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woelen
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[*] posted on 11-5-2009 at 13:22
A remarkable nickel(I) complex


I just prepared a nickel(I) complex in aqueous solution. This complex has a beautiful red color. The oxidation state +1 is very uncommon for nickel.

Making the complex can be done as follows:

- Dissolve some nickel(II) sulfate in a small amount of water.
- Dropwise add a concentrated solution of KCN. After each drop shake, and continue until a yellow solution is obtained. Initially you get a precipitate, but on addition of sufficient cyanide, this redissolves again. It is important to have no excess of cyanide. One can best stop adding cyanide, when just a small amount of greenish precipitate remains.
- Wait, until all insolubles settle at the bottom and decant the clear yellow liquid.
- To this yellow liquid, add some sodium hydroxide and dissolve this. Now you have a basic solution of a Ni(CN)4(2-) complex.
- Boil this solution briefly. This is needed to expel all air (oxygen) from the solution.
- Let the solution cool down somewhat and then pour a layer of ligroin on it. This prevents oxygen from air to re-enter the solution.
- Now add some zinc powder or small zinc granules to the test tube and then add some solid NaOH.
- In the strongly alkaline environment of zinc and NaOH, the zinc dissolves and gives nascent hydrogen. This reduces the light yellow nickel(II) complex to a very dark red nickel(I) complex. Swirl the test tube every few seconds to bring fresh solution of the nickel(II) complex around the zinc granules.
- The final result is a deep red solution under a layer of ligroin.

The red complex is Ni(CN)4(3-).

I made some pictures of this neat experiment, but I still have to make a write-up in a webpage. I hope to have some time for that next Thursday, so later this week there will be pics and a webpage. I now already want to mention this, I find nickel(I) sufficiently remarkable to let you know about this.

The information about the nickel(I) complex is from one of my old 1950's chemistry books.




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[*] posted on 12-5-2009 at 02:23


More about preparation of this Ni(I) complex can be found in these articles/procedures:
http://dx.doi.org/10.1021/ja01257a053
http://dx.doi.org/10.1002/zaac.19140860105
http://dx.doi.org/10.1002/anie.197000711
http://dx.doi.org/10.1002/9780470132364.ch57
X-ray investigations show that this ion is dimeric ,with Ni-Ni bond.
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[*] posted on 12-5-2009 at 04:21


Square planar complexes of nickel(II) are red/brown in colour. Please note that I am not accusing you of being mistaken here; what I am asking is, besides citing the original reference, what evidence can you gather together yourself (possibly via further chemical tests) to show that what you have really is Ni(I)?

Now I'm not trying to yank your chain - actually, I'd really be interested to know how one tells such similarly coloured complexes apart definitively: educate me!:)




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[*] posted on 12-5-2009 at 06:29


One thing which really convinces me that this is a nickel(I) complex, is that on heating, the color fades again, while at the same time a fine dark gray (almost black) precipitate is formed. This precipitate is metallic nickel and this can only be formed by means of disproportionation of nickel(I) in nickel(II) and nickel metal.

Another thing which makes me really think that this is nickel(I) is that the red color is formed at the zinc particles. This color does not appear at other places, the zinc particles are the source of the color.




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[*] posted on 12-5-2009 at 06:40


OK, both explanations sound reasonable to me - thanks for explaining.



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[*] posted on 12-5-2009 at 12:37


It's good that you asked that question. In my webpage I certainly will add an explanation why it is quite certain that the red compound indeed is a nickel(I) compound. This kind of things must be mentioned explicitly and I did not do it here.



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[*] posted on 12-5-2009 at 14:27


It's a valid question, because Ni(II) complexes can be deep red, too. A well known one is Ni(dppp)Cl2 which is a common catalyst for Kumada couplings.



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[*] posted on 13-5-2009 at 10:18


This is a [(CN)3Ni-Ni(CN)3]4- ion with Ni-Ni bond not Ni(CN)4]3-...
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[*] posted on 13-5-2009 at 19:43


What sort of bonding would the Ni atoms in that dimeric compound have, assuming that it is dimeric with a Ni-Ni bond, and not [Ni(CN)4]3- which BTW has a different stoichiometry? In Ni(0) compounds and complexes such as the tetracarbonyl, in which the 3d orbitals are full, the 4 donated electron pairs occupy the formerly empty 4s and the three 4p orbitals.

