Chromium Revisited

Chromium Revisited

A stroll through the oxidation states and colors of Chromium. Pt1.

About 25 yrs. ago I did a series of simple experiments to show my teenaged son how varied this fascinating element’s chemistry was. Recently I felt a yen to repeat these and did so. I thought it might be worth while to suggest some ideas, and chromium seems to be rarely mentioned, so here goes.

Note 1: All chromium compounds are poisonous, in the order Cr[II] < Cr[III] < CrO4 - - and Cr2O7- - .Figures for LD50 rat oral are approx. 1900,1800,180, 25 mg/kg. NACN is around 6mg/kg. Hence the chromates are very poisonous, about 1/4 as lethal as cyanide. They can be absorbed though the skin, so wear gloves.

Note 2: The colors quoted here are as I perceive them, in daylight under partly cloudy skies. In general many of the Cr compounds show quite different colors in artificial light, different grades of fluorescent and especially incandescent, due to illuminating spectrum. Eyes differ – my wife sees blues where I see greens. Photo sensors also vary – the only way is to see them for yourself and decide. Do the experiments!

I will run through the oxidation states of Cr, starting at 0, the metal itself (and, due to the definition used, that of a few organic Cr compounds such as carbonyl, Cr(CO)6).

(1) Metallic Cr, Cr[0], can be made by the amateur quite easily by thermite reduction of the oxide Cr2O3 (lots of data on this site, UTSE). Another method is by electrolysis of acidified Cr2(SO4)3 on to a copper cathode. Good hard shiny coatings are hard to do and usually the coat can be peeled off. A mercury cathode can be used to make an amalgam. An iron cathode can be used (if inserted with current on) and will produce a black powered Cr metal easily scraped off.

The Cr[I] oxidation state does not exist as far as I know (AFAIK), nor do distinct hydrides, although the metal can occlude hydrogen like all transition metals.

(2) For the following experiments, prepare a saturated solution of about 50-100cc of a chromic Cr[III] salt, usually sulphate or chloride. Chrome Alum for growing crystals is often available and this is what I use, KCr(SO4)2.12H2O. This and the sulphate are both dark reddish-violet crystals. DO NOT HEAT to assist solution for reasons below, dissolve at RT, first grinding the alum or sulphate to a fine powder. {If you use the chloride, CrCl3.6H2O it will probably come as dark green crystals rather than violet}. In solution the alum acts the same as the sulphate, the K+ are merely spectator ions.

(3) Aqua Complexes:

(3a) Heat a small amount of the violet (purple by transmitted light) solution of Cr[III] sulphate to boiling for a few minutes. The solution turns from violet to deep green. On cooling it preserves the green color. It will stay green for several days to weeks but finally return to the original violet. The green form is chemically different from the violet form as can be shown by precipitation of the sulphate ion with a barium salt, and also by electric conductivity or freezing point measurement. The green salt does not form a precipitate. The mystery was solved in the early 1900s; the violet ion is the complex Cr(H2O)6+++; the dark green one maybe [Cr(H2O)4(SO4)]+; or perhaps [Cr2(H2O)l0(SO4)]++++ or the covalent [Cr2(H2O)6(SO4)3], or a mixture of all three.

Only the violet form will crystallize. If the green form is heated to dryness, only green scales result. Further cautious heating produces an orange-yellow product which is the anhydrous form with perhaps some decomposition. Strong heating (bright red heat) causes decomposition and SO3 evolution: Cr2(SO4)3  Cr2O3 + 3SO3

For the chloride of Cr[iii], Wiki has:

“Chromium(III) chloride (also called chromic chloride) is a violet coloured solid with the formula CrCl3. The most common form of CrCl3 sold commercially is a dark green hexahydrate with the formula [CrCl2(H2O)4]Cl.2H2O. Two other hydrates are known, pale green [CrCl(H2O)5]Cl2.H2O and violet [Cr(H2O)6]Cl3. This unusual feature of chromium(III) chlorides, having a series of [CrCl3−n(H2O)n]z+, each of which is isolable, is also found with other chromium(III) compounds.”

This indicates the facility with which Cr forms complexes, in this case with water. It is also interesting to note that both Cr[III] sulphate and chloride, if anhydrous, are virtually insoluble in water. The crystalline form of sulphate is Cr2(SO4)3.12H2O or, according to the above, [Cr(H2O)6]2(SO4)3.6H2O, and is very soluble.

