d-Block and Transition Elements

Electron arrangements of the d-block elements

The elements in between Group 2 and Group 13 are correctly referred to as the d-block elements. This is because the highest energy subshell (i.e. the subshell being filled) is a d-subshell.

Recall that:

  • the 3d subshell and the 4s subshell are actually very close together in energy, but 4s is slightly lower in energy, so the 4s fills before 3d
  • a d-subshell contains five d-orbitals, so can hold up to 10 electrons
  • electrons go one into each orbital first, before any orbital gets a second electron

When we examine the electron arrangements of the first row of d-block elements we find two anomalies:

tm-electron-arrgt

A rather simplistic and inadequate explanation for this is to say that having all the d-orbitals half-filled is an especially stable arrangement, so Cr has 3d and 4s all half filled rather that the expected 3d4 4s2 arrangement. Mo shows the same effect in Period 5. In the same way it can be suggested for Cu that having the 3d orbitals all full is more stable than having the expected 3d9 4s2 arrangement. In other periods, Ag and Au show the same effect. A more satisfactory explanation is beyond the scope of the A-level course.

Electron arrangements of the ions of d-block elements      

Putting electrons into 3d shields the 4s orbital from the nuclear charge, and the 4s electrons now become easier to remove than the 3d i.e. 3d is now the lower energy subshell. As a result, when we start to remove electrons from the d-block atoms to form ions, we remove the 4s electrons before we remove any from 3d.

e.g. iron forms green Fe2+ and orange Fe3+ ions

Fe        1s2 2s2 2p6 3s2 3p6 3d6 4s2

Fe2+     1s2 2s2 2p6 3s2 3p6 3d6          – 2 × 4s electrons removed

Fe3+     1s2 2s2 2p6 3s2 3p6 3d5          – 2 × 4s electrons and 1 × 3d electron removed

e.g. vanadium forms violet V2+ ions and green V3+ ions

V          1s2 2s2 2p6 3s2 3p6 3d3 4s2

V2+       1s2 2s2 2p6 3s2 3p6 3d                          – 2 × 4s electrons removed

V3+       1s2 2s2 2p6 3s2 3p6 3d2                      – 2 × 4s and 1 × 3d electron removed

BUT

Zinc only forms colourless 2+ ions,

Zn        1s2 2s2 2p6 3s2 3p6 3d10 4s2

Zn2+     1s2 2s2 2p6 3s2 3p6 3d10             – 2 × 4s electrons removed

Scandium only forms colourless 3+ ions,

Sc        1s2 2s2 2p6 3s2 3p6 3d1 4s2

Sc3+      1s2 2s2 2p6 3s2 3p6                 – 2 × 4s and 1 × 3d electron removed

We might propose that colour in these ions is associated with having a partially-filled d-subshell, and experimentation would provide a physical basis for this suggestion.

We also have the basis for the definition of a transition element – which is NOT the same as a d-block element :

A transition element is a d-block element that forms at least one stable ion with a partially filled d sub-shell. Therefore in Period 4, Zn and Sc are d-block elements but NOT transition elements, whereas Ti – Cu are both d-block elements AND transition elements.

Transition Metal Chemistry

The partially-filled d-subshell in transition elements, and the ease with which electrons can be removed from, or added to these orbitals, gives them their characteristic properties:

  • forming coloured compounds and complexes
  • acting as catalysts
  • having compounds with multiple oxidation states

All transition metals form compounds with ions with +2 oxidation number; in most cases this is due to losing the two electrons from the 4s orbital. However, the 3d electrons can also be lost allowing transition metals to form stable ions with higher oxidation numbers. This happens because the 3d and 4s energy levels are so close in energy.

Ions of transition metals also readily change from one oxidation state to another, by accepting or donating electrons from the 3d subshell.

An illustration of colour and multiple oxidation states

A solution containing vanadate(V) ions is yellow, but when a reducing agent such as zinc is added, the oxidation state can be reduced from +5 in steps down to +2 with a corresponding change in colour at each stage:

VO2+(aq)                        Oxidation state: +5  Colour: yellow

VO2+(aq)                        Oxidation state: +4  Colour: blue*

* usually an intermediate green solution is seen containing a mixture of VO2+ and VO2+ ions before the blue appears.