For [(CN)3Ni-Ni(CN)3]4- , if it exists, the charge would indicate that it is a complex of Ni(II), assuming that the Ni-Ni bond is a single one. Three pairs of electrons from the negatively-charged C atoms of the CN- ligands would occupy, on each Ni(II) atom, the 4s and three of the four 4p orbitals. It is not clear whether the Ni-Ni bond is a pi bond, or a delta bond, or an hybrid, but it would utilize a further orbital. That would still leave one unutilized orbital, either 3d or 4p, unless the Ni-Ni bond is a double bond.
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[*] posted on 13-5-2009 at 22:27


Are you going to try and obtain crystals of your nickel complex? Perhaps it would crystallize as a potassium or sodium salt, in the same way as, for example, the ferricyanide complex?



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[*] posted on 13-5-2009 at 23:08


My old 1950's book propably has it wrong and indeed the complex seems to be a dimer, given the large amount of references which are talking about the dimer.

Crystallizing the compound is extremely difficult. I need to isolate it from the zincate and NaOH, and it must be isolated from remaining Ni(CN)4(2-) and free CN(-). Besides that, it is very air sensitive. I took some of the red liquid from under the ligroin layer and then it quickly looses the red color. I think that oxygen from the air quickly oxidizes the red complex. A funny thing is if you put a few ml of the red liquid in a test tube and allow it to stand. The upper few ml then soon become light yellow, while lower parts remain red. This yellow layer slowly becomes more and more high and it has a fairly sharp boundary with the underlying red layer.

I do not have the equipment to perform crystallization of this compound, it makes no sense to try that without special air-free apparatus.




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[*] posted on 18-5-2009 at 12:38


When will the webpage be ready woelen? i'm really looking forward to it.

I had never heard of a red nickel complex before, except for this nickel(I)cyanide complex. Every nickel complex I have seen so far ranged from green to blue tot purple, with an exception being the yellow Ni(II)cyanide complex. I know otherwise now, after having isolated the nickel compound dinitrotetraamminenickel(II) or [Ni(NH3)4(NO2)2], wich forms red crystals.
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[*] posted on 18-5-2009 at 14:10


Quote: Originally posted by Jor  
I had never heard of a red nickel complex before, except for this nickel(I)cyanide complex. Every nickel complex I have seen so far ranged from green to blue tot purple, with an exception being the yellow Ni(II)cyanide complex. I know otherwise now, after having isolated the nickel compound dinitrotetraamminenickel(II) or [Ni(NH3)4(NO2)2], wich forms red crystals.

The blue/green colour is due to octahedral Ni(II), usually with H2O as the ligand, but you can also make a whole series of square-planar (ie. tetra-coordinated) Ni(II) complexes where H2O is not the ligand and these tend to be red/brown in colour.




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[*] posted on 18-5-2009 at 21:14


Quote:
The colour of nickel (II) complexes is a very specific feature of them; the square–planar complexes can be yellow, orange or red, whereas the tetrahedral and octahedral complexes can be green or blue of different intensities


doi:10.1016/S1387-7003(02)00446-X

Inorganic Chemistry Communications
Volume 5, Issue 7, July 2002, Pages 464-467

From isothiocyanato– to silyl–nickel complexes

Hieronim Maciejewskia, Bogdan Marciniec, Jacek Gulinskia, Aleksandra Karolaka and Nikolay K. Skvortsov

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[*] posted on 23-5-2009 at 13:19


Finally it's there, the webpage with pictures. It took a little longer than planned ;)

http://woelen.homescience.net/science/chem/exps/nickelI/inde...




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[*] posted on 23-5-2009 at 14:53


Very nice again woelen, I am impressed! Maybe you could filter this zinc salt of the complex. I know you have no access to nitrogen or a glove box (who has :P), but mayeb you could put a tube just above the filter funnel, and pass a continuous, fast stream of CO2 throught it. Do this is a non-ventilated area, and you should be able to reduce the O2-concentration to minimum around the filter. Then quickly pour the whole mixture in the funnel, and make sure the precitipate is covered with liquid at all all times. Then wash with water a few times (boiled). and VERY QUICKLY transfer the wet mass to a reagent bottle, and put the open bottle in a dessicator over P4O10 (the dessicator should be CO2 atmosphere), and when you open the dessicator when it's dry quickly cap the bottle. The best thing would be to also put a compound in the dessicator wich quickly abosrbs oxygen, so the concentration is recued to a absolute minimul. Maybe pyrogallol?
I think this can be done, you just need a fast CO2 generator, with a flexible tube attached so you can stream CO2 in anywhere you want a inert atmosphere.
This would be difficult, and won't yield a very pure product, but I think it can be done, if prepare really well for all steps.

