The Chemistry of CHROMIUMCHROMIUM(III)oxidation state chemistryChromium forms thestable green(greyish dark green almost violet sometimes?)chromium(III) ion, [Cr(H2O)6]3+(aq).

Aqueous solutions of chromium(III) chloride are suitable for investigating the aqueous chemistry of the chromium(III) ion.

With aqueous ammonia (alkaline) or sodium hydroxide, chromium(III) ions form a green gelatinous precipitate of chromium(III) hydroxide.

Cr3+(aq)+ 3OH(aq)==>Cr(OH)3(s)(but the structures can be quite complex)

or[Cr(H2O)6]3+(aq)+ 3OH(aq)==>[Cr(OH)3(H2O)3](s)+ 3H2O(l)

The hydroxide readily dissolves in acids to form salts,

Cr(OH)3(s)+ 3H+(aq)==>Cr3+(aq)+ 3H2O(l)

or more elaborately:[Cr(OH)3(H2O)3](s)+3H3O+(aq)[Cr(H2O)6]3+(aq)+ 3H2O(l)

or more simplyCr(OH)3(s)+3H+(aq)Cr3+(aq)+ 3H2O(l)thus showingamphoteric behaviour, since thehydroxide ppt. also dissolves in excess strong alkali to give a dark green solutionand the hydroxide ppt. doesnot dissolve in the weak baseaqueous sodium carbonate. However, it will dissolve in excess ammonia because a new green complex ion is formed. (more details on these reactions below)

The whole sequence of each theoretical step of chromium(III) hydroxide precipitation and its subsequent dissolving in strong basealkali is shown the series of diagrams below.

All are, for simplicity, treated as octahedral complexes of 6 ligands either water H2O or hydroxide ion OH

[Cr(H2O)6]3+=> [Cr(OH)(H2O)5]2+=> [Cr(OH)2(H2O)4]+=> [Cr(OH)3(H2O)3](s)precipitate

dissolving => [Cr(OH)4(H2O)3]=> [Cr(OH)5(H2O)]2=> [Cr(OH)6]3

The sequence of chromium(III) hydroxide precipitate formation and its subsequent dissolving in excess strong alkali. Each step is essentially one of proton removal from each complex (from 3+ to 3).

1234From 1 to 7 happen as you add more alkali, increasing pH and the OHconcentration, removing protons from the chromium(III) complex.

567*From 7 back to1 represents what happens when you add acid, decreasing pH, increasing H+/H3O+concentration and protonating the chromium(III) complex.

Chromium(III) ions with aqueous sodium carbonateform the greenhydroxide precipitate(as above) andcarbon dioxidebecause of the acidity of the hexaaquachromium(III) ion (see Appendix 1.):*initially2[Cr(H2O)6]3+(aq)+ CO32(aq)==>2[Cr(H2O)5(OH)]2+(aq)+ H2O(l)+ CO2(g)

this process of proton donation (deprotonation) continues until[Cr(OH)3(H2O)3](s)is formed

No Cr2(CO3)3is formed because of the acidbase reaction above, due to the acidity of the chromium(III) ion. Note the similarly highly charged small ionsAl3+andFe3+behave in the same way.

*the acidity of a the hexaaquachromium(III) ion can be expressed as ...[Cr(H2O)6]3+(aq)+ H2O(l)rev[Cr(H2O)5(OH)]2+(aq)+ H3O+(aq)

With excess sodium hydroxide or ammonia, further complex ions are formed from chromium(III) ions by ligand displacement/replacement reactions:[Cr(H2O)6]3+(aq)+ 6OH(aq)==>[Cr(OH)6]3(aq)+ 6H2O(l) (from original hexaaqua ion)

or[Cr(OH)3(H2O)3](s)+ 3OH(aq)==>[Cr(OH)6]3(aq)+ 3H2O(l)(from hydroxide ppt.)

or more simplyCr(OH)3(s)+ 3OH(aq)==>[Cr(OH)6]3(aq)

showing amphoteric behaviour, since the hydroxide ppt. also dissolves in acid.

[Cr(H2O)6]3+(aq)+ 6NH3(aq)==>[Cr(NH3)6]3+(aq)+ 6H2O(l) (from original hexaaqua ion)

or[Cr(OH)3(H2O)3](s)+ 6NH3(aq)==>[Cr(NH3)6]3+(aq)+ 3OH(aq)+ 3H2O(l)(from hydroxide ppt.)

or more simplyCr(OH)3(s)+6NH3(aq)==>[Cr(NH3)6]3+(aq)+ 3OH(aq)

The uncharged ligand molecules ammonia NH3and water H2O are similar in size and ligand exchange occurs without change in coordination number. They all octahedral complexes with a coordination number of 6.

Chromium(III) complexes are extremely numerous and varied, including many examples of isomerism.

