10.6. Chemistry of Chromium Cr, Z=24, 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ 

data comparison of chromium with the other members of the 3d–block and transition metals

Extended data table for CHROMIUM

• Uses of CHROMIUM

  • Chromium is a hard bluish–white metal that is extremely resistant to chemical attack at room temperature e.g. very resistant to oxidation.

  • Chromium is used in the production of extremely hard steel alloys e.g. ball bearings.

  • Chromium metal is an important component in 'stainless steel'.

  • Chromium is used to electroplate other metals like steel because of its anti–corrosion properties ('chrome/chromium plating').

  • Chromium(III) oxide, Cr₂O₃ is used in stained glass and a catalyst in the chemical industry.

  • Chromium(IV) oxide is used in magnetic tapes for sound/video recording.

• Biological role of chromium

  • Chromium is an essential trace element, but its role in the body is unknown?

• Extraction of chromium

  • Chromium ore is processed and purified into chromium(III) oxide. This is reacted, very exothermically, in a thermit style reaction, with aluminium (see reactions of aluminium) to free the chromium metal.

  • Cr₂O_3(s) + 2Al_(s) ==> Al₂O_3(s) + 2Cr_(s) 

  • The chromium(III) oxide is reduced to chromium by O loss, the aluminium is oxidised to aluminium oxide by O gain, and the aluminium is the reducing agent i.e. the O remover.

• These are examples of metal displacement reactions e.g. the less reactive chromium or titanium are displaced by the more reactive sodium, magnesium or aluminium.

The Chemistry of CHROMIUM

CHROMIUM(III) oxidation state chemistry

• Chromium forms the stable 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 simply Cr(OH)_3(s) + 3H⁺_(aq) Cr³⁺_(aq)  + 3H₂O_(l)

      • thus showing amphoteric behaviour, since the hydroxide ppt. also dissolves in excess strong alkali to give a dark green solution and the hydroxide ppt. does not dissolve in the weak base aqueous 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 base–alkali 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–

  • VIEW ppts. with OH^–, NH₃ and CO₃^2–, & complexes, if any, with excess reagent.

			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–).

• Chromium(III) ions with aqueous sodium carbonate form the green hydroxide precipitate (as above) and carbon dioxide because of the acidity of the hexaaquachromium(III) ion (see Appendix 1.):

  • *initially 2[Cr(H₂O)₆]³⁺_(aq) + CO₃^2–_(aq) ==> 2[Cr(H₂O)₅(OH)]²⁺_(aq) + H₂O_(l) + CO_2(g)   

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

  • No Cr₂(CO₃)₃ is formed because of the acid–base reaction above, due to the acidity of the chromium(III) ion. Note the similarly highly charged small ions Al³⁺ and Fe³⁺ behave in the same way.

  • * the acidity of a the hexa–aquachromium(III) ion can be expressed as ...

    • [Cr(H₂O)₆]³⁺_(aq) + H₂O_(l) [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 hexa–aqua ion)

    • or [Cr(OH)₃(H₂O)₃]_(s) + 3OH^–_(aq) ==> [Cr(OH)₆]^3–_(aq) + 3H₂O_(l) (from hydroxide ppt.)

      • or more simply Cr(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) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 6H₂O_(l)   (from original hexa–aqua ion)

    • or [Cr(OH)₃(H₂O)₃]_(s) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 3OH^–_(aq) + 3H₂O_(l) (from hydroxide ppt.)

      • or more simply Cr(OH)_3(s) + 6NH_3(aq) ==> [Cr(NH₃)₆]³⁺_(aq) + 3OH^–_(aq)

  • The uncharged ligand molecules ammonia NH₃ and water H₂O are similar in size and ligand exchange occurs without change in co–ordination number. They all octahedral complexes with a co–ordination number of 6.

• Chromium(III) complexes are extremely numerous and varied, including many examples of isomerism. (see Appendix 2 and Appendix 3 for an introduction to complexes)

• Ionisation isomerism in chromium(III) chloride based on Cr³⁺, 3Cl^– and 6H₂O 

  • [Cr(H₂O)₆]³⁺(Cl^–)₃  (violet or grey–blue?)

  • [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? ^*the ⁰ to 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.

  • 

  • 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.

