Asked by MAD

I have trouble finding out the oxidation number of each element in the :
[Cr(N2h4CO)6]4[Cr(CN)6]3
Answered by DrBob222
I would start this way.
This is a zero charged compound so the cation must be +3 and the anion part must be -4.
We know CN is -1.
I always go with +1 for H and -2 for O unless I know better. A value of +4 for C is always a good choice (unless it interferes with my +1 for H atoms). I usually assign -3 for N atoms especially if they are affiliated with H because I want +1 for that.
For the anion part, -6 for CN and -4 overall must leave Cr with +2.
For the cation.
-6 for N, +4 for H, +4 for C and -2 for O is zero. For an overall charge of +3 on the cation means Cr must be +3.
As I've written before, these oxidation states are nothing more than bookkeeping. Cr is about the only one of those above you worry about while assigning numbers. Cr can be +3 or +6 commonly but +2 especially in coordination complexes. Those are the ones that require care.
Answered by MAD
So , on the reactant site you assigned two Chrome in two different oxidation number , but really and basically , is such something possible ? how is it possible one element in a compound on a site of reactant or products has two different oxidation number ??? !!
Answered by DrBob222
I'm confused by your use of the terms reactant/products side.
From the original question I thought that was a compound and if it is a compound there is no reason why you can't have an element exhibiting two different oxidation states. In coordination compounds that is fairly common. Another example is Fe4[Fe(CN)6]3. This is ferric ferrocyanide (of course the IUPAC name isn't that anymore). The Fe outside the coordination sphere (the left one as I've written it) is +3 oxidation state and the Fe inside the coordination sphere is +2.
Answered by MAD
Ok , I really thank you , but if it's so , can we balance this equation in half reaction method ? :
[Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 ---> K2Cr2O7 + MnSO4 + CO2 + KNO3 +K2SO4 + H2O

Answered by DrBob222
It's not simple but here is what I found on the internet. I don't vouch for its authenticity.


10[Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176KMnO4 + 1399H2SO4 = 35K2Cr2O7 + 420CO2 + 660KNO3 + 1176MnSO4 + 223K2SO4 + 1879H2O
Answered by MAD
Not only I don't think so , but I believe it's impossible to balance that , but there is no such reaction and such authentic and valid rxn and equation ! because how possible in chemistry science world that there be an substance has two different oxidation numbers before reaction , and has another different oxidation number after reaction ???!!! I think it's absolutely impossible
Answered by DrBob222
You don't think so or you don't know so. Did you check it? I checked it and all atoms appear to balance. I even found a site in which someone did a lot of work to do it with half reactions. Here it is if you are interested. The numbers match those from another site which I gave to you initially. The only thing I did not check was if the electrons balanced.
https://answers.yahoo.com/question/index?qid=20071024151845AAczaBb

Although I don't vouch for the authenticity of this work I can tell you that having a compound in which an element exhibits more than one oxidation state and going through a reaction to some other oxidation state is not that common but it does happen. Furthermore, having a compound in which an element exhibits more than one oxidation state is NOT that uncommon. KFe[Fe(CN)6] is one example of that. The first Fe is +3 and the second one is +2. This is prussion blue used in painting as well as a qualitative test for Fe.

There are no AI answers yet. The ability to request AI answers is coming soon!