chemical compound
Factsheet
Copper(II) sulfate
7758-99-8 (pentahydrate) Y
16448-28-5 (trihydrate) N
19086-18-1 (heptahydrate) N
Copper(II) sulfate
7758-99-8 (pentahydrate) Y
16448-28-5 (trihydrate) N
19086-18-1 (heptahydrate) N
My copper sulphate crystals suddenly turned green. Any theories?
"What is the colour of copper sulphate?"
inorganic chemistry - Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron? - Chemistry Stack Exchange
What is the colour of copper sulphate crystals? What happens when its heated?
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Let's compare the two compounds, here I plotted the 
centers of 
and 
from their crystal structure data.
As you can see, it changes from a 
to a let's call it 
. So all effects that are typical for 
should remain, it is still an octahedron, there is still Jahn-Teller present, we only got the change from four 
ligands to four 
ligands.
Those two are similar in their ligand field splitting, where the latter one is slightly weaker, so we expect a red-shift of the absorption energy. And that is also what Nagase et al. investigated on (Ref.1):
As shown partly in Fig. 1, the
band maxima of the hydrated and anhydrous copper sulfates were observed at
(
), [...] and
(
), respectively, in their reflectance spectra. This is in agreement with the expectation from the spectrochemical series, and indicates that, in a series of complexes of comparable tetragonalities, the substitution of a coordinated water molecule with a sulfate oxygen weakens the apparent ligand field strenght about
. In addition to this shift of the absorption maxima, the following spectral changes were observed on dehydration: (i) the width of the
band becomes gradually narrower with the advance of dehydration; (ii) a broad and very weak absorption appears near
in the spectrum of
, and, with somewhat increased intensity, in that of
; and (iii) a strong absorption (probably charge-transfer band) appears in the ultraviolet, and shifts remarkably to the lower frequency side with the advance of dehydration, indicating that the
ions are getting more and more strongly bonded, as more and more sulfate oxygens are getting coordinated in the
plane of the complex and shared among many
ions. [...] Both the slight red-shift and the narrowing of the
band contribute to the disappearance of the blue color [...]
Or, as it is summarized in the book by K. Sone and Y. Fukuda (Ref.2):
Therefore, the disappearance of the blue color of
caused by dehydration is only an apparent phenomenon, which is brought about by a relatively small shift of the
band to the red end of the visible spectrum (and partially into the near infrared) and perhaps also by the spontaneous pulverization of the dehydrated product; if we had eyes which could register radiation from near IR to near UV, we should certainly feel that
is much more darkly colored than
!
Since the ligand field strengths of
and
are seemingly only slightly different from each other, these changes will only bring about a small shift in the
band towards the red. On the other hand, the
ions, which are weakly held in the axial positions of
, are gradually brought into the equatorial coordination sphere, so that their CT interaction with the
ion will be strongly enhanced. The observed spectral changes thus come about. Somewhat similar spectral changes are also observed when the double sulfates
(where
and
) are heated.
References:
- K. Nagase, H. Yokobayashi, K. Sone, “Spectrophotometric and thermal analytical studies on the dehydration of copper(II) sulfate and its double salts,” Thermochimica Acta 1978, 23(2), 283-291 (https://doi.org/10.1016/0040-6031(78)85070-9).
- Kozo Sone, Yutaka Fukuda, In Inorganic Thermochromism (Inorganic Chemistry Concepts, Volume 10); Christian K. Jørgensen, Michael F. Lappert, Stephen J. Lippard, John L. Margrave, Kurt Niedenzu, Heinrich Noth, Robert W. Parry, Hideo Yamatera, Eds.; Springer-Verlag: Berlin, Germany, 1987 (ISBN 978-3-642-51019-9).
The 
configuration is an unsymmetrical field, with the 
orbitals being completely filled while the 
orbitals have only three electrons. So, according to the Jahn Teller theorem, these octahedral shapes will undergo slight distortions and will not be a perfect octahedron.
It is easy to see why it is so. Due to this unsymmetrical filling up of 
electrons, some ligands will be repelled more as compared to other when they approach to form the complex. In this case, the 
orbitals (which point directly at the approaching ligands) have two electrons in one while one electron in the second 
orbital. So, the ligand along either 
or 
will be repelled more as compared to the other.
Hence it will experience some sort of elongation towards that side, and more importantly, to remove the asymmetry, further sub-splitting will occur in the 
level, so even the restrictions of spin will be lifted. Clearly, the Laporte rule is no longer valid, and so, it can undergo 
transitions. While in the anhydrous form, we can clearly see that a centre of symmetry will be present, and so transitions will be Laporte-forbidden.
I'm a highschool student and we had a lab experiment about hydrates. The teacher noted that we need to explain in the Lab report about the blue color of the Copper sulfate pentahydrate.
After some research the only thing I got is that when Anhydrous copper sulfate is a white powder which I already knew and that when you add water to the white powder it returns to it's hydrated state. My question is why whenever water is mixed into the Copper sulfate structure it becomes blue and when none is mixed it is white.
As the title asks, I am confused as to why my crystals of copper sulfate are this colour (See first photo). I dissolved some copper in sulfuric acid and then boiled down the solution and let it crystallize over about a day (with some time overnight in a fridge). It's difficult to accurately convey the colour on my phone's camera, but it is a very light baby blue with some patches appearing nearly white rather than the dark, rich blue I am used to seeing online (Or perhaps it is normal for crystals this size, and that deep colour is more so for bigger crystals?). Could it be that I boiled it down too much and the extremely saturated solution led to crystals of lower hydration forming (some monohydrate perhaps)?
Also, the crystal shapes look a lot more scraggly and messy than I was hoping for, but that's probably because the solution was VERY saturated and it crystallized a bit too quickly.
Overall, I'll have to recryst them probably to get a prettier product, but is that very light colour normal or indicitive of some sort of contaminant? (First time growing crystals, please have patience with my incompetence)
Thanks