  • 

• 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, ethane–1,2–diamine (ethylenediamine), is often represented in shorthand by en,

    • 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 L–L represents H₂NCH₂CH₂NH₂

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

• Both the hexa–aqua ions of chromium(II) and chromium(III) readily complex with EDTA

  • [Cr(H₂O)₆]²⁺_(aq) + EDTA^4–_(aq) ===> [Cr(EDTA)]^2–_(aq) + 6H₂O_(l)

    • K_stab = {[Cr(EDTA)₃]^2–_(aq)} / {[Cr(H₂O)₆]²⁺_(aq)} [EDTA^4–_(aq)]

    • K_stab = 1.0 x 10¹³ mol^–1 dm³ [lg(K_stab) = 13.0]

  • [Cr(H₂O)₆]³⁺_(aq) + EDTA^4–_(aq) ===> [Cr(EDTA)]^–_(aq) + 6H₂O_(l)

    • K_stab = {[Cr(EDTA)₃]^–_(aq)} / {[Cr(H₂O)₆]³⁺_(aq)} [EDTA^4–_(aq)]

    • K_stab = 1.0 x 10²⁴ mol^–1 dm³ [lg(K_stab) = 24.0]

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

• –

• The electrode potential chart highlights the values for various oxidation states of chromium.

  • It shows the weak oxidising or weak reducing power of chromium(III), Cr³⁺(aq), the strong reducing power of chromium(II), Cr²⁺(aq) and the strong oxidising power of chromium(VI), Cr₂O₇^2–

• Summary of some complexes–compounds & oxidation states of chromium compared to other 3d–block elements

CHROMIUM(VI) oxidation state chemistry

• The '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, CrO₃ and 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 non–metal.

  • If the oxidation state of the central metal ion is over +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 the oxyanion.

    • ie the theoretical [Cr(H₂O)₆]⁶⁺ ends up in reality as Cr₂O₇^2– or CrO₄^2– depending on pH.

    • The reason for this situation is that the 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 M–OH₂ and M–OH ligand bonds in two stages to leave a M=O bond in the oxyanion

        • ie considering one co–ordinated water molecule (and ignoring the charge on intermediate complexes) ...

          • M–OH₂  ==proton loss==>  M–OH  ==proton loss==>  M=O

  • There is, theoretically, always an equilibrium between the chromate(VI) ion and the dichromate(VI) ion.

    • 2CrO₄^2–_(aq) + 2H⁺_(aq) Cr₂O₇^2–_(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 the yellow chromate(VI) ion CrO₄^2– . 

  • 2Cr³⁺_(aq) + 3H₂O_2(aq) + 10OH^–_(aq) ==> 2CrO₄^2–_(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 H₂O₂ and Cr(VI) compounds are oxidising agents but in alkaline solution H₂O₂ is the stronger oxidising agent.

    • E^Ø = +?V details to add???

  • When the resulting solution from above is acidified with dilute sulphuric acid, the orange dichromate(VI) ion Cr₂O₇^2–  is 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 <= more yellow (net change R to L).

  • 2CrO₄^2–_(aq) + 2H⁺_(aq) Cr₂O₇^2–_(aq) + H₂O_(l) (no change in ox. state)

• The dichromate(VI) ion is reduced in two stages by a zinc/dilute sulphuric acid mixture.

  • Cr(VI, +6) ==> Cr(III, +3): Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– 2Cr³⁺_(aq) + 7H₂O_(l)

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

  • Cr(III, +3) ==> Cr(II, +2): Cr³⁺_(aq) + e^– Cr²⁺_(aq)

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

  • Note the  E^Ø_{Zn(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:

    • Cr₂O₇^2–_(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 by–product of the zinc–acid reaction but the reductions take place on the surface of the zinc.

• Potassium dichromate(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.

  • Cr₂O₇^2–_(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 of redox 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.

• Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6I^–_(aq) ==> 2Cr³⁺_(aq) + 3I_2(aq) + 7H₂O_(l)

  • The released iodine can be titrated with standard sodium thiosulphate solution using starch indicator.

  • 2S₂O₃^2–_(aq)  +  I_2(aq)  ==>  S₄O₆^2–_(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 dm^–3) using a few drops of starch solution as an indicator. Iodine gives a blue colour with starch, so, the end–point is very sharp change from the last hint of blue to colourless.

      • See also fully worked examples of redox 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) + CrO₄^2–_(aq) ==> PbCrO_4(s) and 2Ag⁺_(aq) +  CrO₄^2–_(aq) ==> Ag₂CrO_4(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

  • K_sp = [Ag⁺_(aq)][Cl^–_(aq)] = 2 x 10^–10 mole²dm^–6

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

  • K_sp = [Ag⁺_(aq)]²[CrO₄^2–_(aq)] = 3 x 10^–12 mole³dm^–9, until all the chloride is precipitated. The next drop of silver nitrate causes the precipitation of brownish–red silver chromate, so the end point is the formation of the dark red ppt

• CHROMIUM(II) oxidation state chemistry:

  • The blue 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^Ø = –0.26V, O₂+H⁺/H₂O E^Ø = +1.23V , see Redox Electrode Potential Chart