V3+(aq)               Oxidation state: +3  Colour: green

V2+(aq)               Oxidation state: +2  Colour: lilac

In each stage, the vanadium has been progressively reduced. At the same time, the zinc has been oxidized   (Zn(s)  → Zn2+(aq)) so this is a series of redox reactions.

Catalytic behavior in transition metal chemistry

Catalysts provide an alternative reaction pathway with lower activation energy. We classify catalysts as homogeneous (in the same physical state as the reactants) or heterogeneous (in a different physical state to the reactants). 

Examples of heterogeneous catalysis

  • Iron (solid) is used as the catalyst for the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g)
  • Solid vanadium(V) oxide, V2O5, is used as the catalyst for the step in the Contact process to manufacture sulphuric acid that oxidises sulphur dioxide to sulphur trioxide: SO2(g) + ½O2(g) ⇌ SO3(g)
  • Nickel is used in the catalyst in the hydrogenation of margarines, saturating (some of) the alkene functional groups in polyunsaturated vegetable oils: -CH=CH- + H2 → -CH2-CH2
  • Manganese(IV) oxide, MnO2, is used to catalyse the decomposition of hydrogen peroxide into oxygen and water: H2O2(aq) → H2O(l) + ½O2(g)

The reaction between solid zinc and acids (producing a zinc salt and hydrogen) is catalyzed by the presence of copper(II) ions in the solution. The copper(II) ions react with the surface of the solid zinc in a displacement reaction, forming copper metal. The copper metal then acts as the catalyst for the reaction between the zinc and the acid.

Examples of homogeneous catalysis

In solution, transition metal compounds can provide an alternative reaction pathway for redox reactions by reacting with one reactant, then being regenerated in the reaction with the other reactant.   e.g. if A reacts with B in a redox reaction, with A being oxidised and B reduced, the alternative pathway might consist of A being oxidized by the catalyst, which is itself reduced, then the catalyst reducing B, and being oxidized back to its original form.

  • Hydrogen peroxide oxidises tartrate ions to carbon dioxide gas and methanoate ions. The reaction proceeds very slowly if there is no catalyst, even at elevated temperatures.rochelle-decomp-eqnThe presence of cobalt(II) ions in solution catalyses the reaction.  The hydrogen peroxide initially oxidises the pink Co2+, to green Co3+, and is reduced to water.            The cobalt(III) ion causes oxidiation of the tartarate ion, and as a result of which the Co3+ is reduced back to Co2+ and the pink colour returns.
  • The reaction between iodide ions, I, and peroxydisulphate ions, S2O82- forms sulphate ions and iodine: 2I(aq) + S2O82-(aq)  →  2SO42-(aq) + I2(aq)

This formation of iodine can be highlighted by the addition of starch, which forms a blue-black complex with the iodine. The reaction is catalyzed by the presence of Fe2+(aq), and when a small amount of this is added the blue-black colour forms much more quickly, with the Fe2+ ions left unchanged at the end of the reaction.

        Step 1:    S2O82-(aq) + 2Fe2+(aq) → 2SO42-(aq) + 2Fe3+(aq)    reaction involving catalyst

        Step 2:    2I(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)                    regeneration of catalyst

Complexes and Colour

The solid compounds and aqueous ions of transition metals have characteristic colours:

  • Cu2+(aq) ions are blue
  • Co2+(aq) ions are pink
  • Fe2+(aq) ions are pale green
  • Fe3+(aq) ions are yellow-orange
  • Mn2+(aq) ions are pale pink

By contrast, compounds of Zn and Sc are white (colourless in solution). This is because Zn2+ ions have a 3d10 electron arrangement, and Sc3+ have no 3d electrons, so neither has the incompletely filled d-orbitals necessary for colour in transition metal compounds.

However, consider copper(II) sulphate. It is blue when hydrated crystals or in solution, but white when anhydrous. The copper ions are [Ar]3d9 in every case. We must conclude that the unfilled d-subshell is only one of the necessary requirements for colour.

 When in solution and in the hydrated crystal lattice, the transition metal ions are not merely dissolved in water, but they are actually bonded to a number of water molecules, by dative covalent bonds formed from a lone pair on water’s oxygen atom to the metal ion.

e.g.         the blue Cu2+(aq) ion is actually  [Cu(H2O)6]2+(aq) cu6h2o-complex

This is an example of a complex. Because it has an overall charge, we call it a complex ion.

