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CharlieA
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[*] posted on 16-2-2016 at 16:18


You are a busy beaver! Good work.
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[*] posted on 16-2-2016 at 23:24


Quote: Originally posted by crystal grower  
These are my first bismuth crystals:
http://postimg.org/image/jcr0ngeev/full/
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bismuth is my no.1 metal :)

[Edited on 3-1-2016 by crystal grower]


Very nice, Bismuth is definitely my #1 as well. Here's a few of mine. I figure at least one of these phone photos will look good on the big screen.

IMAG1303.jpg - 1.6MBIMAG1304.jpg - 1.6MBIMAG1305.jpg - 1.3MBIMAG1306.jpg - 1.3MBIMAG1307.jpg - 1024kBIMAG1308.jpg - 1MB
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[*] posted on 17-2-2016 at 08:10


Quote: Originally posted by alexleyenda  
Results of 20 hours of inorg chemistry in my lab classes.


PbCl6(pyr)2, PbI2, recristalised PbI2, (Ni(NH3)6)Cl2, Ni(NH3)4(NO2)2, (tren)Ni(NO3)2, Bn3SnCl, SnCl4(DMSO)2

[Edited on 16-2-2016 by alexleyenda]

How did you do this one Ni(NH3)4(NO2)2?
Isn't the Ni hexacoordinator? --> Ni(NH3)6(NO2)2?
Like in your Ni(NH3)6)Cl2 and (tren)Ni(NO3)2?
With N2H4, Ni(NO3)2 fixes 3 N2H4 --> Ni(H2N-NH2)3(NO3)2




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[*] posted on 17-2-2016 at 08:47


Haha the Ni(NH3)4(NO2)2 was quite a long process. First, NiSO4 hexahydrate is reacted with KHCO3 to make the carbonate, then acetic acid is added to form Ni(OAc)2(H2O)4 (both reactions bubble CO2). Then you mix NaNO2 and NH4CH3COO in large excess to make a good NO2 donor (NH4NO2) and add it to the nickel complex with conc. NH3 (aq) to give Ni(NH3)4(NO2)2. After mixing 10 min at room temp, you let it precipitate for around 3 hours, which gives the red solid that precipitates out of the blue solution. You must be careful not to add too much water, or the complex will not precipitate.
Metal centers do not always hexacoordinate. I guess the complex is tetrahedric or square planar, which is common when metal centers are d8 (8 electrons in the d orbitals) (search about it, it is very common). For exemple d8 paladium is almost always square planar.




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[*] posted on 17-2-2016 at 11:38


Quote: Originally posted by alexleyenda  
Haha the Ni(NH3)4(NO2)2 was quite a long process. First, NiSO4 hexahydrate is reacted with KHCO3 to make the carbonate, then acetic acid is added to form Ni(OAc)2(H2O)4 (both reactions bubble CO2). Then you mix NaNO2 and NH4CH3COO in large excess to make a good NO2 donor (NH4NO2) and add it to the nickel complex with conc. NH3 (aq) to give Ni(NH3)4(NO2)2. After mixing 10 min at room temp, you let it precipitate for around 3 hours, which gives the red solid that precipitates out of the blue solution. You must be careful not to add too much water, or the complex will not precipitate.
Metal centers do not always hexacoordinate. I guess the complex is tetrahedric or square planar, which is common when metal centers are d8 (8 electrons in the d orbitals) (search about it, it is very common). For exemple d8 paladium is almost always square planar.

I expect that it is octahedral, with the two nitrites coordinated along with the ammonias. Nitrite is a much better ligand than nitrate (I have a batch of K4[Ni(NO2)6] on my shelf).




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[*] posted on 17-2-2016 at 13:37


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by alexleyenda  
Haha the Ni(NH3)4(NO2)2 was quite a long process. First, NiSO4 hexahydrate is reacted with KHCO3 to make the carbonate, then acetic acid is added to form Ni(OAc)2(H2O)4 (both reactions bubble CO2). Then you mix NaNO2 and NH4CH3COO in large excess to make a good NO2 donor (NH4NO2) and add it to the nickel complex with conc. NH3 (aq) to give Ni(NH3)4(NO2)2. After mixing 10 min at room temp, you let it precipitate for around 3 hours, which gives the red solid that precipitates out of the blue solution. You must be careful not to add too much water, or the complex will not precipitate.
Metal centers do not always hexacoordinate. I guess the complex is tetrahedric or square planar, which is common when metal centers are d8 (8 electrons in the d orbitals) (search about it, it is very common). For exemple d8 paladium is almost always square planar.

