction half-equations use H⁺ ions:

MnO₄^-(aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O(l)

• More H⁺ pushes this equilibrium to the right, so manganate(VII) is a stronger oxidising agent in acid.
• With too little acid, manganate(VII) is only reduced to brown MnO₂ instead: MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O.
• The reverse is also true: lower oxidation states are more easily oxidised in alkaline solution. For example, green iron(II) hydroxide turns brown in air as oxygen oxidises it to iron(III).

Ligand. Changing the ligand changes how easily the metal changes oxidation state:

• [Co(H₂O)₆]²⁺ is not oxidised by air.
• [Co(NH₃)₆]²⁺ is oxidised by oxygen in the air to [Co(NH₃)₆]³⁺ (pale yellow to dark brown).

Tollens' reagent. [Ag(NH₃)₂]⁺ (linear) is reduced to metallic silver by aldehydes but not by ketones. This is used to distinguish between aldehydes and ketones: an aldehyde gives a silver mirror.

[Ag(NH₃)₂]⁺(aq) + e^- → Ag(s) + 2NH₃(aq)

Variable oxidation states: redox titrations with manganate(VII)

Potassium manganate(VII), KMnO₄, is used to find the amount of a reducing agent such as Fe²⁺ or C₂O₄^2-.

The half-equations:

Combining them gives the titration equations:

MnO₄^-(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

2MnO₄^-(aq) + 16H⁺(aq) + 5C₂O₄^2-(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

So the mole ratios are MnO₄^- : Fe²⁺ = 1 : 5 and MnO₄^- : C₂O₄^2- = 2 : 5.

How the titration is done:

• KMnO₄ goes in the burette; the reducing agent goes in the conical flask with excess dilute sulfuric acid.
• It is self-indicating. While the reducing agent is in excess, purple MnO₄^- is decolourised to almost colourless Mn²⁺.
• The end point is the first permanent pale pink colour: colourless to pale pink, because MnO₄^- is now in excess.
• The purple solution is hard to see through, so read the top of the meniscus in the burette.
• For ethanedioate, warm the flask (to about 60 °C), because the reaction starts very slowly (see autocatalysis below).

Why the acid must be dilute sulfuric acid:

• Not hydrochloric acid. MnO₄^- oxidises Cl^- to Cl₂ (E^⦵ MnO₄^-/Mn²⁺ = +1.51 V is more positive than E^⦵ Cl₂/Cl^- = +1.36 V). Extra manganate(VII) is used up, so the titre is too large.
• Not too little acid. Brown MnO₂ forms instead of Mn²⁺, which hides the end point and changes the titre.

Catalysts: heterogeneous catalysis

Transition metals and their compounds can act as heterogeneous and homogeneous catalysts.

A heterogeneous catalyst is in a different phase from the reactants, and the reaction occurs at active sites on the surface.

How it works, in order:

1. Reactants are adsorbed onto active sites on the catalyst surface (they form weak bonds to the surface).
2. Bonds in the reactants are weakened, and the reactants are held close together in a suitable orientation.
3. The reaction happens on the surface.
4. The products are desorbed (they leave the surface), freeing the active site.

Support medium. The catalyst is spread thinly over a support (or used as a powder, pellets or a fine mesh) to maximise its surface area and minimise the cost, since less of the expensive catalyst is needed.

Poisoning. Impurities can block the active sites (they adsorb strongly and are not desorbed). The catalyst then has reduced efficiency, which has a cost implication: the catalyst must be cleaned or replaced, and the reactants purified first.

Two industrial examples:

How V₂O₅ acts as a catalyst relies on variable oxidation states of vanadium (+5 ⇌ +4):

V₂O₅(s) + SO₂(g) → V₂O₄(s) + SO₃(g) (vanadium reduced from +5 to +4)

V₂O₄(s) + ½O₂(g) → V₂O₅(s) (vanadium oxidised back from +4 to +5)

Add the two equations and V₂O₅ cancels, leaving SO₂ + ½O₂ → SO₃. The catalyst is regenerated.