Because transition metal ions are small they have a strong electric field around them that attracts electron-rich species. Species that have a lone pair available to donate can form dative covalent bonds to the transition metal ion. We call such species ligands.

A ligand is a molecule or ion that can donate a lone pair to form a dative covalent bond to a transition metal ion.

It is the presence of these ligands around the central transition metal ion that is the other necessary requirement for colour.

Examples of ligands:

  • water    H2O:              
  • ammonia  :NH3               
  • chloride  :Cl                  
  • cyanide  :CN                 
  • thiocyanate  :SCN               
  • hydroxide  :OH

The dative covalent bonds around the transition metal ion repel one another, and so are arranged around the transition metal ion in geometric arrangements positioning them as far apart as possible (like in VSEPR Theory). The number of dative covalent bonds arranged around a transition metal is called the co-ordination number.

  • Complex ions with a co-ordination number of 6 tend to be octahedral
  • Complex ions with a co-ordination number of 4 are tetrahedral or occasionally square planar

Drawing complexes, and writing formulae

Formulae for complex ions are written in square brackets with the overall charge on the brackets. No individual charges on ligands or the metal ion are shown. Ligands are in brackets unless single atoms, with the number of each ligand after the bracket if more than one.

Complex ions are drawn in 3D using shaded and wedge bonds to represent bonds into and out of the plane of the paper. The bonds between ligands and the transition metal ion are dative covalent bonds, so we sometimes use an arrow to indicate the direction in which the electron pair has been donated (towards the transition metal ion), but this is optional.

As with the formula, square brackets are used and the overall charge written at the top right.

Examples:cu6h2o-complex

1)        [Cu(H2O)6]2+           octahedral

pale blue                  Name: hexa aqua copper II ion

Aqueous solutions of all the transition metal ions (with no other ligands present) have a co-ordination number of 6 and are octahedral.

2)         [CuCl4]2-              tetrahedralcucl4-complex-ion

yellow                      Name: tetra chloro cuprate II ion

It is the size of the ligands that has the most significant effect on the geometry of the complex ions – chloride ions are much larger than water molecules, so only four chloride ions will pack around the small transition metal ion, where six water molecules would otherwise fit.

3)         [Fe(H2O)5SCN]2+         octahedralfescn-complex

blood red         Name: penta aqua thiocyanato iron III ion

The ligands don’t all have to be the same, here five water ligands and a thiocyanate ligand form dative bonds to the iron(III) ion. Notice also that the overall charge on the complex ion is 2+ even though the iron is in +3 oxidation state, because of the 1- charge on the SCN ion.

Crystallising complexes

Complex ions are not found on their own – there is always an oppositely-charged ion to balance the charges although we may ignore it much of the time. We may focus on the [Cu(H2O)6]2+ ion when we dissolve copper(II) sulphate in water, forgetting the SO42- ion is also present in equal concentrations. Similarly when we form [CuCl4]2- by the addition of hydrochloric acid to copper sulphate solution, we have two H+ ions present for every [CuCl4]2- ion.

This is of no consequence when we are considering solutions – the “other” ions are spectators, but when we evaporate the water and form crystals, these other ions become part of the giant ionic lattice along with the complex ions.

e.g. K3Fe(CN)­6 contains the complex ion [Fe(CN)6]3- and three K+ ions to balance the charges.   Name: potassium hexacyanoferrate(III)

Typical ions we may find in this role include Na+, K+ or NH4+. These other ions are NOT ligands and are NOT bonded to the central metal ion.

Ligand Substitution

When a new ligand is added to a solution containing a complex ion, the new ligand can replace an original ligand to form a different complex. This is known as ligand substitution.

cunh4complexExamples:

1) Addition of excess concentrated ammonia solution to Cu2+(aq) results in the formation a solution containing a dark blue complex ion.  Both complexes are octahedral.

[Cu(H2O)6]2+ + 4NH3  [Cu(NH3)4(H2O)2]2+ + 2H2O

 

2) If concentrated hydrochloric acid is added to a solution cucl4-complex-ion
containing Cu2+(aq), the solution turns green and progressively more yellow as chloride ligands replace the water molecules. The new complex is yellow, but the solution containing both complexes appears green.

[Cu(H2O)6]2+ + 4Cl [CuCl4]2- + 6H2O

Multidentate ligands

All the ligands we have seen so far have formed a single dative covalent bond to the transition metal ion. We refer to these as monodentate (literally: one-toothed) ligands.

