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

Question

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?

Answer 1

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.

Answer 2

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.

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$^1$ 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)

Answer 3

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.

Answer 4

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.