(4) Cr[III] Compounds

{The temptation to insert a Pourbaix or Eo/pH diagram here was great – if you understand them they tell a lot at a glance. But many do not and it might merely confuse.}

The salts of Cr+++ with strong acids all give an acid reaction in solution – in other words Cr(OH)3 is a weak base; being a bit more technical, the ion Cr+++ has a pKa of about 4 whereas a strong base ion like Na+ has pKa~14.

(4a) Take a test-tube of the violet sulphate solution and drip in some fairly concentrated solution of NaOH (or KOH). Until the hydroxide neutralizes some of the acidity nothing happens, then a flocculent precipitate of a light gray-green color settles out. If too much hydroxide is added, this precipitate re-dissolves. The precipitate is hydrated Cr[III] hydroxide, Cr(OH)3.xH2O. If instead the green form of sulphate is used, the formation is slower but the same precipitate obtained. If done carefully, the liquid becomes colorless and depleted of Cr[III] ions. If the precipitate is left long enough (or centrifuged) the color becomes a very distinct green color and compacts, probably due to loss of water.

Reaction is simply Cr+++ + 3(OH)- + xH2O  Cr(OH)3. xH2O (s); x is approx. 3 when dried carefully. In alkaline solutions Cr[III] always exists as the hydroxide, except when the pH is high.

(4b)Addition of excess hydroxide causes the precipitate to redissolve giving a light green solution. This is usually stated to be due to chromite ion Cr(O2)-, or a hydrated form of this. This is still a Cr[III] ion, probably O=Cr—O-. Reducing the pH with acid re-precipitates the hydroxide.

(4c) Using NH4OH a similar precipitate with a grayish color is first formed. As further hydroxide is added carefully, again the liquid will clear. If excess ammonia is added, and the solution left for an hour or so, this gelatinous ppt. will turn a light purple, quite different in color from the sodium hydroxide case. On standing for a few days, or on prolonged boiling, the purple color disappears as ammonia is lost, leaving a gray-blue hydroxide. My guess is that the purple is a pentamminechromium complex; [Cr(NH3)5.(OH)x](3-x)(OH). yH2O.

{The colors of the Cr[III] ammine chloride complexes are; tri+++, yellow; tetraCl2+, green; pentaCl++, purple; non ionic tri Cr(NH3)3Cl3, violet and the ion Cr(NH3)2Cl4-, orange-red (Pauling, General Chem.)}.

See Brauer, p1345ff, for more on hydroxides and the organic complexes of Cr[III].

(4d) The hydroxide has a very low solubility product in water, about 6x10^-31. It can be purified by washing with large amounts of water and can be dried as a dull green powder, hydrated, and is a useful source of Cr[III] salts with common mineral acids. On heating to red heat it loses water to become a bright green oxide, 2Cr(OH)3 - -> Cr2O3 + 3 H2O. This oxide, the most stable form of Cr[III] under mild conditions, is usually insoluble in acids but can be reacted with fused alkali hydroxides to form chromates – see later on. Structure is O=Cr-O-Cr=O.

Most of the Cr[III] salts exist as blue-violet or green types when hydrated. As large crystals the violet forms look black; as powder, reddish. Chromium and potassium sulphate are isomorphous and crystallize together in all proportions, giving crystals that vary from lightest pink to almost black. Chrome alum is the equimolecular mixture. It can form massive octahedral crystals with patience – I have grown them up to 7cms. long looking as black as coal.

Insoluble Cr[III] salts are the normal phosphate {blue} CrPO4, can be hydrated; sodium carbonate solution produces a similar precipitate to hydroxide that may be a basic carbonate or Cr2(CO3)2.xH2O. The color is purple – Cr[III] also produces CO2 complexes.

Correction to Part 1 Above.

……sodium carbonate solution produces a similar precipitate to hydroxide that may be a basic carbonate or Cr2(CO3)2.xH2O. The color is purple – Cr[III] also produces CO2 complexes…..

I re-ran this one today and now find that the color quoted is nearer lavender or light blue gray. Nor does it dissolve in excess Na2CO3. A short boiling does not convert the color, except to maybe reduce the bluish tint to slightly more gray. Try it and see!

But it is not green as several texts have it. (But I have not yet tried the green form of the sulphate) All texts also say this is not a carbonate but a hydroxide and on further effort I tend to agree. If the precipitate is washed many times with water, to eliminate carbonate contamination, and treated with acid, very little CO2 is evolved. It dissolves to give the violet form of Cr[III]+++ aqua complex.

