# What is the oxidation state of the iron center in ["Fe"("H"_2"O")_5"NO"]"SO"_4?

## a) $+ 1$ b) $+ 2$ c) $+ 3$ d) $0$

May 14, 2016

CHARGE BINDING AT THE SECOND COORDINATION SPHERE

In ["Fe"("H"_2"O")_5"NO"]"SO"_4, recall that the charge on sulfate is $2 -$. That is a good starting point. It implies the charge on $\left[\text{Fe"("H"_2"O")_5"NO}\right]$ is $2 +$ to balance it out.

LIGAND CHARGES IN THE FIRST COORDINATION SPHERE

We know that water is a neutral, weak-field ligand. Thus it does not contribute to the overall charge.

The tricky part is that $\text{NO}$ could either bind as ${\text{NO}}^{+}$ (linear, triple-bonded) or ${\text{NO}}^{-}$ (bent, double-bonded)... and we don't really know which. This is known as a "non-innocent ligand", because it messes with us when we are trying to find the oxidation state of the transition metal.

Suppose that it was the linear ${\text{NO}}^{+}$. Then the oxidation state on iron would be $+ 1$. If it was the bent ${\text{NO}}^{-}$, then the oxidation state on iron would be $+ 3$. Unfortunately, both answers are available.

NARROWING DOWN WHICH OXIDATION STATE IS MORE LIKELY

Consider the electron configuration of iron, then. The atomic number is $26$, so its configuration is $\left[A r\right] 3 {d}^{6} 4 {s}^{2}$.

Since the sixth $d$ electron is paired, it is sensible for iron to have a $\textcolor{b l u e}{+ 3}$ oxidation state by losing two $4 s$ and one $3 d$ electron. This gives a $\textcolor{b l u e}{\left[A r\right] 3 {d}^{5}}$ configuration and a total electron spin of 5*"1/2" = color(blue)("5/2").

Although, it could lose one $4 s$ electron and still be favorable according to Hund's rule for maximizing total electron spin when possible, giving it a $\textcolor{b l u e}{\left[A r\right] 3 {d}^{6} 4 {s}^{1}}$ configuration, a $+ 1$ oxidation state, and a total electron spin of 5*"1/2" + ("1/2" + (-"1/2")) = color(blue)("5/2").

That is indeed still the same total electron spin, but the $3 d$ orbitals are higher in energy than the $4 s$ orbitals for iron, AND there is one set of electrons paired, increasing the "pairing energy" of this configuration relative to the $+ 3$ oxidation state (the energy associated with pairing two electrons is repulsive and thus positive).

If this configuration could be stabilized, then it would be more likely...

STABILIZING THE NORMALLY LESS-LIKELY CONFIGURATION?

If the nitrosyl ligand is ${\text{NO}}^{+}$, then it acts as a good $\setminus m a t h b f \left(\pi\right)$-acceptor ligand (isoelectronic with $\text{CO}$), accepting electron density into its antibonding $\pi$ orbitals from a ${d}_{x z}$ or ${d}_{y z}$ orbital on iron via a phenomenon called "$\pi$-backbonding".

That would indeed stabilize the $d$ orbitals by decreasing the amount of electron density that needs to get spread around, thus decreasing the "pairing energy" (making pairing more favorable).

On the other hand, ${\text{NO}}^{-}$, which is bent, is basically not as suitable a $\pi$-acceptor; technically, its antibonding $\pi$ orbital is not the right "symmetry" to accept electrons from iron's ${d}_{x z}$ or ${d}_{y z}$ orbitals.

So, I would expect the nitrosyl ligand to be $\textcolor{b l u e}{{\text{NO}}^{+}}$ and for iron to be the (normally less-likely) $\textcolor{b l u e}{+ 1}$ oxidation state.