Catalysts: homogeneous catalysis and autocatalysis

A homogeneous catalyst is in the same phase as the reactants. When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species.

Why variable oxidation states matter. A transition metal ion can be oxidised in one step and reduced back in the next (or the other way round). It provides an alternative route with a lower activation energy and is regenerated at the end.

Fe²⁺ catalysing the reaction between I^- and S₂O₈^2-

Overall: S₂O₈^2-(aq) + 2I^-(aq) → 2SO₄^2-(aq) + I₂(aq)

• Uncatalysed, it is slow: both reactants are negative ions, so they repel, giving a high activation energy.
• With Fe²⁺, each step is between oppositely charged ions, which attract, so each step has a lower activation energy:

S₂O₈^2-(aq) + 2Fe²⁺(aq) → 2SO₄^2-(aq) + 2Fe³⁺(aq)

2Fe³⁺(aq) + 2I^-(aq) → 2Fe²⁺(aq) + I₂(aq)

Note the 2Fe in both steps: one S₂O₈^2- takes two electrons. Fe³⁺ would catalyse it too, because the same two steps can happen in the other order.

Autocatalysis by Mn²⁺ in the reaction between C₂O₄^2- and MnO₄^-

In autocatalysis, a product of the reaction acts as the catalyst. Here the catalyst is Mn²⁺, a product.

Overall: 2MnO₄^-(aq) + 16H⁺(aq) + 5C₂O₄^2-(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

The catalysed route goes through Mn³⁺:

4Mn²⁺(aq) + MnO₄^-(aq) + 8H⁺(aq) → 5Mn³⁺(aq) + 4H₂O(l)

2Mn³⁺(aq) + C₂O₄^2-(aq) → 2Mn²⁺(aq) + 2CO₂(g)

Mn²⁺ is oxidised by MnO₄^- to Mn³⁺. The Mn³⁺ then oxidises C₂O₄^2- and is reduced back to Mn²⁺. Each step is again between oppositely charged ions.

Reading the autocatalysed (S-shaped) curve:

1. Slow at first: no catalyst yet, and the reaction is between two negative ions that repel (high activation energy).
2. Speeds up: Mn²⁺ forms, and the catalyst concentration increases.
3. Slows and levels off: MnO₄^- is being used up.

With Mn²⁺ added at the start, there is no slow start: the rate is greatest at the beginning and falls as the reactants are used up.

The reaction can be followed with a colorimeter, because MnO₄^- is purple and the products are effectively colourless.

Worked examples

Example 1: energy gap from the wavelength absorbed

A complex ion absorbs visible light most strongly at a wavelength of 475 nm. Calculate the energy difference, ΔE, between the ground state and the excited state of its d electrons.

h = 6.63 × 10^-34 J s; c = 3.00 × 10⁸ m s^-1

Step 1: convert the wavelength to metres.

$\lambda = 475\ \text{nm} = 475 \times 10^{-9}\ \text{m} = 4.75 \times 10^{-7}\ \text{m}$

Step 2: substitute into ΔE = hc/λ.

$\Delta E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}\ \text{J s}) \times (3.00 \times 10^{8}\ \text{m s}^{-1})}{4.75 \times 10^{-7}\ \text{m}}$

$\Delta E = \frac{1.989 \times 10^{-25}\ \text{J m}}{4.75 \times 10^{-7}\ \text{m}} = 4.187 \times 10^{-19}\ \text{J}$

Answer: ΔE = 4.19 × 10^-19 J (3 significant figures, matching the data).

If you are asked for the frequency instead:

$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}\ \text{m s}^{-1}}{4.75 \times 10^{-7}\ \text{m}} = 6.32 \times 10^{14}\ \text{s}^{-1}$

Where marks go: leaving λ in nm gives an answer 10⁹ times too small (a classic wrong option in multiple-choice). ΔE is the energy for one electron, in J, not kJ mol^-1. Frequency units are s^-1 or Hz.

