# What is the number of unpaired electrons in (Fe(H2O)5NO)2+? [closed]

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 $\mathrm{sp^{3}d^{2}}$.

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 The idea that Fe(I) is present in this famous brown ring test is outdated.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.
• 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). – Oscar Lanzi Mar 5 '18 at 10:39