I think the difference in color is probably due to crystal size, a frequent cause of such phenomena.As for CO2 complexes, this too is ruled out. I think this is a Der Alte aberration; I must have been thinking of the carbonyl, Cr(CO)6.

Remembering that the (violet) ion is [Cr(H2O)6]+++ and is acidic, we have in water

[Cr(H2O)6]+++(aq) + H2O(liq) < -- > [Cr(H2O)5(OH)]++(aq) + H3O+(aq);

So one would expect Na2CO3 + Cr+++ salt to produce CO2 provided the pH is low enough. It does (best seen in conc. Cr salt solution, else the CO2 remains dissolved). As the salt is gradually neutralized by OH- ions from the dissolved Na2CO3 or NaOH a sequence of exchanges between the H2O ligands and the OH- ligands can be assumed to take place:

[Cr(H2O)6]+++(aq) => [Cr(OH)(H2O)5]++(aq) => [Cr(OH)2(H2O)4]+(aq)

=> [Cr(OH)3 (H2O)3] (Solid precipitate);

And if the pH exceeds ~11 afforded by Na2CO3 in solution, as in the case of NaOH solution, this precipitate dissolves to form negative aqua complex ions:

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

as the pH goes from 11+ to 14+.

Whether the so-called chromites like NaCrO2 really exist I do not know. Mellor has:

“...when hydrated chromic oxide is treated with sodiumhydroxide, the formation of chromite precedes the formation of the hydroxide.With sodium hydroxide below 102V, primary sodium chromite is formed, whilstabove 102V the soln. contains also tertiary sodium chromite. Potassium chromiteis similarly produced ; below 82V-alkali only the primary chromite is formed, whilstabove 82V the soln. contains also secondary chromite. From soln. of potassiumchromite which have stood for a long time, needle-shaped crystals of the formulaCr2O3.3K2O.8H2O have been obtained.”

(V stands for the inverse of concentration, liters/mol, IIRC). Best of luck if you want to try it!

The precipitated hydroxide aqua complex does dissolve in strong NH3 solution, in spite of its weakness as an alkali. It switches water for NH3 in the hexa coordinated complex,

[Cr(H2O)6]+++ + 6NH3 => [Cr(NH3)6]+++ + 6H2O etc.

For such beauties as Hexaamminechromium (III) Chloride, etc., see Brauer– you are on your own.

The overall message is that Cr[III] preferentially forms hexa-coordinated compounds, ionic and covalent, unless anhydrous. And anhydrous CrCl3, eg, is insoluble if pure – and covalent.

I performed the original experiments over a few months last year. These notes come from that record.

Chromium Revisited – Part II

A bit more about Cr2O3 and Cr(OH)3:

In the first part of this series the attention was paid to Cr[III] compounds and a few of their complexes. The most stable form of Chromium under normal conditions is in fact Cr2O3 which to my surprise has a higher heat of formation (lower enthalpy) than even the fluoride (that entropy term at work, no doubt!). Again I am severely tempted to add a Pourbaix diagram at this point, but again I resist for two reasons; one, it may confuse and two, I cannot find a suitable simple one at the 1M Cr+++ concentration indicative of typical experimental conditions. For those familiar with the Eo/pH diagrams, the one in Wiki under Chromium can be consulted, but it is for 10^-5M Cr+++ concentration. Change of concentration changes the position of vertical boundaries especially but moves all boundaries except, of course, the (dashed) water lines.

You will, for instance, never see Cr(OH)3 on Pourbaix diagrams (OK, I wouldn’t bet on it) because they show the most thermodynamically stable form under the Eo and pH condition of the axes at STP. That is Cr2O3; it covers a wide area.

2Cr(OH)3 -- > Cr2O3 + 3H2O + 2.13E05 j/mole

(see a discussion of issues re Eo/pH diags for Cr/water: it has several but they cannot be copied – secured pdf)

In other words, CR(OH)3 is metastable at RT and the equilibrium lies well to the right in the above equation. You can show this experimentally as follows. {There, I knew I ought never to have mentioned Pourbaix diagrams….}

(4e) This takes a bit of time to carry out. One must first produce a working quantity of Cr2(OH)3.xH2O from a Cr+++ compound as outlined previously – aim to get 5-10 g reckoned as unhydrated. Wash copiously with water several times. The let settle and decant, finally filtering through a fine mesh filter paper. Dry carefully in an oven at ~ 120-140C for several hours, until the mud becomes a powder of a dull green color.