Example 2: concentration from a colorimetry calibration graph

Excess ammonia was added to a series of copper(II) solutions of known concentration to intensify the colour, and the absorbance of each was measured. A 10.0 cm³ sample of an unknown copper(II) solution was treated the same way and made up to 50.0 cm³. Its absorbance was 0.33.

Step 1: read the concentration of the measured solution from the graph. Go across from absorbance 0.33 to the line, then down:

$[\text{Cu}^{2+}]_{\text{measured}} = 0.055\ \text{mol dm}^{-3}$

Step 2: correct for the dilution. The original 10.0 cm³ was made up to 50.0 cm³, a dilution factor of 5.00:

$[\text{Cu}^{2+}]_{\text{original}} = 0.055\ \text{mol dm}^{-3} \times \frac{50.0\ \text{cm}^3}{10.0\ \text{cm}^3} = 0.275\ \text{mol dm}^{-3}$

Answer: 0.28 mol dm^-3 (2 significant figures, because the absorbance 0.33 is only given to 2 s.f.).

Where marks go: stopping at the graph reading (0.055) and forgetting the dilution is the commonest error. If the question gives the concentration of a salt such as Fe₂(SO₄)₃, remember there are two Fe³⁺ ions per formula unit.

Example 3: percentage of iron in a tablet (manganate(VII) with iron(II))

An iron tablet of mass 1.50 g was dissolved in excess dilute sulfuric acid and the solution made up to 250 cm³. 25.0 cm³ portions were titrated with 0.00500 mol dm^-3 KMnO₄. The mean titre was 21.40 cm³. Calculate the percentage by mass of iron in the tablet. (A_r Fe = 55.8)

MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Step 1: moles of MnO₄^- used.

$n(\text{MnO}_4^-) = 0.00500\ \text{mol dm}^{-3} \times \frac{21.40}{1000}\ \text{dm}^3 = 1.070 \times 10^{-4}\ \text{mol}$

Step 2: moles of Fe²⁺ in 25.0 cm³ (ratio 1 : 5).

$n(\text{Fe}^{2+}) = 5 \times 1.070 \times 10^{-4} = 5.350 \times 10^{-4}\ \text{mol}$

Step 3: scale up to the 250 cm³ flask (factor 250/25.0 = 10).

$n(\text{Fe}^{2+})_{\text{total}} = 5.350 \times 10^{-4} \times \frac{250}{25.0} = 5.350 \times 10^{-3}\ \text{mol}$

Step 4: mass of iron.

$m(\text{Fe}) = 5.350 \times 10^{-3}\ \text{mol} \times 55.8\ \text{g mol}^{-1} = 0.2985\ \text{g}$

Step 5: percentage by mass.

$\%\ \text{Fe} = \frac{0.2985\ \text{g}}{1.50\ \text{g}} \times 100 = 19.9\ \%$

Answer: 19.9 % (3 significant figures, matching 1.50 g and 0.00500 mol dm^-3).

Where marks go: using a 1 : 1 ratio instead of 1 : 5; forgetting the ×10 scale-up (or using 250/1000 instead); dividing the tablet mass by the iron mass instead of the other way round; using 56 instead of the 55.8 on the Periodic Table.

Example 4: Mr of hydrated ethanedioic acid (manganate(VII) with ethanedioate)

1.50 g of hydrated ethanedioic acid, H₂C₂O₄·xH₂O, was dissolved in water and made up to 250 cm³. 25.0 cm³ portions were acidified with dilute sulfuric acid, warmed, and titrated with 0.0200 mol dm^-3 KMnO₄. The mean titre was 23.80 cm³. Find M_r and the value of x.

2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 8H₂O + 10CO₂

Step 1: moles of MnO₄^-.

$n(\text{MnO}_4^-) = 0.0200 \times \frac{23.80}{1000} = 4.760 \times 10^{-4}\ \text{mol}$

Step 2: moles of ethanedioate in 25.0 cm³ (ratio 2 : 5).

