How can the electronic structure of pentaaquanitrosyliron be explained?

The magnetic moment of $\ce{[Fe(H2O)5NO]^2+}$ is 3.87 Bohr magneton, which implies that there are 3 unpaired electrons. Since iron is in $+1$ oxidation state (due to charge transfer with nitrosyl group), it's electronic configuration is $[\ce{Ar}]\ \mathrm{3d^7}$. Having 3 unpaired electron is only possible if the 4th and 5th electron enter the higher energy orbital $\mathrm{e\_g}$ (crystal field theory). The $\mathrm{t\_{2g}}$ or the lower energy orbital has 5 electrons. Why do the 4th and 5th electron enter $\mathrm{e\_g}$ despite a strong field ligand (nitrosyl group) being present?




As explained in [Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited](http://pubs.acs.org/doi/pdf/10.1021/ic010628q) Inorg. Chem. 2002, 41, 4-10, the $\ce{Fe}$ is really 3+ oxidation state. 


The $\ce{Fe^3+}$ is spin 5/2 and antiferromagnetically couples to spin 1 $\ce{NO-}$ to yield a S=3/2 state.


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The presence (or absence) of *a single* strong field ligand is rarely enough to tip the entire complex’ balance from high to low spin (or vice-versa). I have often come across questions here that attemt to argue ‘but $\ce{X}$ is a strong field ligand, why is complex $\ce{Y}$ not low-spin?’, and the answer is always that the *entire* picture must be considered — merely looking at just one ligand is never enough.


To add to the confusion, some complexes are known in *both* the high and low spin state, indicating that there is a tipping point that can be crossed but which also needs to be reached; and if you only barely reach it you might tip this way or the other. So *do not* take the presence of a single cyanido or carbonyl ligand and immediately assume that the complex should be low-spin.


Finally, the actual electronic structure of pentaaquanitrosyliron $\ce{[Fe(H2O)5(NO)]^2+}$ must be considered. This complex is a very tricky one because iron is very redox active and nitrosyl is what is known as a very non-innocent (promiscuous?) ligand: it can occur as all three of $\ce{NO+, NO^.}$ and $\ce{NO-}$. To add to the confusion, these can also bond in a linear or bent fashion — and one cannot simply look at a crystal structure and infer the charge of nitrosyl from its bent or linear configuration. And lastly, $\ce{NO-}$ which is isoelectronic to $\ce{O2}$, can adopt both a singlet and a triplet configuration. Depending on which nitrosyl variant you decide to use, the oxidation state of iron will be $\mathrm{+I}$, $\mathrm{+II}$ or $\mathrm{+III}$ in the same order.


For IUPAC nomenclature purposes — which are based on observing a certain configuration and deriving a systematic name without knowledge of the intrinsic chemistry — nitrosyl is always considered a neutral three-electron donor ($\ce{NO^.}$; donating the radical *and* a lone pair) leading to a formal IUPAC oxidation state of $\mathrm{+II}$. The physical reality was hard to determine. The spin is known to be $\frac32$ which is consistent with a number of different states. The currect state of science is to assume iron(III) ($S = \frac52$) and a singlet nitrosyl anion ($S=1$; hyponitrite?). These then couple antiferromagnetically to give an overall spin state of $S=\frac32$ as DavePhD’s answer already indicates.


What seems to be a very simple problem is, in fact, rather complicated! Indeed, the notation $\ce{\{FeNO\}^7}$ has been proposed for nitrosyliron complexes where the superscripted $7$ denotes the number of electrons in iron’s d orbitals and the ligand’s coordinating orbitals. This allows one to write a structure without having to imply any given actual electronic configuration.


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