I expect that it is octahedral, with the two nitrites coordinated along with the ammonias. Nitrite is a much better ligand than nitrate (I have a batch of K4[Ni(NO2)6] on my shelf).


It is true that it might be, I did not take time to think about it and made the too rapid asumption that because NO2* is negatively charged it was only the counter ion, but it can be coordinated too. The nickel is d8 tho, so it "likes" to have a coordination of 4. I'll have to sit and think about it for longer :p

[Edited on 18-2-2016 by alexleyenda]




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[*] posted on 17-2-2016 at 14:40


Quote: Originally posted by alexleyenda  

It is true that it might be, I did not take time to think about it and made the too rapid asumption that because NO3 is negatively charged it was only the counter ion, but it can be coordinated too. The nickel is d8 tho, so it "likes" to have a coordination of 4. I'll have to sit and think about it for longer


NitrIte, not nitrAte. Nitrate is a lousy ligand; nitrite is a fairly good one.

Nickel(II) is d8, but it will only be square planar with ligands that are strong field splitters (cyanide, or ones like DMG). Apart from low-spin complexes, it has no more tendency to be 4-coordinate than cobalt(II) or iron(III). With halides, it will form NiCl4<sup>2-</sup> or NiBr4<sup>2-</sup>, but that's more to do with the size of the halides than the electron configuration.




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[*] posted on 17-2-2016 at 17:01


Copper(II) lactate trihydrate

CopperLactate.jpg - 93kB

[Edited on 18-2-2016 by DraconicAcid]




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[*] posted on 17-2-2016 at 23:12


Quote: Originally posted by DraconicAcid  
Quote: Originally posted by alexleyenda  

It is true that it might be, I did not take time to think about it and made the too rapid asumption that because NO3 is negatively charged it was only the counter ion, but it can be coordinated too. The nickel is d8 tho, so it "likes" to have a coordination of 4. I'll have to sit and think about it for longer


NitrIte, not nitrAte. Nitrate is a lousy ligand; nitrite is a fairly good one.

Nickel(II) is d8, but it will only be square planar with ligands that are strong field splitters (cyanide, or ones like DMG). Apart from low-spin complexes, it has no more tendency to be 4-coordinate than cobalt(II) or iron(III). With halides, it will form NiCl4<sup>2-</sup> or NiBr4<sup>2-</sup>, but that's more to do with the size of the halides than the electron configuration.


Well you are right, it makes sense. This inorganic theory is quite fresh for me, I have to really take my time to really think about it and not forget anything to get to the good conclusion :p And yeah, i'm aware of the difference between nitrate and nitrite as ligands (I just wrote NO3 by mistake), my error was to forget that it had to be low spin to favorise the coordination of 4. Now the question is on which atom is it coordinated even tho (i'm not sure but if my memory is good) most of the time it is on the oxygen. I'll have to look at my IR spectrums, maybe tomorrow ^^ Nice cristals by the way.


[Edited on 18-2-2016 by alexleyenda]

[Edited on 18-2-2016 by alexleyenda]




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[*] posted on 18-2-2016 at 07:44


Quote: Originally posted by alexleyenda  
Haha the Ni(NH3)4(NO2)2 was quite a long process. First, NiSO4 hexahydrate is reacted with KHCO3 to make the carbonate, then acetic acid is added to form Ni(OAc)2(H2O)4 (both reactions bubble CO2). Then you mix NaNO2 and NH4CH3COO in large excess to make a good NO2 donor (NH4NO2) and add it to the nickel complex with conc. NH3 (aq) to give Ni(NH3)4(NO2)2. After mixing 10 min at room temp, you let it precipitate for around 3 hours, which gives the red solid that precipitates out of the blue solution. You must be careful not to add too much water, or the complex will not precipitate.
Metal centers do not always hexacoordinate. I guess the complex is tetrahedric or square planar, which is common when metal centers are d8 (8 electrons in the d orbitals) (search about it, it is very common). For exemple d8 paladium is almost always square planar.

Thank you for the process.