By the way, you really should mention that HCN will be formed when the liquid is acidified. ;)
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[*] posted on 23-5-2009 at 20:40


Nice write-up, but - how concentrated is the KCN solution, it doesnt say? Is it saturated (KCN dissolves to 70gms in 100ml at RT), 1M, 0.1M?
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[*] posted on 24-5-2009 at 03:12


Len1, I did not really measure the concentration of KCN and it is not critical. I made the complex two times to check repeatability and both times I just put a spatula full of KCN in appr. 2 ml of water. If you really want me to give a rough estimate, I think that the concentration might be somewhere between 10% and 20% by weight, but more accurate figures I cannot give, but it also is not important. Just drip in solution of KCN until the precipitate of Ni(CN)2 redissolves.

Jor, I doubt that the experiment with CO2 will work. The solutions, involved in this experiment are strongly alkaline, and the CO2 will be absorbed. A possible 'inert' gas suitable for this might be butane or propane from a gas cylinder but this is fairly dangerous due to buildup of concentrations of highly flammable gas.
I'll add the remark about the formation of HCN. The amount of HCN formed, however, is not very high, and no bubbles of gas are formed. Fortunately, this gas dissolves in water very well. But it is good to add a warning for this.

[Edited on 24-5-09 by woelen]




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[*] posted on 24-5-2009 at 04:51


I guess you could use argon as the gas? Should be easily available.

On another note, I happened across your zinc-plating of a copper coin: very interesting! Especially when it turns to brass at room temp over time.




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[*] posted on 24-5-2009 at 15:31


Thanks for that. It seems with weak cyanide solutions this does not work. I was using a few ml of 0.01M KCN I brought from the lab (I did this at home with the kids), the white/light blue precipitate formed immediately (I think this must be a mixture of the hydroxide and cyanide of Ni), but would not dissolve fully on subsequent CN addition. After an hour or so we did get a yellow solution above the precipitate - it was decanted - but no red colouration could be obtained with nascent hydrogen.
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[*] posted on 25-5-2009 at 07:42


Len1, did you add the additional KOH or NaOH, together with the zinc? This really is important and you also need to drive off oxygen. Especially if the solutions are so dilute the dissolved oxygen from air weighs in extra strong, relatively speaking.



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[*] posted on 25-5-2009 at 15:15


Yes of course I added about 0.5gms NaOH, we got a good stream of hydrogen bubbles given out - I dont think KCN is of itself basic enough to liberate hydrogen from zinc. Plus its always good when working with cyanides to know the solution is strongly basic, so that no HCN is given off. We boiled the solution as you described. Cooled, added NaOH and a small heap of Zn powder, poured some parafin on last, because using powder, I was concerned it might get coated. Lots of small H2 bubbles - but solution remained yellow. Sad kids (:

[Edited on 25-5-2009 by len1]
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[*] posted on 26-5-2009 at 10:22


Len1, I will try your experiment as well with very dilute KCN. I'll let you know one of these days.



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[*] posted on 30-5-2009 at 05:37


Len1, I tried the experiment with very dilute KCN and I also failed. I'm not completely sure why, but I have the impression that the making of the nickel(II)/cyanide complex failed. The complex is too light to be observed in the aqueous solution, I just obtained a colorless liquid with a very little amount of precipitate of Ni(OH)2 and/or Ni(CN)2 in it. When this liquid is used for the experiment, it does not work.

I added slightly more KCN and retried the same experiment and with that, I obtained a light orange/brown liquid when the Zn and NaOH were added.

From my experiments, I draw the conclusion that with very dilute solutions, the experiment is very critical and you really need to carefully adjust the amounts of nickel(II) and cyanide.

With more concentrated solutions of KCN the experiment really always works. I now tried it three times and it worked all the time again.




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[*] posted on 7-6-2009 at 10:21


I decided to make a few grams of crystalline Na2[Ni(CN)4]x3H2O and make some experimens with it.
Indeed - Al/NaOH converts (during some time) yellow solution to red one - no heating and ligroin is needed. Adding few drops of 3% H2O2 gives again yellow solution. If NaOH is exhausted, sol. slowly turns yellow in the air and adding another portion of NaOH causes yellow--> red change (Al is present all the time).
I will try to make a sample of solid product of reduction. In a paper mentioned earlier (first reference) it is simply done by adding ethanol to reduced solution. They used potassium salt and I do not know if it works also with Na salt.
BTW.
It is interesting that this Ni(I) complex cannot be reduced (at least in a simple way) to Ni(0) but Ni(0) can be obtained by direct (two-electron) reduction of Ni(II).
It is very possible that Ni(I) complex originates from reaction:
(monomeric)Ni(II)+(monomeric)Ni(0) --> (dimeric)Ni(I) (it is stated also by another paper).
It may explain why Ni(I) cannot be obtained from diluted solutions - reaction of Ni(0) with water is faster than reaction with Ni(II).


[Edited on 7-6-2009 by kmno4]
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