# What is an innocent ligand vs. a non-innocent ligand?

Mar 30, 2016

An innocent ligand is one that binds to a transition metal in such a way that the oxidation state of the metal is able to be determined with clarity, i.e. the method of binding and the contributed charge are both unambiguous.

On the other hand, a non-innocent ligand can bind in multiple ways in which the contributed charge must be stated first before being able to determine the oxidation state of a transition metal with clarity.

Some examples of innocent ligands:

• acetylacetonato (a bidentate that bonds via the $\pi$ system established by the two oxygens)
• ammine (${\text{NH}}_{2}$) (simple monodentate ligand)
• ethylenediamine (${\text{H"_2"N"-("CH"_2)_2-"NH}}_{2}$; cis) (bidentate ligand that binds in a similar way to the ammine ligand)

Some examples of non-innocent ligands:

• Nitrosyl (${\text{NO}}^{+}$ vs. ${\text{NO}}^{-}$)
• Dioxo (${\text{O}}_{2}^{-}$ vs. ${\text{O}}_{2}^{2 -}$)
• Dithiolene (three different possible bonding modes!!!)

For the rest of this answer I go into more depth on the nitrosyl ligand, which is apparently "simple" (at least, according to Jørgensen).

THE NITROSYL LIGAND IS "NON-INNOCENT"

A "simple" example of a non-innocent ligand that we can go more in-depth on is the nitrosyl ligand, which can either bind as ${\text{NO}}^{-}$ (bent geometry) or ${\text{NO}}^{+}$ (linear geometry).

Two examples that illustrate a difference in bonding mode for the nitrosyl ligand are:

• $\setminus m a t h b f \left({\left[\text{Co"("en")_2("NO")"Cl}\right]}^{+}\right)$ (with a bent nitrosyl)
• \mathbf(["Fe"("CN")_5("NO")]^(3-)) (with a linear nitrosyl).

One of the isomers for each compound are shown below.

In the iron compound, we have five ${\text{CN}}^{-}$ and one ${\text{NO}}^{+}$ each contributing to the overall charge, so with $+ 1 - 5 \cdot 1 + 1 = - 3$ charge, we know that iron is ${\text{Fe}}^{+}$.

So, we have accounted for an ${\text{NO}}^{+}$, linear, bonding mode. FYI, iron in this case is a ${d}^{7}$ metal. For this compound we have $6$ $\pi$-compatible (${t}_{2 g}$) electrons, and the oxidation state of the metal is $\textcolor{b l u e}{+ 1}$.

In the cobalt compound, we have ${\text{NO}}^{-}$ and ${\text{Cl}}^{-}$ contributing to the overall charge (ethylenediamine contributes $0$), so with a $+ 3 - 1 - 1 = + 1$ charge), we know that cobalt is ${\text{Co}}^{3 +}$.

So, we have accounted for an ${\text{NO}}^{-}$, bent, bonding mode. FYI, cobalt in this case is a ${d}^{6}$ metal. For this compound we have $7$ $\pi$-compatible (${t}_{2 g}$) electrons, and the oxidation state of the metal is $\textcolor{b l u e}{+ 3}$.

Not the same oxidation state for the metal atom!

According to the Enemark-Feltham approach, the $7$ $\pi$-compatible electrons on the cobalt compound favors a bent ${\text{NO}}^{-}$, whereas the $6$ $\pi$-compatible electrons on the iron compound favors a linear ${\text{NO}}^{+}$.

WHY THE DIFFERENCE IN BONDING MODE FAVORABILITY?

If you're curious, the difference in the bonding modes can be explained by the extent of $\setminus m a t h b f \left(\pi\right)$ back-bonding from the metal to the ligand.

This is a stabilizing interaction between the compatible metal $d$ orbitals and ${\pi}^{\text{*}}$ ligand LUMOs because the former is lower in energy and the latter is higher in energy (the orbitals lower in energy become stabilized).

LINEAR NITROSYL

Basically, ${\text{NO}}^{+}$ has a ${\pi}^{\text{*}}$ orbital aligned with a metal ${t}_{2 g}$ orbital (such as $3 {d}_{x z}$), so it can accept electron density from a compatible $d$ bonding MO from the metal to stabilize this linear bonding configuration.

If we take the plane of the screen to be the $x z$ plane, then we have this metal-to-ligand ${\pi}_{x}^{\text{*}}$ acceptance:

A similar $\pi$ acceptance can work with the $3 {d}_{y z}$ orbital and the ${\pi}_{y}^{\text{*}}$ MO.

BENT NITROSYL

On the other hand, ${\text{NO}}^{-}$ is bent away from the vertical axis, so the orbitals that contain the lone pairs of electrons are not compatible with the $3 {d}_{x y}$, $3 {d}_{x z}$, or $3 {d}_{y z}$ orbitals of the metal.

They are compatible with only the ${d}_{{z}^{2}}$ and ${d}_{{x}^{2} - {y}^{2}}$ orbitals via $\setminus m a t h b f \left(\sigma\right)$ donation, rather than $\pi$ acceptance.

So, we have the issue that the nitrogen contributes no $\pi$-compatible orbitals required to accept electrons; they're aligned for $\sigma$ donation instead, which conflicts with the "desire" of the metal to $\pi$-donate:

As a result, electron density cannot be donated into the ${\pi}^{\text{*}}$ MOs of ${\text{NO}}^{-}$ in a $\pi$ fashion, and a linear configuration is not stabilized enough. Thus, its bonding mode stays bent.

Usually textbooks will specify which bonding mode they want to use for a given non-innocent ligand; mine specifies ${\text{NO}}^{+}$ but doesn't use ${\text{NO}}^{-}$, because it is nice to analogize ${\text{NO}}^{+}$ with $\text{CO}$, as they are isoelectronic.