Animated Solution for Chemistry - Coordination Compounds: The most stable ion is

When we look at a coordination complex, its overall stability is governed by a delicate dance between the central metal ion and the surrounding ligands. To determine which complex is the most stable, we must evaluate two primary factors.

First, we look at the charge on the central metal ion. A higher positive charge means the metal ion has a greater charge density. This allows it to exert a stronger electrostatic pull on the electron-rich ligands, drawing them in closer and forming a more stable complex.

Second, we must consider the field strength of the ligands. Not all ligands are created equal. Some are weak and form relatively loose bonds, while others are strong and bind tightly to the metal center. The stronger the ligand, the more stable the resulting complex.

Let's apply our first rule to the four complexes given in the problem: $[Fe(OH{)}_6{]}^{3-}$, $[FeC{l}_6{]}^{3-}$, $[Fe(CN{)}_6{]}^{3-}$, and $[Fe(H_{2}O{)}_6{]}^{3+}$. We need to determine the oxidation state of the central iron ($Fe$) atom in each case.

For $[Fe(OH{)}_6{]}^{3-}$, we have six hydroxide ions, each with a $-1$ charge. Setting the oxidation state of iron to $x$, we get:

For $[FeC{l}_6{]}^{3-}$, we have six chloride ions, also with a $-1$ charge:

For $[Fe(CN{)}_6{]}^{3-}$, we have six cyanide ions, which carry a $-1$ charge:

Finally, for $[Fe(H_{2}O{)}_6{]}^{3+}$, we have six neutral water molecules:

In every single complex, the iron ion is in the $+3$ oxidation state! Because the central metal ion and its charge are identical across all options, this factor cannot help us differentiate their stabilities. We must move on to the second factor.

Since the metal ion is a tie, the stability of these complexes rests entirely on the shoulders of the ligands: $O{H}^{-}$, $C{l}^{-}$, $C{N}^{-}$, and $H_{2}O$. To compare them, we turn to the spectrochemical series.

The spectrochemical series is an experimentally determined list that ranks ligands based on their field strength—their ability to split the d-orbitals of the central metal ion. The order of field strength for our specific ligands is:

Chloride is the weakest ligand in this group, followed by hydroxide, then water. Cyanide, however, stands out as a exceptionally strong field ligand.

Because $C{N}^{-}$ is the strongest ligand among the choices, it will form the most robust and stable bonds with the $F{e}^{3+}$ ion.

But why is cyanide so strong? It comes down to a phenomenon called synergic bonding. Cyanide doesn't just donate a lone pair of electrons to the metal to form a standard $\sigma$-bond. It also has empty ${\pi }^{\ast }$ antibonding orbitals that can accept electron density back from the filled d-orbitals of the metal. This two-way street of electron sharing creates an incredibly tight and stable bond.

Therefore, thanks to the formidable field strength of the cyanide ligand, the hexacyanoferrate(III) ion, $[Fe(CN{)}_6{]}^{3-}$, is the most stable complex among the given options.