Ionisation isomerism in chromium(III) chloridebased on Cr3+, 3Cland 6H2O[Cr(H2O)6]3+(Cl)3(violet or greyblue?)

[CrCl(H2O)5]2+(Cl)2.H2O(pale green)

[CrCl2(H2O)4]+Cl.2H2O(dark green)

[CrCl3(H2O)3]0*.3H2O?(brown?, this I found reference to on a Russian website, doesn't seem to be in textbooks?*the0to signify an overall electrically neutral complex can be omitted)

and this is not all, the 3rd one down with two chloride ligands can exist as cis (1) or trans (2) geometric isomers (Z/E isomers) illustrated below, and also serve as models for representing the other octahedral complexes which exhibit cis/trans or Z/E isomerism.

(c) doc b

With excess chloride ion you get the formation of the tetrahedral tetrachlorochromate(III) ion[Cr(H2O)6]3+(aq) + 4Cl(aq) ==> [CrCl4](aq) + 6H2O(l)

You also get geometrical cis/trans isomers (Z/E) with tetraamminedichlorochromium(III) complexes.

(c) doc b

A similar case of isomerism occurs with the chromium(III) complexes with ammonia and chloride ligands shown above. All the complex ions above have a plane of symmetry and cannot exhibit optical isomerism.Again, these are all octahedral complexes with a coordination number of 6.

[Cr(H2NCH2CH2NH2)3]3+, H2NCH2CH2NH2, ethane1,2diamine (ethylenediamine), is often represented in shorthand byen,and the complex simply written as[Cr(en)3]3+.

This complex has mirror image forms i.e. enantiomers of optical isomers.This optical isomerism can be illustrated thus

where LL represents H2NCH2CH2NH2

The ligand bonds via the lone pairs of electrons on the nitrogen which are donated to form the metalligand dative covalent bonds.

Both the hexaaqua ions of chromium(II) and chromium(III) readily complex withEDTA[Cr(H2O)6]2+(aq)+ EDTA4(aq)===> [Cr(EDTA)]2(aq)+ 6H2O(l)Kstab= {[Cr(EDTA)3]2(aq)} / {[Cr(H2O)6]2+(aq)}[EDTA4(aq)]

Kstab= 1.0 x 1013mol1dm3[lg(Kstab) = 13.0]

[Cr(H2O)6]3+(aq)+ EDTA4(aq)===> [Cr(EDTA)](aq)+ 6H2O(l)Kstab= {[Cr(EDTA)3](aq)} / {[Cr(H2O)6]3+(aq)}[EDTA4(aq)]

Kstab= 1.0 x 1024mol1dm3[lg(Kstab) = 24.0]

From the Kstabvalues, you can see that the more highly charged Cr3+(aq)ion complexes more strongly than the Cr2+(aq)ion.

CHROMIUM(VI)oxidation state chemistryThe 'simple' hexaaquachromium(VI) cation, [Cr(H2O)6]6+, cannot exist in aqueous media.In fact, I doubt if the 'simple' Cr6+ion can exist in any ionic compound.

Note that chromium(VI) oxide, CrO3and chromium(VI) fluoride, CrF6, (both compounds have Cr in +6 oxidation state), are covalent compounds, despite the relatively large electronegativity difference between the metal and nonmetal.

If the oxidation state of the central metal ion isover +3, it appears that deprotonation via proton transfer to water is so facilitated that in most cases (there may be exceptions?) all protons are 'theoretically' lost to give theoxyanion.ie the theoretical [Cr(H2O)6]6+ends up in reality asCr2O72orCrO42depending on pH.

The reason for this situation is thatthe high charge density of the 'theoretical' central metal ion, gives it a high polarising power.You can theoretically conceive the situation of imagining the central metal ion pulling on the electrons of the MOH2and MOH ligand bonds in two stages to leave a M=O bond in the oxyanionie considering one coordinated water molecule (and ignoring the charge on intermediate complexes) ...MOH2 ==proton loss==> MOH ==proton loss==> M=O

There is, theoretically, always an equilibrium between the chromate(VI) ion and the dichromate(VI) ion.2CrO42(aq)+ 2H+(aq)revCr2O72(aq)+ H2O(l)

Therefore from Le Chatelier's principle, high pH (alkaline) favours the formation of the chromate(VI) ion and low pH (acid) favours dichromate(VI) ion formation.

When hydrogen peroxide is added to an alkaline chromium(III) solution, oxidation occurs to give theyellow chromate(VI) ion CrO42.2Cr3+(aq)+ 3H2O2(aq)+ 10OH(aq)==>2CrO42(aq)+ 8H2O(l)

Redox changes: oxidation 2Cr(+3) ==> 2Cr(+6), reduction 6 O(1) in 3H2O2==> 6(2) in 6 of the 8H2O, total of 6 'units' oxidation state change.

