Properties and Uses of Transition Metals
Senior 6 · Chemistry
Properties and Uses of Transition Metals
Lesson 2 of 5
25 min
Dissolve a little ammonium vanadate(V) in dilute sulfuric acid and the solution is yellow. Drop in some granulated zinc and swirl: over a few minutes the solution turns blue, then green, then violet. Nothing has been added except zinc, which is a reducing agent. Each colour is vanadium in a different oxidation state. Calcium could never do this; its compounds are always calcium(II). This lesson explains why transition metals can, and why that makes them such good catalysts.
Definition
Variable oxidation states: the ability of an element to exist in compounds in more than one oxidation state. Example: manganese is +2 in , +4 in and +7 in .
The 4s and 3d electrons are close in energy, so an atom can lose or share different numbers of them with no sudden, very large jump in the energy needed. Compare successive ionisation energies:
Table 2.1: Successive ionisation energies / kJ mol⁻¹
| Element | 1st | 2nd | 3rd | 4th |
|---|---|---|---|---|
| Calcium, [Ar] | 590 | 1145 | 4912 | 6491 |
| Iron, [Ar] | 762 | 1561 | 2957 | 5290 |
Calcium's third electron must come from the inner 3p subshell, so its third ionisation energy is about four times its second: calcium stops at +2. Iron's third electron comes from 3d, and the rise is gradual, so iron forms both +2 and +3 compounds. The extra energy is repaid by stronger lattice or hydration energies of the more highly charged ion.
Figure 2.1: Oxidation states of the first-row d-block elements in their compounds.
Figure 2.1 shows the pattern. From Sc to Mn the highest oxidation state equals the total number of 4s and 3d electrons: +3 for Sc () up to +7 for Mn (). After manganese the rising nuclear charge holds the 3d electrons more tightly, so fewer can be used, and the highest state falls. High oxidation states are found in oxoanions such as and , where the metal is covalently bonded to oxygen rather than existing as a bare ion.
Worked example
Problem. Find the oxidation state of the metal in (a) , (b) , (c) , (d) .
Solution. Oxygen is −2 and cyanide is −1; the sum of the oxidation states equals the charge on the ion.
(a)
(b)
(c)
(d)
Answer: chromium(VI), manganese(VII), vanadium(V), iron(III).
Because they change oxidation state easily, transition-metal compounds make good oxidising and reducing agents. You met two in S5 titrations. In acid, purple manganate(VII) is reduced to almost colourless , and orange dichromate(VI) to green :
The vanadium colours in the photograph come from the reduction (+5) (+4) (+3) (+2). The first step is
Worked example
Problem. 25.0 cm³ of a solution of iron(II) sulfate, acidified with dilute sulfuric acid, needs 22.40 cm³ of 0.0200 mol dm⁻³ potassium manganate(VII) to reach the end point. Find the concentration of .
Solution. Each gives one electron and each takes five, so
Answer: 0.0896 mol dm⁻³. No indicator is needed: the first drop of excess manganate(VII) turns the solution pale pink.
Definition
Catalyst: a substance that increases the rate of a reaction by providing a route with a lower activation energy, and is chemically unchanged at the end.
Heterogeneous catalyst: in a different phase from the reactants, usually a solid with gases or liquids reacting on its surface. Example: iron in the Haber process.
Homogeneous catalyst: in the same phase as the reactants, usually all in aqueous solution. Example: ions in the reaction between peroxodisulfate and iodide ions.
Transition metals have partly filled d orbitals that can form weak bonds with reactant molecules. The surface adsorbs the reactants, which weakens the bonds inside them and holds them close together in a good orientation. The product then desorbs, freeing the site for the next molecules (Figure 2.2). The adsorption must be just right: too weak and nothing is held long enough; too strong and the product never leaves.
Figure 2.2: How an iron surface catalyses the Haber process. A simplified model: in reality millions of sites work at once.
Table 2.2: Two industrial processes that depend on transition-metal catalysts
| Haber process | Contact process | |
|---|---|---|
| Product | ammonia, for fertilisers such as urea and ammonium nitrate | sulfur trioxide, then sulfuric acid |
| Catalysed step | ||
| Catalyst | finely divided iron, with potassium oxide and aluminium oxide as promoters | vanadium(V) oxide, |
| Typical conditions | about 450 °C and 200 atm | about 450 °C and 1–2 atm |
| How it works | adsorption on the iron surface (Figure 2.2) | vanadium changes oxidation state (below) |
In the contact process, vanadium(V) oxide first oxidises sulfur dioxide and is itself reduced to vanadium(IV) oxide; oxygen then re-oxidises it:
Doubling the first equation and adding the second gives : the catalyst cancels out, as it must.
Peroxodisulfate ions oxidise iodide ions only slowly, because both are negative and repel each other:
A few drops of iron(II) sulfate speed it up greatly. Each step is now between ions of opposite charge:
The is regenerated. works equally well, because it simply starts the cycle at the second step. A product can even catalyse its own reaction: in the titration of ethanedioate with manganate(VII), the formed speeds the reaction up. This is autocatalysis.
Warning
Catalyst poisons. Impurities that adsorb strongly block the active sites. Sulfur compounds poison the iron catalyst in the Haber process, so the feed gases are purified first. Platinum was once used in the contact process but was replaced by vanadium(V) oxide, which is cheaper and much less easily poisoned by arsenic impurities.
Tip
In an exam answer on transition-metal catalysis, name the mechanism: heterogeneous means adsorption, weakened bonds, reaction and desorption on a surface; homogeneous means the metal ion changes oxidation state and is then regenerated. Always write both steps, and check that the catalyst cancels.
Chemistry Student: Catalysts and Homogeneous and Heterogeneous Catalysis (11 min). Watch for the adsorption steps on a solid surface and for the two-step iron catalysis of the peroxodisulfate and iodide reaction.
Practice
Sources and credits
