Which complexes are stable: ammine complexes or carbonyl complexes?

Question

According to Wikipedia,

A spectrochemical series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and its identity.

$$\ce{I- < Br- < SCN- < Cl- < S^2- < F- < OH- < C2O4^2- < H2O < NCS- < edta^4- < NH3 < en < CN- < CO}$$

Ligands arranged on the left end of this spectrochemical series are generally regarded as weaker ligands and cannot cause forcible pairing of electrons within 3d level and thus form outer orbital octahedral complexes. On the other hand ligands lying at the right end are stronger ligands and form inner orbital octahedral complexes after forcible pairing of electrons within 3d level.

According to above passage $\ce{CO}$ is a stronger ligand than $\ce{NH3}$. So, carbonyl complexes are more stable than ammine complexes.

Now, there are many other factors which determine the stability of complex:

1. Nature of central metal atom
2. Nature of ligand
3. Crystal field effects

Nature of ligand has further classification:

1. Basic character of ligands
2. Charge on ligands
3. Chelate effects

Classification source: see here and here

Basic character of ligands states that the more basic a ligand is, the greater the ease of ligation with the central metal atom is and the more stable the complex is. Now, $\ce{NH3}$ is more or less a basic compound and $\ce{CO}$ is a neutral molecule.

So, according to above assumption, $\ce{NH3}$ is a stronger ligand than $\ce{CO}$. So, ammine complexes are more stable than complex complexes.

So, which one is the stronger ligand: $\ce{CO}$ or $\ce{NH3}$? or which complexes are stable: ammine complexes or carbonyl complexes?

Answer 1

That quotation from Wikipedia is bad and should be reworded (I can’t think of good ways to edit it, though). Ligands in a spectrochemical series are not ordered by bonding strength or anything, they are ordered solely by the field split they induce.

Both $\ce{CO}$ and $\ce{NH3}$ are great at forming complexes — as is iodide. The main difference is that they won’t form similarly strong complexes with the same cores. Just to give a few examples:

• Carbon monoxide is neutral and a π-acid: It has an empty orbital with π symmetry which can take part in metal-to-ligand back-bonding. As such, it stabilises low oxidation states of metals. The neutral carbonyl complexes such as $\ce{[Fe(CO)5]}$ are well known, but also complexes such as $\ce{[Fe(CO)4]^2-}$ (yes, that’s iron($\mathrm{-II}$)!) are stable.

• Ammonia is also neutral and a very weak π-base. However, it is a much stronger Brønsted base than carbonyl is. Therefore, it is much better at stabilising medium to high oxidation states such as in $\ce{[Cu(NH3)4(H2O)2]^2+}$.

• The halides, finally, are anionic and good π bases and thus able to stabilise high oxidation states. They are classed as weak-field ligands because they generally generate a very low field split. An example of a halide complex would be the bright yellow $\ce{[FeCl6]^3-}$.

You cannot take ligand $\ce{X}$ and call it strong or weak and thus expect it to make stable or unstable complexes. You cannot even take the same ligand and expect it to make inert or labile complexes all the time. One ligand-metal combination may be stable, another may be very unstable — $\ce{[Fe(CO)6]^2+}$ can only be synthesised under a $\ce{CO}$ atmosphere in $\ce{SbF5}$ from $\ce{FeF2}$ where the antimony pentafluoride abstracts any free fluoride ions from the solution leaving a ‘naked’ iron ion which will grab the only ligands it can reach. Iron pentacarbonyl is stable. The same goes for ammonia: Some will readily accept it as a ligand, others won’t. There is no black and no white, only shades of grey.

(Although since coordination compounds are generally colourful let me rephrase that: There is no black and no white, only a wheel of colours.)