Larger molecules may have more than one site that is able to donate a lone pair, and so may be able to form more than one dative covalent bond to the transition metal ion.

Examples:salicylate-ion

salicylate (2-hydroxybenzoate) ion

Salicylic acid (a benzene ring with an adjacent –OH and –COOH group) does not act as a ligand, but the salicylate ion does (presumably because it is more electron-rich).

The O of the –OH group and the –O of the –COO group each have a lone pair which can form a bond to the transition metal ion, so two dative covalent bonds are formed. This as a bidentate ligand.

When salicylate ions are added to a solution containing yellow aqueous iron(III) ions, the six water ligands are replaced with three salicylate ligands to form a deep purple complex of iron(III) salicylate. (Note that this complex has no charge – it is not a complex ion, just a complex).

[Fe(H2O)6]3+(aq) + 3HOC6H4COO(aq)[Fe(HOC6H4COO)3](s) + 6H2O(l)

 

EDTA (ethylenediaminetetraacetic acid)edta-ion   

This ligand exists in complexes as the EDTA4- ion. It has six lone pairs which can form
dative covalent bonds, and is hence a hexadentate ligand.

EDTA is used to bind metal ions, removing them from solution, and is referred to as a chelating agent. Uses include binding calcium and magnesium ions to reduce water hardness, or being added to blood to treat patients suffering from lead or mercury poisoning.

ethane-1,2-diamine    NH2CH2CH2NH2               en-ion

This has two N atoms each with a lone pair that can form a bond to the transition metal ion. This is therefore a bidentate ligand.

For example, nickel forms an octahedral complex with co-ordination number 6 when it bonds with three of these ligands:     [Ni(NH2CH2CH2NH2)3]2+

 

ethanedioate ions  C2O42-  oxalate-ion

The ethanedioate ion is a bidentate ligand. Each O can form a dative covalent bond to the transition metal ion.

Cr3+ forms an octahedral complex with two cr-oxalate-complexethanedioate ions and two water molecules: [Cr(C2O4)2(H2O)2]

 

 

We might be given the structure of a complex and asked to work out and draw the ligands present, indicating how they act as ligands. To do this we need to break the dative covalent bonds to the transition metal ion and restore the lone pair to the atoms in the ligand that formed the bond:pttestcplx

e.g.   [Pt(C2O4)( HOC6H4OCOO)]  

 

1) Take the complex ion apart.

pttestapart

2) Look for atoms whose valency is not satisfied with bonds – one too few bonds indicates a negative ion.

pttestions

3) Add in the lone pairs which formed the dative covalent bonds to the metal ions.

pttestlps

Biological role of ligand substitution

Haemaglobin in blood is responsible for the transport of oxygen to cells for respiration, and the transport away of carbon dioxide.

heam-unit
Haemaglobin is a complex protein, containing four non-protein components called haem groups, each of which has an Fe2+ ion at its centre, and four dative covalent bonds between the iron and four N-atoms in the haem structure, and a further dative covalent bond to the protein globin.

haemoglobinOxygen can bind to the iron in the haem group as a ligand, giving a co-ordination number of 6, in order to be transported. The colour of the complex when oxygen is bound as a ligand is the rich red of oxygenated blood.

Carbon monoxide can also bind to the iron at the same binding site as the oxygen would, the stability constant for this complex is greater (see later) so the bond between the iron and the CO is stronger and much less likely to break. CO will therefore bind preferentially if both carbon monoxide and oxygen are present in the lungs; the binding is irreversible, so that haemaglobin becomes useless. Since tobacco smoking produces carbon monoxide, and this is one of the reasons why smokers become short of breath.

Isomerism in complex ions

Stereoisomerism arises when the ligands can be arranged in different spatial arrangements. In complexes, as with organic substances, we will see both cis-trans isomerism and optical isomerism.

Cis – Trans isomerism

When a complex contains four of one ligand and two of another, we can have cis and trans isomers.

e.g. cobalt forms an octahedral complex of formula [Co(NH3)4Cl2]+. The cobalt is in oxidation state +3 here. When salts containing this complex ion were crystallised, it was found that a green salt and a purple salt both with the same formula could be crystallized. This was clear evidence for two different isomers:

complex-cis-trans

Cis-trans isomerism is also possible in some complexes with co-ordination number 4 if they have a square-planar configuration.

