5.3.1 Transition Elements

For the Period 4 d-block, fill 4s before 3d when you write the atom. For example, Fe is [Ar] 3d⁶ 4s².

Two atoms are worth learning as exceptions to the simple pattern: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. If the question gives you an atomic number, count the electrons carefully before you write the subshells.

When a Period 4 d-block atom forms a positive ion, remove the 4s electrons before the 3d electrons. This is the part students most often reverse.

Fe is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶ and Fe³⁺ is [Ar] 3d⁵. Cu is [Ar] 3d¹⁰ 4s¹, so Cu²⁺ is [Ar] 3d⁹.

OCR defines a transition element as a d-block element that forms at least one ion with an incomplete d subshell.

The important phrase is “at least one ion”. You are classifying the element by an ion it can form, not by whether the neutral atom happens to have a partly filled d subshell.

Scandium and zinc sit in the d-block, but they fail OCR’s transition-element definition. Sc³⁺ is 3d⁰ and Zn²⁺ is 3d¹⁰, so neither element forms the required ion with an incomplete d subshell.

That leaves Ti to Cu as the Period 4 transition elements for this course.

Transition elements commonly form compounds in more than one oxidation state. Two familiar examples are Fe²⁺ and Fe³⁺, and chromium in Cr³⁺ and Cr₂O₇²⁻, where chromium is +6.

If an unfamiliar species is given, calculate the oxidation number from the overall charge. That is usually quicker and safer than trying to remember every example.

You should be comfortable with examples of coloured transition-metal ions: aqueous Cu²⁺ is blue, Fe²⁺ pale green, Fe³⁺ yellow and Mn²⁺ very pale pink. Chromium(III) depends on its exact complex and conditions, so use the colour stated for the species you have been given.

For OCR, recognise the species and use the observation. You do not need to explain how the colour arises.

Transition elements and their compounds often act as catalysts. OCR gives Cu²⁺ catalysing the reaction of zinc with acids and MnO₂ catalysing the decomposition of H₂O₂ as practical examples.

Here, you need the catalytic role: the catalyst increases the reaction rate without being used up overall. OCR does not ask you to learn detailed catalytic cycles for these examples.

Catalysts matter in industry because they let useful reaction rates be reached at lower temperatures, cutting energy demand and cost. Iron in the Haber process and V₂O₅ in the Contact process are familiar transition-metal examples.

If a question asks about the wider benefit, connect the lower energy demand to improved sustainability and reduced CO₂ emissions from fuel use. You do not need a detailed industrial catalyst mechanism for this OCR point.

A ligand is a species that donates a lone pair of electrons to a metal ion or metal atom to form a coordinate bond.

I would keep both parts in the definition: electron-pair donation and formation of the coordinate bond. H₂O, NH₃ and Cl⁻ are common OCR examples.

A monodentate ligand forms one coordinate bond to the metal ion. H₂O, NH₃ and Cl⁻ are examples. A bidentate ligand forms two coordinate bonds using two donor atoms.

For ethanediamine, H₂NCH₂CH₂NH₂, both nitrogen atoms donate a lone pair. When you draw a complex, make each bond go to the actual donor atom.

A complex ion contains a central metal ion surrounded by ligands joined to it by coordinate bonds. Put the whole complex inside square brackets and write its overall charge outside.

The coordination number is the number of coordinate bonds to the central metal ion. It is not necessarily the number of ligands: three bidentate ligands give coordination number 6.

A complex with six coordinate bonds is commonly octahedral. The six bonds point towards the corners of an octahedron, giving 90° angles between neighbouring bonds.

Typical examples are [Cu(H₂O)₆]²⁺ and [Fe(H₂O)₆]³⁺. When you draw one, show all six metal–ligand bonds clearly and include the square brackets and overall charge.

Some four-coordinate complexes are tetrahedral. Four coordinate bonds point towards the corners of a tetrahedron, with bond angles of about 109.5°.

Tetrachloro complexes such as [CuCl₄]²⁻ and [CoCl₄]²⁻ are the familiar OCR examples. Use wedges and dashed bonds when the question wants the 3-D shape shown clearly.

A four-coordinate complex can also be square planar: all four ligands lie in one plane around the metal, with 90° angles.

Platinum complexes are the important OCR example. Pt(NH₃)₂Cl₂ is square planar, which is why it can form distinct cis and trans isomers.

For cis–trans isomerism, the same ligands are arranged differently in space. In the cis form the matching ligands are next to each other; in the trans form they are opposite each other.

With square-planar Pt(NH₃)₂Cl₂, put the two Cl ligands at 90° for cis and 180° for trans. A rotation of the whole drawing does not create a new isomer.

Optical isomers are non-superimposable mirror images. OCR’s standard example is [Ni(en)₃]²⁺, where three bidentate ethanediamine ligands wrap around the metal in two mirror-image arrangements.

When you compare two drawings, imagine rotating one in 3-D. If rotation can make every ligand line up, they are the same structure; if the mirror images still cannot be superimposed, they are optical isomers.

Cisplatin is the cis isomer of Pt(NH₃)₂Cl₂. Its geometry places the two replaceable chloride ligands next to each other, allowing the platinum complex to bind at two nearby sites on DNA.

That geometry lets it form the DNA cross-links associated with the drug’s anti-cancer action. The trans isomer does not form the same effective cross-links.

Inside cells, ligand substitution makes the platinum centre able to form coordinate bonds to donor atoms on DNA. Cisplatin can then bind at two nearby positions and distort or cross-link the DNA.

The result you need for OCR is that this prevents normal DNA replication and cell division. Keep the chemistry centred on ligand substitution, coordinate bonding and the effect on DNA.

With excess aqueous ammonia, water ligands around Cu²⁺ are replaced by NH₃ to form [Cu(NH₃)₄(H₂O)₂]²⁺, a deep blue solution.