$n(\text{C}_2\text{O}_4^{2-}) = 4.760 \times 10^{-4} \times \frac{5}{2} = 1.190 \times 10^{-3}\ \text{mol}$

Step 3: scale up to 250 cm³.

$n_{\text{total}} = 1.190 \times 10^{-3} \times 10 = 1.190 \times 10^{-2}\ \text{mol}$

Step 4: M_r of the hydrated acid.

$M_r = \frac{m}{n} = \frac{1.50\ \text{g}}{1.190 \times 10^{-2}\ \text{mol}} = 126\ \text{g mol}^{-1}$

Step 5: find x. M_r(H₂C₂O₄) = 2(1.0) + 2(12.0) + 4(16.0) = 90.0

$x = \frac{126 - 90.0}{18.0} = 2.00 \approx 2$

Answer: M_r = 126, so the formula is H₂C₂O₄·2H₂O (x = 2).

Where marks go: multiplying by 2/5 instead of 5/2. Check the direction: there is more ethanedioate than manganate(VII), so the number of moles must go up.

Example 5: a "similar redox reaction" with two reducing parts (iron(II) ethanedioate)

1.200 g of impure FeC₂O₄·2H₂O was dissolved in excess dilute sulfuric acid and made up to 250 cm³. 25.0 cm³ portions, warmed, needed 26.10 cm³ of 0.0150 mol dm^-3 KMnO₄. Find the percentage purity. (M_r FeC₂O₄·2H₂O = 179.8)

Step 1: find the ratio by counting electrons (no full equation needed).

• MnO₄^- gains 5 electrons (Mn +7 → +2).
• One FeC₂O₄ loses 3 electrons: 1 from Fe²⁺ → Fe³⁺, plus 2 from C₂O₄^2- → 2CO₂.
• Electrons must balance: 3 × 5 = 5 × 3, so 3 MnO₄^- react with 5 FeC₂O₄.

Step 2: moles of MnO₄^-.

$n(\text{MnO}_4^-) = 0.0150 \times \frac{26.10}{1000} = 3.915 \times 10^{-4}\ \text{mol}$

Step 3: moles of FeC₂O₄·2H₂O in 25.0 cm³, then in 250 cm³.

$n = 3.915 \times 10^{-4} \times \frac{5}{3} = 6.525 \times 10^{-4}\ \text{mol}$

$n_{\text{total}} = 6.525 \times 10^{-4} \times 10 = 6.525 \times 10^{-3}\ \text{mol}$

Step 4: mass, then percentage.

$m = 6.525 \times 10^{-3}\ \text{mol} \times 179.8\ \text{g mol}^{-1} = 1.173\ \text{g}$

$\%\ \text{purity} = \frac{1.173\ \text{g}}{1.200\ \text{g}} \times 100 = 97.8\ \%$

Answer: 97.8 % (3 significant figures, matching 0.0150 mol dm^-3).

Where marks go: using 1 : 5 (forgetting the ethanedioate also reacts) or 2 : 5 (forgetting the Fe²⁺); using M_r of FeC₂O₄ without the water; a "purity" over 100 % is a sign the ratio is wrong.

