Oxidation states of iron in Roussin's salts

Question

For preservation of meat, sodium nitrite is usually added and as a result $\ce{NO}$ is then formed. Consequently, $\ce{NO}$ reacts with the sulfur and iron atoms from decomposition of proteins, forming $\ce{[Fe4S3(NO)7]-}.$ X-ray crystallography shows that the complex anion has a structure as shown below:

a) Blacken all the circles corresponding to iron atoms and add symbols Fe(A), Fe(B), Fe(C) and Fe(D) beside the circles in the sequence of top → left → right.

b) The configuration of 3d electron shell of the iron atoms has been studied with modern structural analysis. Knowing that the mean oxidation number of the four iron atoms is $–0.5$, give their configurations of 3d shell, respectively. Assume that each iron atom adopt sp hybridization.

$\ce{[Fe4S3(NO)7]^{–}}$ anion can be reduced and a new complex $\ce{[Fe2S2(NO)4]^{2-}}$ is formed

c) Give the oxidation state of each iron atom with Arabic numerals.

d) $\ce{[Fe2S2(NO)4]2-}$ can be converted into $\ce{[Fe2(SCH3)2(NO)4]^{n}}$, a carcinogen. Which of the following three species is added to $\ce{[Fe2S2(NO)4]^{2-}}$ : $\ \ce{CH3+}$,$\ \ce{•CH3}$ or $\ \ce{CH3−}$?

My Attempt

a)

I don't have any real reason to why I chose those 4 atoms. I chose them since they were the 4 biggest atoms in the ring and made sense they would alternate with the sulfur atoms in the ring. Is there a more solid explanation?

b) Firstly, why do the iron atoms have a negative oxidation state, doesn't it 'give' its electrons to the sulfur and nitrogen atoms since it is more electropositive? I look up negative oxidation states of iron and I found out when bonded to carbonyl compounds, the iron atom usually has a negative oxidation state. So is the $\ce{NO}$ ligand in this case acting like $\ce{CO}$?

The answer is Fe(A) has 3d7 configuration while the others have a 3d9 configuration. Could someone please give an explanation for this.

c)

I have drawn what I think is the structure above, but I have the same problem as part b)

d) I have no idea. I am guessing the methyl cation since the iron atom has a negative oxidation state (I am guessing from part c) hence it will more likely react will a cation rather than an anion or radical.

Answer 1

I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure. For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.

It's only the symmetry that's evident from the crystal structure: the thioferrate cluster $\ce{[Fe4S3(NO)7]-}$ belongs to $C_\mathrm{3v}$ point group, hence the only possible arrangement of four iron atoms is a tetrahedral one. Another conclusion that can be made is that three equatorial iron atoms are equivalent ($\ce{Fe(B)}$, $\ce{Fe(C)}$, $\ce{Fe(D)}$), and the last apical one ($\ce{Fe(A)}$) most likely deviates in terms of oxidation state.

Supposing formal oxidation number ($\text{O.N.}$) of three equivalent irons and the last one are $y$ and $x$, respectively, and the fact that the average $\text{O.N.}$ is $-0.5$, the following holds true:

$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$

As formal oxidation numbers are integers, there are several pairs giving the solution:

$$ \begin{array}{l|rrrrr} \hline x & \ldots & -5 & -2 & \color{green}{1} & 4 & 7 & \ldots \\ \hline y & \ldots & 1 & 0 & \color{green}{-1} & -2 & -3 & \ldots\\ \hline \end{array} $$

among which the least "extreme" (e.g. the one with the absolute lowest negative $\text{O.N.}$) for a metal would be the $-1;+1$ pair; so that for $\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$ your answer gives two configurations, $(\mathrm{3d})^7$ for $\overset{+1}{\ce{Fe}}\ce{(A)}$ and $(\mathrm{3d})^9$ for $\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$.