11K4[Fe(CN6)] is colourless, Cu2[Fe(CN6)] is chocolate brown and Zn2[Fe(CN6)] is bluish white. I don't think these colours are different intensities of the same colour.
Colours in complexes are decided by the constituents of the coordination sphere, right? So why do K4[Fe(CN)6] and Fe2[Fe(CN)6] differ in colour though coordination sphere is intact?
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11So no zabardast reason for this? Anyway, thanx everyone for your replies[1] and for making this thread the second largest[4]! (i think so)
P.S:
This was not a question in any exam. It was a doubt I'd had ever since i learnt about complexes.
1i dont know much about coordination complexes... !!
the color of a complex depends on the ionization sphere as well as upon the counter ion.
The intensity of the color is decided by the counter ion. !
i m not sure... but most probably it is correct. .!!
:)
111in case of Fe+2 case there is no unpaired electron while in +3 oxidation state there is one that will change the color
i dont know whether done Zin and Zout cases in octahedral complexes but if u have done
in case of Fe+3 the geometry will change to Zout there ligands effecting the co-ordination sphere will have some lesser effect
this will decrease the Δ0 values of the sphere hence changing the color
but i don't think that u should worry about these things because
Zin and Zout and determination of color by Δ0 values are not in course of IIT-jee syllabus
1in case of Cu2[Fe(CN6)] the chocolate brown color is of cuprus ion not of co-ordination sphere and in case of Zn2[Fe(CN6)] the bluish white is not the color it is the tingity ( if u look the water in a deep vessel it will look blue it is not the color it is tingity due to higher concentration in line of view as in the end point it is again clear)
11I thought colour in complexes were due to the constituents of the coordination sphere and that changes outside it were immaterial.[7] (i.e) Only [Fe(CN)6]4- determines the colour of the complex and not anything else!
1sorry i have given u the reasoning for K4[Fe(CN6)] andK3[Fe(CN6)] but in this case the reasoning is very easy the color in the case of Fe2[Fe(CN6)]
the color is due to un paired electrons of Fe2+ outside the co-ordination sphere so idon't think there is a question of confusion in this case
11[7][7][7][7]
what about K4[Fe(CN)6] and Zn2[Fe(CN)6]?
[7][7][7]
1i told u thats tingity but if u will stick to what are u saying then sorry i might be of no help
11What is tingity? You mean brown and bluish white are different tinges of the same colour
1OK JUST TELL ME WHY IS THAT THE COLOR OF WATER SEEMS TO BE BLUISH IN A VESSEL OF LONG HEIHGT AND AGAIN COLORLESS WHEN WE TAKE THAT OUT IN OUR HANDS
1look color determination of the co-ordination compounds is not in the syllabus and there are many-more factors who determine the colors it is more and more a practical fact thats why i didnt take part in this section earlier otherwise there must have been a reason given in WICI u can check that
1if u haven't heard the term TINGITY then check the H2O2 and u will find the term
11I'm not discussing this as a question. I'm discussing it as a doubt.
1it is because not only the corordination sphere that decides the colour but also the oxidation number that decides the colur of the compound.
1but i have a doubt that:
i think the compound is Fe4[Fe(CN)6]3 if it is so then their is charge transfer b/w fe2+ and fe3+ and hence this can be the reason for ur question
11For your first reply,
Oxidation state of what?
For your second reply,
[7][7][7][7][7][7][7]
1the coordination sphere is different in both. one contains Fe +2 and the other has Fe +3
1orMaybe the colour is due the counter ion. In an aqueous solution maybe the water acts as ligands for the counter ion and causes the colour of that complex to add to the mix. U kno, Fe2+ -light green in aq soltn.
1PRIMARY & SECONDARY VALENCY BOTH DETERMINES THE CHARECTERRISTICS OF THE COMPLEX COMPOUNDS AS SAID BY WERNER HERE THEY HAV SAME SECONDARY VALENCY BUT DIFFERENT PRIMARY VALENCY & HENCE THEY HAV DIFFERENT COLOURS
11I thought Werner theory was rejected and is no longer followed?
11By being followed I meant being followed by everyone and not only by you.
Btw this is not a question. It is a doubt [4]
1the nature of the ligand, oxidation state of the metal n the co-ordination of the ion r sum of the factor which affect the colour of a transition metal complex ion.
n as v 9 dat the main coz of colour in transition metal iz its partially filled d-orbital.
When the ligands bond wid the transition metal ion, there iz repulsion between the electrons in the ligands nd the electrons in the d orbitals of the metal ion. dat raises the energy of the d orbitals.but the way the d orbitals are arranged in space, it doesn't raise all their energies by the same amount. Instead, it splits them into two groups.
in octahedral compound,
6 ligands r arranged around a transition metal ion, the d orbitals r split into 2 groups - 2 with a higher energy than the other 3. and
in tetrahedal compound,
d orbitals split into two groups, three of dem hav a greater energy, nd the other two a lesser energy .
nd the difference of the energy gap between dese two group, dependce upon d factor (i wrote it in top).
When white light iz passed through solution of the ion, sum of the energy in the light is used to promote an electron from the lower set of orbitals into a space in the upper set.nd hence colour arises , depending on its wavelength.
as here in the question , K has no d-orbital, nd d- electron , hence deir iz no repulsion in between electron of ligand nd its own, nd hence it apart colour only coz of its ion nd not coz of the repulsion between d-orbital as in Fe2[Fe(CN)6].
..................................................................................... :) hehehe...i think deir 'll no more post . aftr my etta badda lecture. :P