Try reacting some of this with dilute acid – it should react fairly rapidly and give the violet form of the Cr[III] salt. Scrape the powder from the filter paper and place in an evaporating dish and heat in the oven to around 250C. This still does not totally dehydrate the hydroxide but also does not seriously convert it to oxide. This can be proven by reaction with acid or strong alkali. Now take the powder and heat slowly on an iron spoon over a propane flame. Somewhere around a dull red heat (500-600C, I would guess) if you are lucky you may see it caloresce – suddenly brighten and then dim. {This did not work the first time I tried it – it needs a fair quantity (at least 5g) to work, IMHO}. Now heat strongly. On cooling the power still has a dull green color but it almost totally non-reactive to strong acid or alkali solutions, showing that Cr2O3 is very unreactive.

The experiment demonstrates that Cr(OH)3 is changed to the oxide by heating strongly enough, but is not exactly unstable. In fact it can be kept indefinitely at RT in the dried state without losing reactivity. The calorescence is probably due to a change of state (entropy change) as the amorphous structure left on ignition of the hydroxide changes to crystalline Cr2O3 at some transition temperature..

Other ways to make chromic oxide are to heat ammonium dichromate (NH4)2Cr2O7 or K2Cr2O7 with sulphur. Both ways tend to produce a bright green product, especially the second (which is fun to perform!).

(NH4)2Cr2O7 -- > N2 + 2H2O + Cr2O3

K2Cr2O7 + S - -> K2SO4 + Cr2O3

(4d)Heat some solid potassium hydroxide (or NaOH) with about the same quantity of potassium nitrate (or NaNO3; chlorates can be used) in a nickel (or SS) crucible or dish until melted together. Add some Cr2O3 made by one of the above methods or otherwise available. It reacts to produce Chromate, K2Cr2O4, a yellow substance. The Cr[III] state is oxidized to the Cr[VI] state easily under the extreme alkaline conditions – even air will do this, slowly. More of this later.

It is worth noting that Chromium is quite a reactive metal, slightly less so than Zinc but more so than iron, yet chromium is plated on to iron (over Ni or Cu) and is very resistant to corrosion. This is believed to be because of Cr2O3 formation. The metal does dissolve in fairly dilute HCl but nitric acid passivates it, and sulphuric also has little effect, and even when concentrated has to be heated to react with any speed.

Chromium Revisited – Part IIIa

IUPAC Gold Book defines a transition element as one whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell]. The electronic structure of Cr in the ground (unionized gaseous) state is [Ar].3d5.4s1; it has six valence electrons outside the argon electron structure.

For technical reasons (as the physicists are wont to say whenever the explanation is too involved in a cloud of Quantum Mechanics) the 4s level has a lower energy level than 3d in the atoms K, atomic number (Z) =19 and Ca Z=20. That is, 4s is bound more strongly to the nucleus. And hence the d levels are not occupied first. The simplified explanation is that electrons in the s orbitals penetrate the Ar shells and hence see a higher part of the nuclear charge, while the d orbitals do not. As Z increases the Ar core decreases in size due to the increased nuclear charge and the 4s levels begin to approach the 3d levels.

The first appearance of the 3d orbitals is at Sc, [Ar].3d1.4s2; past Zn, [Ar].3d10.4s2, the 3d and 4s shells are filled and the 4p subshell starts at Ga. At vanadium Z=23 the 4s and 3d levels approach equal energy, and at Cr Z=24 the 4s electron actually has a higher energy by 0.96 eV (Pauling). This means the 4s electron is lost first in compounds of Cr.

The 3d block elements have no chemically similar elements of lower atomic weight and represent a new series with new chemical properties. Neither zinc nor scandium ions have any colored compounds AFAIK (unless with transition metal anions, of course!) although they are part of the d block elements. They can form complexes, of course.

I don’t consider them true transition elements, only those from Ti to Cu. These elements form at least one chemically stable ion with a partly filled d sub-shell. The d orbitals can, due to the Pauli exclusion principle, each have two electrons with opposed spin, but the d shell first fills up each with only one in each, due to the fact this is the lowest energy state and these are parallel rather than opposed.