If ammonia is added gradually you may first see a blue Cu(OH)₂ precipitate. In excess ammonia that precipitate dissolves as the deep blue ammine complex forms.

Concentrated chloride ions replace the water ligands around Cu²⁺ to form tetrahedral [CuCl₄]²⁻.

OCR describes [Cu(H₂O)₆]²⁺ as blue and [CuCl₄]²⁻ as yellow. During the change a mixture of the two complexes can look green.

Excess ammonia forms [Cr(NH₃)₆]³⁺, which OCR describes as a purple solution. The overall ligand-substitution equation replaces all six water ligands with NH₃.

In the test tube, adding ammonia first gives a grey-green Cr(OH)₃ precipitate; with further ammonia the precipitate dissolves and the purple ammine complex forms.

Haemoglobin contains Fe²⁺ in a complex. Oxygen acts as a ligand and forms a coordinate bond to the Fe²⁺ centre, allowing haemoglobin to carry O₂ reversibly.

The oxygen can later be replaced by another ligand, so you can describe loading and unloading oxygen as ligand substitution at the iron centre.

Carbon monoxide also acts as a ligand, but it binds more strongly to the iron centre in haemoglobin than oxygen does.

CO therefore occupies sites that would otherwise carry O₂ and is not displaced easily. Less oxygen can be transported around the body, which is why carbon monoxide is toxic.

These five precipitates are worth learning as a set: Cu(OH)₂ blue; Fe(OH)₂ dark green; Fe(OH)₃ orange-brown; Mn(OH)₂ pale brown; Cr(OH)₃ dark green or grey-green.

You can form them with aqueous NaOH, or initially with aqueous NH₃ because ammonia produces enough OH⁻ in water to precipitate the metal hydroxide.

For a simple precipitation equation, combine the metal ion with enough OH⁻ to make the neutral hydroxide. The charge tells you the number of hydroxide ions you need.

So Cu²⁺ needs 2OH⁻, while Fe³⁺ and Cr³⁺ need 3OH⁻. Check the charge before you finish; it catches most mistakes quickly.

Of the five OCR transition-metal hydroxides here, only Cr(OH)₃ dissolves in excess NaOH. It reacts with extra OH⁻ to form the soluble complex [Cr(OH)₆]³⁻, giving a dark green solution.

Cu(OH)₂, Fe(OH)₂, Fe(OH)₃ and Mn(OH)₂ remain as precipitates in excess NaOH.

In excess aqueous ammonia, Cu(OH)₂ dissolves to give the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺ solution and Cr(OH)₃ dissolves to give purple [Cr(NH₃)₆]³⁺.

The Fe²⁺, Fe³⁺ and Mn²⁺ hydroxide precipitates do not dissolve in excess NH₃. That difference is very useful when you are identifying an unknown ion.

Treat the two reagents as a small decision tree. First record the precipitate colour with a small amount of reagent, then add the reagent in excess and see whether the precipitate dissolves.

Cu²⁺ is especially distinctive because its blue precipitate dissolves in excess NH₃ to give a deep blue solution. Cr³⁺ can dissolve in excess NaOH and in excess NH₃. The iron and manganese precipitates stay insoluble in both excess reagents.

The core redox pair is Fe³⁺ + e⁻ ⇌ Fe²⁺. Aqueous Fe²⁺ is pale green and aqueous Fe³⁺ is yellow, so oxidation of Fe²⁺ moves pale green → yellow and reduction of Fe³⁺ reverses that change.

OCR may use acidified MnO₄⁻ to oxidise Fe²⁺ or I⁻ to reduce Fe³⁺. You are not expected to memorise every full equation: use the supplied chemistry and balance the electron transfer.

For the reduction direction, orange Cr₂O₇²⁻ becomes Cr³⁺, which is observed as green in the usual acidified dichromate reduction. The useful half-equation is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

For oxidation, OCR may use H₂O₂ in alkaline solution: Cr³⁺ is oxidised to yellow chromate(VI), CrO₄²⁻; acidifying then gives orange dichromate(VI). Follow the reagents and colours supplied in the question.

Iodide ions can reduce Cu²⁺ to Cu⁺. In water, the Cu⁺ is seen as white CuI precipitate, while iodide is oxidised to iodine.

The balanced equation is useful because it shows both changes at once: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. If the question gives observations, look for the white precipitate together with the brown colour from iodine.

In aqueous solution, Cu⁺ readily disproportionates: one Cu⁺ ion is oxidised to Cu²⁺ while another is reduced to Cu metal.

The equation is 2Cu⁺ → Cu²⁺ + Cu. If electrode-potential data are supplied, you can use them to show that this overall reaction is feasible; you do not need a separate memorised mechanism for the instability.

For an unfamiliar ligand-substitution reaction, identify the metal centre, incoming ligand and leaving ligand first. Then keep the metal oxidation state unchanged and make sure the final complex has the right overall charge.

Draw bonds to the donor atoms, not just to whichever end of the ligand is easiest to sketch. OCR often gives enough structural information for you to work this out without having seen the exact complex before.

If the reaction is unfamiliar, use the same rules as the named tests. Work out the metal ion charge, identify the precipitating ligand or OH⁻ source, then balance the formula and charge of the solid.

Use the observations in the question as evidence. A stated colour, solubility change or coordination number can identify the species, so make that clue part of your answer.

Start an unfamiliar transition-metal redox problem by assigning oxidation numbers. Decide what is oxidised and what is reduced, then use half-equations or electron changes to balance the reaction.

After that, connect the chemical species to the observation you were given. OCR deliberately tests whether you can use clues such as a green Cr³⁺ solution, an orange dichromate ion or a precipitate to identify what has happened.