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 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.