Common exam mistakes

• Saying the colour comes from light emitted. Writing that electrons "fall back down and give out" the colour is penalised. The colour is the light remaining after some is absorbed, which is transmitted or reflected.
• Getting ΔE = hν wrong. Calling ΔE "the enthalpy change" is not accepted; it is the energy absorbed by an electron moving from the ground state to an excited state. h is Planck's constant (students often say it is the frequency), and ν is the frequency, not the wavelength.
• Explaining a colour difference without the d levels. When asked why two complexes have different colours, say the d-electron energy gap (ΔE) is different, so a different frequency of light is absorbed. Just explaining why one complex is coloured does not answer the question.
• Giving "shape" as a factor that changes colour. The three factors are oxidation state, co-ordination number and ligand; "shape" is not credited.
• Defining co-ordination number as "the number of ligands". That is only true for monodentate ligands and loses the mark. It is the number of co-ordinate bonds to the central metal ion.
• Describing a co-ordinate bond without calling it covalent. The mark needs both: a covalent bond and both electrons from one atom. For water as a ligand, say the pair comes from the oxygen atom.
• "A bidentate ligand has two lone pairs." Rejected. It must be two atoms that each donate a lone pair.
• Charges in the wrong place. Showing −1 on each Cl or −2 on each C₂O₄ inside the bracket, or leaving off the overall charge (as in [Cu(C₂O₄)₂(H₂O)₂]^2-). Put one net charge outside the square bracket.
• Writing [Cu(NH₃)₆]²⁺. Substitution is incomplete: it is [Cu(NH₃)₄(H₂O)₂]²⁺.
• Wrong co-ordination numbers with bidentate ligands. [Cu(C₂O₄)₂(H₂O)₂]^2- is 6, not 4; [Co(H₂NCH₂CH₂NH₂)₃]²⁺ is 6, not 3.
• Giving an ion as a reagent. "Cl^-" is not a reagent; write concentrated hydrochloric acid.
• Chelate effect with no entropy. "Multidentate ligands make a more stable complex" scores nothing. You need the particle numbers (e.g. 4 → 7), entropy increases, and ΔG negative. For ΔH you need same number AND same type of bonds broken and made.
• Bond angles. Octahedral is 90°, not 120° or 109.5°. Tetrahedral is 109.5°.
• Mixing up cis-trans and optical. Drawing a mirror image and calling it optical when the question is about two identical monodentate ligands: that is cis-trans.
• Unbalanced catalysis equations. Using one Fe²⁺ with S₂O₈^2-. It is 2Fe²⁺, because S₂O₈^2- takes two electrons. Check charges as well as atoms. The equations must be full equations, not half-equations with electrons.
• "The ions repel" with no link. Say both ions are negative and repel, so the activation energy is high. For the catalysed steps, say the ions are oppositely charged and attract.
• Naming MnO₄^- as the autocatalyst. The autocatalyst is Mn²⁺, a product. "Self-catalysing" is not accepted as a definition.
• Missing the desorption step. In heterogeneous catalysis, the most-missed point is that products desorb (leave the surface).
• Wrong poison. The Haber process iron catalyst is poisoned by sulfur; many other substances are wrongly given.
• Titration ratios and scaling. Wrong ratio (Fe²⁺ : MnO₄^- is 5 : 1); wrong scale-up factor (e.g. using 1000/250 when the volumes need 250/25.0); percentage calculated upside down.
• Not saying why HCl is unsuitable. It is because MnO₄^- oxidises Cl^- ions to chlorine. Confusing chlorine with chloride wrecks the answer.
• Vague colorimetry methods. "A colorimeter is more accurate" scores nothing. State: add a ligand to intensify the colour, measure absorbance of known concentrations, plot absorbance against concentration (not just "plot a calibration curve"), then read off the unknown. Draw the line through the origin.
• Answers to the wrong precision. If the data are to 3 significant figures, give the answer to 3 significant figures.

Key definitions

• Transition metal characteristics: arise from an incomplete d sub-level in atoms or ions.
• Ligand: a molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons.
• Co-ordinate bond: a covalent bond in which both electrons in the shared pair come from one atom.
• Complex: a central metal atom or ion surrounded by ligands.
• Co-ordination number: the number of co-ordinate bonds to the central metal atom or ion.
• Monodentate ligand: a ligand that forms one co-ordinate bond (e.g. H₂O, NH₃, Cl^-).
• Bidentate ligand: a ligand with two atoms that each donate a lone pair (forms two co-ordinate bonds), e.g. H₂NCH₂CH₂NH₂, C₂O₄^2-.
• Multidentate ligand: a ligand that forms more than one co-ordinate bond (donates more than one electron pair), e.g. EDTA^4-, which forms six.
• Chelate effect: the replacement of monodentate ligands by bidentate or multidentate ligands, forming a more stable complex (because the number of particles increases, so entropy increases and ΔG is negative).
• ΔE = hν = hc/λ: ΔE is the energy difference between the ground state and the excited state of the d electrons; h is Planck's constant; ν is the frequency of light absorbed; λ is its wavelength.
• Heterogeneous catalyst: a catalyst in a different phase from the reactants; the reaction occurs at active sites on the surface.
• Homogeneous catalyst: a catalyst in the same phase as the reactants; the reaction proceeds through an intermediate species.
• Active site: the place on the catalyst surface where reactants are adsorbed.
• Catalyst poisoning: impurities block the active sites, reducing the catalyst's efficiency.
• Autocatalysis: a reaction in which a product acts as the catalyst.