Seems like Cu also form such a complex: Cu(NH3)4(NO2)2
For your Ni complex I thought the NO2 was nitrite anion, but it takes part into the hexacoordinated complex and this explains why two positions are not available anymore for NH3 ligands.

Those complex may be energetic since NO2 is an oxydiser and NH3 a fuel...
-->Could you test a few mg in contact with a flame?

By experience with Ni(N2H4)3(NO3)2, Cu(N2H4)3(NO3)2, Co(N2H4)3(NO3)2, Ni(NH3)6(NO3)2, Cu(NH3)4(NO3)2, Co(NH3)6(NO3)2, Ni(en)3(NO3)2, Cu(en)2(NO3)2 and Co(en)3(NO3)2...
The addition of some CH3OH, CH3-CH2OH may help isolation and crystalization from watery solutions followed by washing with ether ((Eth)2O).




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[*] posted on 18-2-2016 at 07:49


Quote: Originally posted by wg48  
Quote: Originally posted by PHILOU Zrealone  
Quote: Originally posted by wg48  
Pic of a glycerol complex of copper. Its more purple when viewed directly.

Could you be more specific? What complex and weight ratio?


Probably K2Cu(C3H6O3)2 ref: CHARACTERIZATION AND REACTIONS OF
COPPER (II)
last part of above post



COPPER (II) Probably K2Cu(C3H6O3)2 ref: CHARACTERIZATION AND REACTIONS OF
COPPER (II) GLYCEROL COMPLEX
by
HAZIMAH ABU HASSAN
July 1 998

for synthsis see ref: Preparation of Ultrafine Copper Powders with Controllable Size via Polyol Process with Sodium Hydroxide Addition
Pattanawong Chokratanasombat and Ekasit Nisaratanaporn
ENGINEERING JOURNAL Volume 16 Issue 4, ISSN 0125-8281 (http://www.engj.org/)

quote:In a typical synthesis, copper (II) nitrate trihydrate (Cu(NO3)2∙3H2O, Carlo Erba) was dissolved in the solution of sodium hydroxide (NaOH, Mallinckrodt) and glycerol (C3H8O3, Carlo Erba) with varying the molar ratio of NaOH:Cu(NO3)2∙3H2O in the range of 0:1 to 5:1 and
Cu(NO3)2∙3H2O:glycerol at 0.02:1

I used 3:1 K not Na

Thank you!
Interesting.
I knew that Pb(OH)2 can form a glycerolate with mild energetic properties...
Pb(OH)2 + HOCH2-CHOH-CH2OH -heat/reflux-> HOCH2-CH(-O-*)-CH2-O-Pb-* + 2 H2O
So apparently Cu(OH)2 also form a glycerolate :-)




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[*] posted on 18-2-2016 at 09:50


Quote: Originally posted by PHILOU Zrealone  

Thank you for the process.

Those complex may be energetic since NO2 is an oxydiser and NH3 a fuel...
-->Could you test a few mg in contact with a flame?



By the way each intermediate complexes were filtered and re-diluted in a minimum of water, and each solids were mixed in a minimum of water too.

When I took the melting points, the complexe decomposed to give black crap at around 170-180 degrees celcius, which is the lowest temperature in all the complexes I prepared, so yeah it would probably react energetically with a flame. I'll see if I can test it.




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[*] posted on 19-2-2016 at 08:46


Quote: Originally posted by Noflers  


Very nice, Bismuth is definitely my #1 as well. Here's a few of mine. I figure at least one of these phone photos will look good on the big screen.

Yeah bismuth is awesome.
Btw bismuth crystal is much prettier in real life than in the picture IMO.




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[*] posted on 21-2-2016 at 07:36
NdCl2 +FeCl2



I did a reaction of NdFeB magnet with HCl and it gave nice blue solution.
Then I added some water and the solution suddenly turn to reddish brown. When I added some Hcl it turned red at the bottom and green at the top of test tube. After while the solution turned purple-blue (the reaction was still on).
Do u know what caused the change of the colour? (maybe pH of the solution??)
And by the way could you tell me the simpliest way of separating Fecl2 and NdCl2 ?
Thanks.
P2192461.JPG - 1.6MB P2192464.JPG - 1.2MB P2212477.JPG - 1.7MB P2212480.JPG - 1.6MB P2212490.JPG - 1.4MB

[Edited on 21-2-2016 by crystal grower]




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[*] posted on 21-2-2016 at 08:08


Honestly, I remember running into problems like this during my very first Nd extraction (a very long time ago). The true color of the 'magnet chloride' is supposed to be an incredibly dark purple, and I've only once seen intermediate colors like these (and even then, that was after I boiled down to crystals, which were really a mushy substance that kept changing color between green, blue, purple, etc.).

