In the standard test for the sulfide ion, sodium nitroprusside $(\ce{Na^_2[Fe(CN)5(NO)]})$ is used. The IUPAC name for this compound as stated by Wikipedia is sodium pentacyanidonitrosylferrate(II)

I have three queries related to this:

  1. Does the nitrosyl ligand donate a pair of electrons (through back bonding) to form $\ce{(NO)^+}$ or is $\ce{(NO)^+}$ the ligand?

  2. In both the cases I guess the name should be sodium pentacyanidonitrosoniumferrate(II) instead.

  3. Now the product formed after the sulfide test is $\ce{Na^_4[Fe(CN)5(NO)S]}$ whose IUPAC name is sodium pentacyanidonitrosylsulfidoferrate(III). But the nitrosyl ligand is neutral in this case. So why in this case there is no back bonding?


1 Answer 1


Nitrosyl is the textbook example for a non-innocent ligand — that is, one that can appear in different oxidation states with different numbers of donated electrons and can make determining the bonding situation rather complicated. It can appear as the nitrosyl cation $\ce{NO+}$ (isosteric to $\ce{CO}$ or $\ce{CN-}$), the nitrous oxide radical $\ce{NO^{.}}$ or the ‘hyponitrite’ anion $\ce{NO-}$ (isosteric to either $\ce{{}^1O2}$ or $\ce{{}^3O2}$). I’m not going to start talking about possible donated electron counts yet.


For nomenclature purposes, any form of $\ce{NO}$ bound to a metal is always treated as a neutral three electron donor and is always termed nitrosyl. This is because IUPAC wants rules that can be applied blindly without having to perform complicated experiments to determine how exactly the ligand is bound. And we shall see in a second that it is very non-trivial.

Considering that and checking the $\ce{[Fe(CN)5NO]^2-}$ fragment, we take away 5 $\ce{CN-}$ leaving us with $\ce{[Fe(...)NO]^3+}$ which indicates that the correct IUPAC name for sodium nitroprusside is sodium pentacyanidonitrosylferrate(III). (Not (II).)

The product formed would be, by the same approach, sodium pentacyanidonitrosylsulfidoferrate(III). But we’ll come back to that.

But what is the actual ligand?

As I already stated, nitrosyl is a non-innocent ligand. It can be $\ce{NO+}$ and allow significant π-backbonding (metal to ligand) as exemplified in the tricarbonylnitrosylferrate($\mathrm{-I}$) system $\ce{[Fe(CO)3NO]-}$. By formality, this is considered a neutral donor giving iron a formal oxidation state of $\mathrm{-I}$. But in reality the ligand is $\ce{NO+}$ meaning that we are dealing with iron($\mathrm{-II}$) — analogous and isoelectronic to the tetracarbonylferrate($\mathrm{-II}$) complex.

On the other side of the spectrum we have the complex formed in one nitrate detection method: $\ce{[Fe(H2O)5NO]^2+}$ — pentaaquanitrosyliron(II) according to IUPAC. Approaching this complex is difficult, however most evidence points towards some structure between $\ce{[Fe^{II}NO^{.}]^2+}$ and $\ce{[Fe^{III}NO- ]^2+}$. The $\ce{Fe-N-O}$ unit is still linear (VSEPR would predict $120^\circ$ for a hyponitrite). This can be explained by considering the $\ce{NO-}$ (or $\ce{NO^{.}}$) in a triplet state much like $\ce{^3O2}$ (or a corresponding dublet state in case of $\ce{NO^{.}}$). There is a slight tendency for the unpaired electrons in iron and $\ce{NO}$ to couple; this is usually understood as antiferromagnetic coupling and results in an overall observed spin of $+\frac{3}{2}$.

In the $\ce{NO+}$ case, the bonding picture is rather clear: $\ce{NO+}$ behaves exactly like $\ce{CO}$. We have a σ donating (forwards) bond and significant π back bonding. The other case is rather hazy except for the σ bonding, which is of course ligand to metal. As I wrote, the spin coupling is considered antiferromagnetic coupling and not actual bonding usually, and if anything, it would be a covalent type of bond (each fragment donating one electron). Thus it would be wrong to speak about back-bonding — especially since metals in high oxidation states usually don’t back-bond significantly.

Removing the five cyanido ligands in nitroprusside leaves us with a $+3$ charge as shown above. Thus we are either dealing with $\ce{[Fe^{II}NO+]}$ or $\ce{[Fe^{III}NO^{.}]}$. The extremely short $\ce{N#O}$ distance points strongly towards an $\ce{NO+}$ cation. $\ce{Fe^{II}}$ is again not good at back-bonding and no π forward bonding is to be expected due to the mismatched energy of the π electrons of $\ce{NO}$. Therefore, the $\ce{NO}$ group is rather labile and can be displaced at ambient temperature and pH. This is the mode of action of nitroprusside as a drug: It releases $\ce{NO}$.

And what about the sulfide test?

I didn’t dig into it deeply, but it seems like you are mistaken in the product formed. What I found on the web[1] implies (I couldn’t read the actual article, only the abstract) that sulfur does not coordinate additionally to the iron centre (there would not be much space anyway) but rather attacs the $\ce{N#O}$ triple bond nucleophilicly on the nitrogen atom. Therefore, the final complex is better written as $\ce{[Fe(CN)5(SNO{-}\kappa N)]^4-}$ and better named pentacyanidothionitrito-κN-ferrate(III).


[1] S. L. Quiroga, A. E. Almaraz, V. T. Amorebieta, L. L. Perissinotti, J. A. Olabe, Chem. Eur. J. 2011, 17, 4145. DOI: 10.1002/chem.201002322

Background from Prof. Klüfers’ internet scriptum to his coordination chemistry II course.

  • $\begingroup$ A forum discussion in the depths of the interwebs (no source because just no) seems to confirm my assumption on thionitrite. $\endgroup$
    – Jan
    Oct 27, 2015 at 21:41
  • $\begingroup$ According to "Red Book- IUPAC Recommendations 2005 ", at page 315, $\ce{NO+}$ is oxidonitrogen. $\endgroup$
    – Apurvium
    Oct 21, 2021 at 5:11

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