Both H2O2and Cr(VI) compounds are oxidising agents but in alkaline solution H2O2is the stronger oxidising agent.E= +?V details to add???

When the resulting solution from above is acidified with dilute sulphuric acid, theorange dichromate(VI) ion Cr2O72is formed.

The equilibrium is pH dependent. From 'Le Chatelier's Principle':in more acidic solution, more H+, decrease pH ==> more orange (net change L to R) or in

more alkaline, less H+(removed by OH), increase pH Cr(III, +3):Cr2O72(aq)+ 14H+(aq)+ 6e2Cr3+(aq)+ 7H2O(l)

orange(+6) ==>green(+3), E= +1.33V

Cr(III, +3) ==> Cr(II, +2):Cr3+(aq)+ eCr2+(aq)

green(+3) ==>blue(+2), E= 0.41V, so Cr(II) is readily oxidised by dissolved oxygen and can only be retained in an inert atmosphere.

Note theEZn(s)/Zn2+(aq)is 0.76V,so the reducing power of zinc is sufficient to effect either of the two chromium oxidation state reduction changes.

The full redox equations for the reactions which happen on the surface of the zinc are:

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

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

You will see hydrogen formed as a byproduct of the zincacid reaction but the reductions take place on the surface of the zinc.

Potassiumdichromate(VI), K2Cr2O7, can be crystallised to high purity standard without water of crystallisation, and is a valuable 'standard' redox volumetric reagent.

e.g. It can used to titrate iron(II) ions in solution acidified with dilute sulphuric acid, using a redox indicator like barium diphenylamine sulphonate which is less readily oxidised than iron(II) ions, but once all the iron(II) ions are oxidised the indicator is oxidised to a blue colour.

The iron(III) ions formed affect the indicator to give an inaccurate end point so phosphoric(V) acid is also added at the start to complex the Fe3+ions as they form.

Cr2O72(aq)+ 14H+(aq)+ 6Fe2+(aq)==>2Cr3+(aq)+ 6Fe3+(aq)+ 7H2O(l)

Theoretically, there are actually two simultaneous colour changes, both masked by the redox indicator change.

The orange dichromate(VI) ion changes on reduction to the green chromium(III) ion,

and the pale green iron(II) ion changes on oxidation to the orange iron(III) ion,

so without the indicator I'm not sure exactly how the colour change you would really observe would pan out!See also fully worked examples ofredox volumetric titration calculation questions.

The dichromate(VI) ion is a strong oxidising agent examples of oxidising action:

See above for oxidation of iron(II) ions.

It oxidises iodide ions to iodine.

Cr2O72(aq)+ 14H+(aq)+ 6I(aq)==>2Cr3+(aq)+ 3I2(aq)+ 7H2O(l)The released iodine can be titrated with standard sodium thiosulphate solution using starch indicator.

2S2O32(aq) + I2(aq) ==>S4O62(aq)+ 2I(aq)(black/brown ==> colourless endpoint)

This reaction between the released iodine and sodium thiosulfate can be used to estimate oxidising agents like dichromate(VI) ions.The iodine is titrated with standardised sodium thiosulphate (e.g. 0.10 mol dm3) using a few drops of starch solution as an indicator. Iodine gives a blue colour with starch, so, the endpoint is very sharp change from the last hint of blue to colourless.See also fully worked examples ofredox volumetric titration calculation questions.

Soluble chromate(VI) salts give yellow solutions, but lead(II) ions give a yellow ppt. of lead(II) chromate(VI) and silver ions a dark red ppt. of silver chromate(VI).

Pb2+(aq)+ CrO42(aq)==>PbCrO4(s)and2Ag+(aq)+ CrO42(aq)==>Ag2CrO4(s)A few drops of silver chromate is used as an indicator when titrating chloride solutions with silver nitrate solution in neutral solution. The solubility product for the white ppt. of silver chloride

Ksp= [Ag+(aq)][Cl(aq)] = 2 x 1010mole2dm6

is exceeded before the solubility product of silver chromate(VI)

Ksp= [Ag+(aq)]2[CrO42(aq)] = 3 x 1012mole3dm9,until all the chloride is precipitated. The next drop of silver nitrate causes the precipitation of brownishred silver chromate, so the end point is the formation of the dark red ppt

CHROMIUM(II) oxidation state chemistry:Theblue hexaaquachromium(II) ion, [Cr(H2O)6]2+(aq),can be formed by reducing chromium(III) salt solutions with zinc and hydrochloric acid but it is rapidly oxidised back to green chromium(III) ions by dissolved oxygen unless protected by an inert atmosphere.V3+/V2+E= 0.26V, O2+H+/H2O E= +1.23V