# Consider the molecule [MnX6]2+, where X is a neutral ligand. Assume that X is a strong field ligand?

Dec 1, 2016

A)

THE NUMBER OF 3D VALENCE ELECTRONS IN THE COMPLEX

When you examine the location of $\text{Mn}$ on the periodic table, you should find that the electron configuration ends in $4 {s}^{2} 3 {d}^{5}$.

Since $\text{X}$ is stated to be a neutral ligand, $\text{Mn}$ in ${\left[{\text{MnX}}_{6}\right]}^{2 +}$ by charge conservation is a $+ 2$ oxidation state. Therefore, we doubly ionize neutral manganese to obtain the electron configuration $3 {d}^{5}$, removing the two $4 s$ electrons first.

HIGH-SPIN VS LOW-SPIN, AS IT RELATES TO STRONG-FIELD AND WEAK-FIELD LIGANDS

A ${d}^{5}$ metal such as this can form either a high-spin or low-spin complex, depending on the crystal field strength of the ligand.

We assume strong-field for $\text{X}$, which means that it can either be:

• A $\sigma$ donor (such as $: {\text{NH}}_{3}$, $\text{CO}$, or $\text{^(-)"CN}$)
• A $\pi$ acceptor (such as $\text{CO}$ or $\text{^(-)"CN}$)
• Both (such as $\text{CO}$ or $\text{^(-)"CN}$)

SIGMA DONORS

Let's say we just have a good $\sigma$ donor.

In this case, it destabilizes the ${e}_{g}$ orbitals (there are two of those orbitals), since those interact with the $\sigma$ bonding molecular orbitals brought in by the $\sigma$ donor(s). $\text{CO}$ is an example:

As a result, the two upper-energy (${e}_{g}$) orbitals get higher in energy (relative to the free-ion energy), increasing the crystal-field splitting energy ${\Delta}_{o}$ ($\Delta$ in your diagram), promoting low spin.

PI ACCEPTORS

Or, let's say we have a good $\pi$ acceptor.

In this case, the $\pi$ antibonding molecular orbitals of the ligand accept electrons from the $3$ lower-energy (${t}_{2 g}$) $3 d$ orbitals.

This stabilizes these $3$ lower-energy (${t}_{2 g}$) $3 d$ orbitals, decreasing their energy. This also increases ${\Delta}_{o}$.

Since $\text{CO}$ is both, it is one of the strongest-field ligands there is.

CRYSTAL-FIELD SPLITTING DIAGRAM

When a transition metal complex has a coordination number of $6$ (given that it has six $\text{X}$ ligands bound to it), it forms an octahedral electron geometry.

The diagram you were given is the crystal-field splitting diagram for the five $3 d$ orbitals of manganese once the manganese(II) complex is placed into an octahedral field that splits its $d$-orbital energies.

Here is the filled diagram for a manganese(II) complex in high spin or low spin:

The difference is that for a low spin complex, which corresponds to having a strong-field ligand, ${\Delta}_{o}$ is large.

So the electrons can capably fill the $3$ lower-energy ${t}_{2 g}$ orbitals first... before they can feasibly fill the $2$ higher-energy ${e}_{g}$ orbitals.

B)

PI DONORS

A weak-field ligand contributes little repulsion. Actually, weak-field ligands are typically $\pi$ donors, and weak $\sigma$ donors.

If you have a $\pi$ donor, such as any halide (${\text{Cl}}^{-}$, ${\text{F}}^{-}$, etc):

It has filled $\pi$ bonding molecular orbitals that donate into the three (${t}_{2 g}$) $3 d$ orbitals and destabilize them, increasing their energy.

That actually decreases ${\Delta}_{o}$, and promotes high spin.

So, ${\Delta}_{o}$ would be smaller for a weaker-field ligand.

This diagram summarizes the trends for $\pi$ acceptors, $\sigma$ donors, and $\pi$ donors:

C)

CRYSTAL-FIELD SPLITTING ENERGY

Since the light absorbed will promote electrons from the ${t}_{2 g}$ into the ${e}_{g}$ orbitals, the size of ${\Delta}_{o}$ tells you how much energy this light frequency corresponds to.

Since ${\Delta}_{o}$ becomes smaller when switching to a weaker-field ligand, and ${\Delta}_{o} \propto E \propto \nu$, a smaller frequency of light can be absorbed than before.

(Thus, you can also say that a longer wavelength is absorbed.)