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Given the fact that $\ce{NO+}$ is a strong field ligand and iron is in the +1 oxidation state, the valence orbitals of $\ce{Fe+}$ must undergo rearrangement from $\mathrm{3d^{6}4s^{1}}$ to $\mathrm{3d^{7}4s^{0}}$, which must contain 6 electrons in pairs, leaving one unpaired electron.

So, the hybridisation of iron can be concluded to be $\mathrm{sp^{3}d^{2}}$.

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This is no Iron(I)! The idea that Fe(I) is present in this famous brown ring test is outdated. NO is a non-innocent ligand and will take the form of $NO^-$ here while the Fe is in the oxidation state +III.

Even if you think about how you prepare it. There are no real Fe(I) compounds (I mean simple ones). And Fe(II) oxidizes really easily to Fe(III) and you use a strong oxidizer like nitrate. And still you expect the Fe to be reduced? There are very few complexes with a real Fe(I). One example is the related nitrosyl-complex nitroprusside, so cyanide ligands instead of water. When this is reduced some percent, I think like 25% may actually contain some Fe(I). But with non-innocent ligands it's pretty much useless to asign any oxidation state to the central atom.

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  • $\begingroup$ You just said it, strange oxidation states. As I mentioned above those are the so called non-innocent ligands. They can have various charges and they can take up electrons or donate them back. Nature uses them as well in their porphyrine-systems. The iron is something between +1 and +3. But just thinking about it logically, how would that reaction work? The nitrosyl is formed from the nitrate you add. Nitrate is a strong oxidizer and the nitrogen is already at its highes oxidation state, while Fe(II) is a good reducing agent. Why would this suddenly switch? $\endgroup$
    – Justanotherchemist
    Commented Mar 5, 2018 at 10:04
  • $\begingroup$ Do not forget the other iron that's involved. Simplifying somewhat by removing water ligands: $4\ce{Fe^{2+}}+4\ce{H^+}+\ce{NO_3^-}+\rightarrow \ce{FeNO^{2+}}+3\ce{Fe^{3+}}+2\ce{H_2O}$. You need to oxidize the "other" iron ions to balance oxygen and keep that element where it wants to be (OS=-2). $\endgroup$
    – Oscar Lanzi
    Commented Mar 5, 2018 at 10:39
  • $\begingroup$ That is correct. The actual reaction is the oxidation of the Fe(II) to Fe(III) and the reduction of nitrate to nitrosyl. And when the nitrosyl forms the rest of the Fe(II) will form the brown-purple complex. Nonetheless is there no Fe(I) in this reaction. I think I read about a high-energy form of said complex with Fe(I) but at ground state it's definately an Fe(III). $\endgroup$
    – Justanotherchemist
    Commented Mar 5, 2018 at 11:58
  • $\begingroup$ Now I see what chemists have found, and deleted my first comment. Still want to show the role of additional iron though. $\endgroup$
    – Oscar Lanzi
    Commented Mar 5, 2018 at 13:54
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You need to consider the fact that d-orbitals split in octahedral field. You will have 3 x t2g-type orbitals and 2 x eg-type orbitals. Then, you need to determine the oxidation state of Fe, which you have correctly done - it's +1. Fe normally has 8 valence electrons, minus 1 for the charge gives you 7. Therefore, this is a d7 complex. This can be either low spin or high spin. Low spin d7 has 1 unpaired electron, high spin d7 has 3 unpaired electrons. However, even though NO+ is a very strong field ligand, there's just one, so the complex is most likely high spin, also considering the fact that Fe is first row transition metal.

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