Nickel forms a square planar complex [NiCl2(NH3)2]. This has no overall charge, as the nickel is in the +2 oxidation state.

sqpl-cis-trans

A similar complex of platinum [PtCl2(NH3)­2] is one of the most effective drugs in chemotherapy for cancer. It is the cis-form of the complex which is biologically active. The drug is a liquid, usually administered intravenously as a drip, and goes under the name of cis-platin.cis-platin

The importance of the exact shape and structure of the molecule is emphasized by the fact that the trans-form is ineffective.

The benefit of using cis-platin to treat cancer is that it works by binding to the DNA of fast-growing cancer cells, changing the DNA structure and preventing them from dividing to reproduce. Chemotherapy has unpleasant side effects, however, and a new generation of cancer-treatment drugs requiring lower doses and with fewer side effects than cis-platin have been developed, e.g. carboplatin.

Optical isomerism

Optical isomerism arises with octahedral complexes containing multidentate ligands, where these can be arranged around the central transition metal atom as non-superimposable mirror images.   Recall that optical isomers rotate plane-polarised light differently – one isomer will rotate it to the right, and the other to the left. An equal mixture of isomers will have no effect on plane polarized light because the rotations cancel out.

Examples:

[Ni(en)3]2+   – a complex with three bidentate ligands

ni-en-optical-isomers

[Co(en)2(H2O)2]2+ – a complex with two bidentate ligands and two monodentate ligands. Complexes like this will have cis and trans isomers, but there will be two cis isomer are optical isomers of each other.

co-optical-isomers

 

Case studies in Transition Metal chemistry

Reactions of copper compounds

i) Cu2+(aq) with sodium hydroxide

Blue aqueous copper(II) ions react with hydroxide ions to form a blue precipitate of copper(II) hydroxide. The precipitate is insoluble in excess sodium hydroxide.

Cu2+(aq) + 2 OH(aq) → Cu(OH)2(s)

ii) Cu2+(aq) with ammonia solution

Remember that ammonia forms an equilibrium reacting with water when it dissolves, NH3(aq) + H2O(aq) ⇌ NH4+(aq) + OH(aq) so ammonia solution is both a source of NH3 and OH ligands.

Initially it is the OH ions that react with the aqueous copper(II) ions, forming a blue copper hydroxide precipitate as above.

When excess ammonia solution is added, the copper hydroxide precipitate redissolves undergoing a ligand substitution reaction to form a dark blue solution containing the octahedral trans-complex of [Cu(NH3)4(H2O)2]2+

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

iii) Cu2+(aq) with concentrated hydrochloric acid

When concentrated hydrochloric acid is added to blue aqueous copper(II) ions, a yellow solution containing tetrahedral [CuCl4]2-(aq) is formed in a ligand substitution reaction, although a green colour is often seen when both the aqueous ions and the new complex ion are present in solution. As this is an equilibrium, the colour of the solution changes as the relative concentrations of chloride ions and water present change.

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

iv) Cu2+(aq) with iodide ions

Aqueous copper(II) ions react with iodide ions in a redox reaction that forms brown aqueous iodine and a white precipitate of copper(I) iodide. Note that the copper(I) ion has a 3d10 electron arrangement, so it is not surprising with no partially filled d-subshell that it is white.

cu-and-i2-redox

Copper(I) compounds are unstable in solution. When solid copper(I) oxide reacts with hot dilute sulphuric acid, a brown precipitate of copper and a blue solution of copper(II) sulphate are formed in a disproportionation reaction, where Cu(I) ions are oxidised to Cu(II) and reduced to metallic Cu.

Cu2O(s) + H2SO4(aq) CuSO4(aq) + Cu(s) + H2O(l)

Reactions of iron compounds

i) Fe2+(aq) and Fe3+(aq) with sodium hydroxide

Pale green iron(II) ions in solution react with sodium hydroxide to form a dirty green precipitate of iron(II) hydroxide. On standing the precipitate turns brown at its surface as iron(II) hydroxide is oxidized to iron(III) hydroxide.

Fe2+(aq) + 2OH(aq) Fe(OH)2(s)

Pale yellow iron(III) ions in solution react with sodium hydroxide to form an orange-brown precipitate of iron(III) hydroxide.

Fe3+(aq) + 3OH(aq) Fe(OH)3(s)

Neither precipitate redissolves on addition of excess sodium hydroxide.

