# Why is ligand substitution only partial with copper(II) ions and ammonia?

When studying ligand substitution (at UK year 13 level), the following example has been given:

\begin{align} \ce{[Cu(H2O)_6]^2+ + 2NH3 &<=> [Cu(OH)_2(H2O)_6] + 2NH4^+}\\ \ce{[Cu(OH)_2(H2O)_4] + 4NH3 &<=> [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\ \end{align}

There has been no explanation given of why the substitution is only partial, whereas the substitution of, say, $\ce{[Co(H2O)_6]^2+}$ is complete.

To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? Moreover my textbook states ammonia is a better ligand than water, so why would $\ce{NH3}$ not replace those final two aqua ligands?

Metal ion complexes have stepwise stability constants: \begin{align} \ce{[Cu(H2O)6]^2+ + NH3 &<=>[K_1] [Cu(NH3)(H2O)5]^2+ + H2O}\\ \ce{[Cu(NH3)(H2O)5]^2+ + NH3 &<=>[K_2] [Cu(NH3)2(H2O)4]^2+ + H2O}\\ \ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &<=>[K_3] [Cu(NH3)3(H2O)3]^2+ + H2O}\\ \ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &<=>[K_4] [Cu(NH3)4(H2O)2]^2+ + H2O}\\ \ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &<=>[K_5] [Cu(NH3)5(H2O)]^2+ + H2O}\\ \ce{[Cu(NH3)5(H2O)]^2+ + NH3 &<=>[K_6] [Cu(NH3)6]^2+ + H2O}\\ \end{align} with

\begin{align} K_1 &= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\ K_2 &= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\ K_3 &= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\ K_4 &= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\ K_5 &= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\ K_6 &= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\ \end{align}

The overall stability constant is:

$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$

(In particular, note the exponent 6.)

Furthermore, usually $$K_i/K_{i+1}>1$$; thus, $$K_1 > K_2 > K_3 > K_4 > K_5 > K_6$$. Consequently, it gets more and more difficult to exchange further ligands.

Therefore, the predominating complex strongly depends on the concentration of $$\ce{NH3}$$. And you probably work with dilute aqueous solutions.

Simply speaking, the missing $$\ce{[Cu(NH3)6]^2+}$$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.

To my understanding, $$\ce{Cu(II)}$$ actually prefers four-fold coordination with ligands like water and ammonia, in the square planar configuration. Gas-phase DFT / MD simulations bear this out$$^1$$, at least for $$\ce{H2O}$$, and the pioneering work$$^2$$ of Jannik Bjerrum indicates, apropos Loong's answer, that the first four step-wise stability constants for the substitution of ammonia ligands for water are significantly higher than the fifth (IIRC the 6th was un-measurable at that time). I'm pretty sure this is because the $$\ce{[Ar]}3\mathrm d^9$$ ground state configuration of $$\ce{Cu(II)}$$ just doesn't have enough un-occupied d-orbitals to support 'proper' six-coordination.

You're absolutely correct that $$\ce{NH3}$$ is a better ligand than water -- this explains why the first four molecules substitute readily into the $$\ce{Cu(II)}$$ coordination environment. However, if the remaining "axial coordination sites" are really not coordination sites at all, but instead are part of a non- or weakly-coordinating solvation shell, then it just becomes a statistical competition between excess $$\ce{NH3}$$ solutes and the $$\ce{H2O}$$ solvent, in which case the water will win out until the ammonia concentration is quite high.

On the contrary, the comparatively d-electron-poor $$\ce{Co(II)}$$ $$(\ce{[Ar]}3\mathrm d^7)$$ has plenty of room in the $$3\mathrm d$$ subshell to support six-coordination.

$$^1$$ Bérces et al. J Phys Chem A 103: 9693-9701 (1999), doi: 10.1021/jp992612f

$$^2$$ Bjerrum, J. "Metal Ammine Formation in Aqueous Solution." Haase and son, 1941. (link)

In this complex you're going to see an arrangement with 4 equatorial and 2 axial ligand positions. In such compounds, ligands that are stronger σ-donors and weaker π-acceptors preferentially occupy the axial positions. On the other hand, weaker σ-donors and stronger π-acceptors preferentially occupy equatorial positions. A quick look at the spectrochemical series (which ranks ligands by increasing field splitting effects which have been measured experimentally, which is the same as ranking them by increasing π-acceptor strength) shows that $\ce{NH3}$ is a stronger π-acceptor than both $\ce{H2O}$ and $\ce{HO}$, which means that it will preferentially occupy equatorial positions while the $\ce{HO/H2O}$ preferentially occupies the axial positions. This is because the equatorial plane is electron-rich, which is a more stable environment for a stronger π-acceptor. Conversely, the axial positions are electron-deficient, and thus they attract the stronger σ-donors.

The reason you only see 4 substitutions is because these substitutions are reversible. It is energetically favorable for a stronger π-acceptor to displace equatorial ligands, but it's not energetically favorable for a weaker σ-donor to displace an axially-substituted ligand.

You can sort of see this in the structures of the ligands. $\ce{NH3}$ has one lone pair, $\ce{H2O}$ has 2 lone pairs, and $\ce{HO-}$ has 3 lone pairs. σ-bonding interactions involve free lone pairs on the ligands, so in this case you can see that more lone pairs means a greater ability to σ-bond. Now look up how the spectrochemical series ranks these three ligands in terms of increasing σ-donor/decreasing π-acceptor strength.

• I've never heard of such a theory before and would be very interested to see a book that explains it. – orthocresol Jan 7 at 19:13
• It's a direct consequence of MO theory, any graduate-level text covering coordination chemistry should do the trick. – Stagg C. Jan 8 at 20:13
• Do you have a particular book in mind? Or a website, somewhere out there? I have genuinely never heard of such a thing and it doesn't make sense to me when considering that in an octahedral geometry, all six positions are the same, so there is no such thing as "equatorial" or "axial". If you're talking specifically about the Jahn–Teller distorted geometry of Cu(II), then that's not an issue, but the text seems to imply that it's generally true for everything. Lastly, I've never seen NH3, H2O, or OH− being referred to as π acceptors. – orthocresol Jan 8 at 20:28
• You could also explain the phenomenon based on σ-donor strength if that's a more comfortable way to think about it. It's probably the better way to think about it since the three are σ-donors, perhaps I was generalizing where I shouldn't. You'll still come to the same conclusion as to which bond where. Another term you want to look up is Ligand Field Theory, that will give you resources. – Stagg C. Jan 8 at 20:58
• Honestly this is like 4 years old and I have spent those years studying medicinal organic chemistry, I'm really rusty on MO theory and LF theory. Good luck! – Stagg C. Jan 8 at 21:01

From the question:

To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? (Emphasis mine)

This is a misconception. The shape of the hexacoordinated copper(II) complexes is independent of the type of ligands. Indeed, the hexaaquacomplex $\ce{[Cu(H2O)6]^2+}$ already features the same distorted shape. This distortion is termed Jahn-Teller distortion. The distortion actually results in the distorted complex being more stable than a regular octahedron. In general, you can (and should) assume that the shape a complex assumes (in solution and in crystal structures) is the most stable or at least more stable than other easily accessible variations.

Ammonia, in general, does give a stronger ligand-metal interaction with transition metals than water as a ligand. Therefore, there should be a general trend towards amination to give more stable complexes. However, in the case of axial ligands in a Jahn-Teller distorted complex it is actually slightly better to have a slightly weaker ligand there. The hexaammin complexes can be attained in liquid ammonia but when dissolved in water two ammin ligands redissociate and are replaced by aqua ligands.