Specification

• I can state that transition metal characteristics of Ti to Cu arise from an incomplete d sub-level in atoms or ions, and explain why zinc is not a transition metal.
• I can list the four characteristic properties: complex formation, coloured ions, variable oxidation state and catalytic activity.
• I can define ligand, complex and co-ordination number, and work out the formula and charge of a complex ion.
• I can explain why NH₃ and H₂O swap without a change in co-ordination number, and write equations including the incomplete substitution that forms [Cu(NH₃)₄(H₂O)₂]²⁺.
• I can explain why substituting H₂O by Cl^- can change the co-ordination number, with equations for Co²⁺, Cu²⁺ and Fe³⁺.
• I can describe bidentate (H₂NCH₂CH₂NH₂, C₂O₄^2-) and multidentate (EDTA^4-) ligands, and write substitution equations using them.
• I can state that haem is an iron(II) complex with a multidentate ligand, explain how O₂ is transported by co-ordinate bonding to Fe(II), and explain why CO is toxic.
• I can explain the chelate effect in terms of the balance between the entropy change and the enthalpy change.
• I can draw octahedral, tetrahedral, square planar and linear complexes with their bond angles, and give an example of each (including [Ag(NH₃)₂]⁺ in Tollens' reagent).
• I can draw cis-trans isomers of octahedral and square planar complexes (cisplatin is the cis isomer) and optical isomers of octahedral complexes with bidentate ligands.
• I can explain how colour arises from absorption of visible light and d-electron excitation, and use ΔE = hν = hc/λ.
• I can explain how changes in oxidation state, co-ordination number and ligand change ΔE and the colour.
• I can describe how a colorimeter is used to find the concentration of a coloured ion, and read a concentration from a graph of absorbance against concentration.
• I can describe the reduction of vanadate(V) by zinc in acidic solution to vanadium(IV), (III) and (II), with the colours.
• I can state that redox potentials of transition metal ions are influenced by pH and by the ligand.
• I can describe how Tollens' reagent is reduced to silver, distinguishing aldehydes from ketones.
• I can carry out calculations for redox titrations of Fe²⁺ and C₂O₄^2- with MnO₄^-, and similar redox reactions.
• I can define heterogeneous and homogeneous catalysts, and explain active sites, support media and poisoning (with the cost implications).
• I can name V₂O₅ (Contact process) and Fe (Haber process) as heterogeneous catalysts, and explain with equations how V₂O₅ acts.
• I can explain the importance of variable oxidation states in catalysis.
• I can explain, with equations, how Fe²⁺ catalyses the reaction between I^- and S₂O₈^2-, and how Mn²⁺ autocatalyses the reaction between C₂O₄^2- and MnO₄^-.

Related notes

• Atomic structure (electron configurations of atoms and ions)
• Bonding (co-ordinate bonds, shapes of molecules)
• Kinetics (catalysts and activation energy)
• Oxidation, reduction and redox equations (half-equations, oxidation states)
• Amount of substance (titration calculations)
• Thermodynamics (3.1.8): ΔG = ΔH − TΔS for the chelate effect
• Electrode potentials and electrochemical cells (3.1.11): E^⦵ values in redox and catalysis
• Reactions of ions in aqueous solution (3.2.6): metal-aqua ions with hydroxide, ammonia and carbonate
• Aldehydes and ketones (3.3.8): Tollens' reagent
• Optical isomerism (3.3.7)