 

ii) Fe2+(aq) and Fe3+(aq) with ammonia solution

It is only the hydroxide ions in the ammonia solution that react with iron(II) and iron(III) ions to form iron(II) hydroxide and iron(III) hydroxide exactly as above. There is no further reaction with the ammonia present, and the precipitates do not redissolve.

 

iii) oxidation of iron(II) with acidified managanate(VII)

Pale green aqueous iron(II) ions will decolourise purple managante(VII) ions in acidic solution in a redox reaction. The iron(II) is oxidized to pale yellow aqueous iron(III) ions, while the manganate(VII) is reduced to pale pink aqueous manganese(II) ions.

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

 

iv) reduction of iron(III) with iodide ions

Pale yellow aqueous iron(III) ions can be reduced to pale green iron(II) ions by redox reaction with colourless aqueous iodide ions. In this reaction iodine is formed in aqueous solution, which is brown, and masks the colour change of the iron ions.

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

Reactions of manganese compounds

i) Mn2+(aq) with sodium hydroxide

Pale pink aqueous manganese(II) ions react with sodium hydroxide to form a pale pinky-brown precipitate of manganese(II) hydroxide, which darkens on standing in air. The precipitate is does not redissolve or react further on further addition of sodium hydroxide.

Mn2+(aq) + 2OH(aq) Mn(OH)2(s)

 

ii) Mn2+(aq) with ammonia solution

It is only the hydroxide ions in the ammonia solution that react with manganese(II) ions to form manganese(II) hydroxide exactly as above. There is no further reaction with the ammonia present, and the precipitates do not redissolve.

 

Reactions of chromium compounds

The colour of the complex [Cr(H2O)6]3+(aq) is pale purple. This may be a surprise since we have consistently referred to the colour change when acidified dichromate(VI) ions are used as an oxidizing agent as orange for dichromate(VI) and green for chromium(III). This is because the chromium(III) ion formed in the presence of sulphuric acid is actually not [Cr(H2O)6]3+ but [Cr(H2O)5(SO4)]+ and this is the green ion we have been referring to as Cr3+(aq).

i) Cr3+(aq) with sodium hydroxide

Pale purple aqueous chromium(III) ions react with sodium hydroxide to form a green-grey precipitate of chromium(III) hydroxide.

Cr3+(aq) + 3OH(aq)  Cr(OH)3(s)

On addition of excess sodium hydroxide, this precipitate redissolves to form a dark green solution containing the complex ion [Cr(OH)6]3-(aq).

Cr(OH)3(s) + 3OH(aq) →  [Cr(OH)6]3-(aq)

 

ii) Cr3+(aq) with ammonia solution

Initially it is the OH ions that react with the aqueous chromium(III) ions, forming a green-grey chromium hydroxide precipitate as above.

When excess ammonia solution is added, the chromium hydroxide precipitate redissolves undergoing a ligand substitution reaction to form a purple solution containing the octahedral complex [Cr(NH3)6]3+

[Cr(H2O)6]3+(aq)+ 6NH3(aq)  [Cr(NH3)6]3+(aq) + 6H2O(l)

 

iii) oxidation of chromium(III) with alkaline hydrogen peroxide

Hot alkaline hydrogen peroxide is a powerful oxidizing agent, and reacts in a redox reaction to oxidise aqueous chromium(III) ions (green or pale purple, depending on the ligands bonded to the chromium) to yellow chromate(VI) ions.

3H2O2(aq) + 2Cr3+(aq) + 10 OH(aq) →  2CrO42-(aq) + 8H2O(l)

 

iv) reduction of dichromate(VI) with zinc under acidic conditions

The dichromate(VI) ion is orange in aqueous solution. It can be reduced in a redox reaction with a reducing agent such as zinc metal under acidic conditions, forming chromium(III) ions in solution. The chromium(III) ion formed is likely to be green, as a ligand from the acid will form part of the Cr3+(aq) complex.

Cr2O72-(aq) + 14H+(aq) + 3Zn(s)  2Cr3+(aq) + 3Zn2+(aq) + 7H2O(l)

If an excess of zinc is used, the chromium(III) ions in solution can be reduced in a further redox reaction to pale blue chromium(II) ions.

Zn(s) + 2Cr3+(aq) →  Zn2+(aq) + 2Cr2+(aq)