As for the simplest way of separating iron and neodymium, a good choice is oxalic acid (wood bleach). Add this and some sodium hydroxide to the 'magnet chloride', and the iron will chelate into solution as sodium ferrioxalate, forming a light green solution of Fe2+, while the neodymium crashes out of solution as neodymium oxalate (a pink solid). From there, it can be gently calcined (to gray, if it turns white it becomes basically inert) and redissolved in a very strong acid mixture (something like aqua regia, maybe, or hot sulfuric acid).




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[*] posted on 22-2-2016 at 03:15


Thanks for super procedure.




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[*] posted on 22-2-2016 at 11:24


Awesome pic of my bismuth crystal:):):).

P2222505.JPG - 1.7MB




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[*] posted on 22-2-2016 at 11:32


What is your trick to make great cristals like that? I made a bismuth cristal as great as this one once after many tries, but I don't know what I did and i'm not able to do it again x)



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[*] posted on 22-2-2016 at 12:03


hmmmm......:D
In fact I made this one a year ago and... now it seems i can´t make so pretty crystal again for some reason:(.
Maybe I have contamined bismuth with something or its some curse of bismuth:D:D:D.
I will probably buy some new ingot and try it again. If it will be a success, I will tell u more.




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[*] posted on 22-2-2016 at 20:09


Although I haven't done any home chemistry in months, I've been working in an organic chemistry research lab, synthesizing highly conjugated compounds that have potential use in OLEDs. Here are some pretty pictures involving my first such compound:

Workup of a Cadiot-Chodkiewicz coupling (aqueous layer on top, product dissolved in DCM)
extraction_resized.jpg - 1.3MB

TLC spot of product fluorescing under longwave UV (this is a good sign).
fluorescent TLC.jpg - 2.6MB

[Edited on 2-23-2016 by Metacelsus]




As below, so above.
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[*] posted on 24-2-2016 at 01:35


Details in my recent post in the When chemistry goes wrong thread.

The coppery stuff is what remains of a brass spoon. The bubbles were kind of sparkly looking and were chlorine. The purple is the copper-sodium dichlorocyanurate complex. (Interesting that Na-DCCA was not one of my starting compounds.)



2016-02-24 15.09.10.jpg - 1.2MB2016-02-24 15.09.41.jpg - 1.2MB




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[*] posted on 24-2-2016 at 07:00
Analcime, chabazite from mindat


That's really awesome.
link:
http://www.mindat.org/photo-160295.html



[Edited on 24-2-2016 by crystal grower]

trans.png - 68B




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[*] posted on 27-2-2016 at 08:06
Indigo


Hi guys,

recently I´ve prepared indigo and did some vat dyeing for fun at the university (actually one of the labs :)) ) Thanks everyone for visiting my web! INDIGO PREPARATION


indigo.jpg - 1.4MB
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[*] posted on 27-2-2016 at 08:17


Nice! I'm actually currently working on an indigo synthesis too, however I'm actually synthesizing my own 2-nitrobenzaldehyde, so it's a more lengthy procedure. I have 2-nitrotoluene now, and I'm currently awaiting the arrival of the solvent that I need for the oxidation step. Then it should be smooth sailing from there. A writeup will be posted once I am finished. :)



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[*] posted on 27-2-2016 at 10:27


Quote: Originally posted by zts16  
Nice! I'm actually currently working on an indigo synthesis too, however I'm actually synthesizing my own 2-nitrobenzaldehyde, so it's a more lengthy procedure. I have 2-nitrotoluene now, and I'm currently awaiting the arrival of the solvent that I need for the oxidation step. Then it should be smooth sailing from there. A writeup will be posted once I am finished. :)


Oh man, such a long way. The last step is really smooth sailing besides making your own 2-nitrobenzaldehyde. Did you prepare 2-nitrotoluene?
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