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A ligand exchange reaction involves many part reactions. So, does the mixture contain all the complexes from those part reactions?

Quoting an example from ChemGuide:

This can be written as an equilibrium reaction to show the overall effect:

$$\ce{[Cu(H2O)6]^2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]^2+ + 4H2O}$$

In fact, the water molecules get replaced one at a time, and so this is made up of a series of part-reactions:

$$ \begin{align} \ce{[Cu(H2O)6]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)}(H2O)5]^2+ + H2O}\\ \ce{[Cu\color{blue}{(NH3)}(H2O)5]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)2}(H2O)4]^2+ + H2O}\\ \ce{[Cu\color{blue}{(NH3)2}(H2O)4]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)3}(H2O)3]^2+ + H2O}\\ \ce{[Cu\color{blue}{(NH3)3}(H2O)3]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)4}(H2O)2]^2+ + H2O}\\ \end{align}$$

And if yes, then which complex has the highest concentration?

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    $\begingroup$ The answer is pretty much given in the link you provide. Each equilibrium constant is large , hundreds to thousands, so pretty much complete reaction and as each species is formed it then reacts to form the next. The vary majority of species will be the last one in your scheme. $\endgroup$ – porphyrin Apr 25 '18 at 14:17
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    $\begingroup$ Related: Why is ligand substitution only partial with copper(II) ions and ammonia? $\endgroup$ – Faded Giant♦ Apr 26 '18 at 16:55
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Yes, all those complexes from all those reactions will be present, and some more. Essentially Loong did all the mathematics in his answer to: Why is ligand substitution only partial with copper(II) ions and ammonia?

Let's shorten this a little bit and write the general reaction equation: \begin{align} \ce{[Cu(H2O)_{(7-n)}]^2+ + n NH3 &<=> [Cu(NH3)_n(H2O)_{(6-n)}]^2+ + n H2O};& n&\in\{\mathbb{N}; 1,\dots,6\} \end{align} Therefore we can write the general equilibrium constant: \begin{align} K_n &= \frac{{\left[ \ce{[Cu(NH3)_n(H2O)_{(6-n)}]^2+} \right]}} {{\left[ \ce{[Cu(H2O)_{(7-n)}]^2+} \right]\left[ \ce{NH3} \right]^n}}\\ % K_\mathrm{B} &= % \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}} % {{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} % &&= \prod_{n=1}^{6} K_n \end{align}

Since all reactions are in equilibrium, all possible species will exist, although they might not be quantifiable. The overall equilibrium depends on the concentration of the involved species, especially ammonia.

If $[\ce{NH3}]$ is low, then a complex with smaller $n$ will be more stable, if you go to higher concentrations, larger $n$ will be preferred.

In all of this it should be mentioned that there are also complexes with vacant coordination sites $\square$, like anything in between $$\ce{[Cu(H2O)5\square_1]^2+},\dots, \ce{[Cu(H2O)3(NH3)2\square_1]^2+},\dots, \ce{[Cu(NH3)5\square_1]^2+},\\ \left(\ce{[Cu(H2O)4\square_2]^2+},\dots,\ce{[Cu(NH3)4\square_2]^2+}\right),$$ and possibly others.

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  • $\begingroup$ Thanks Martin! :D "In all of this it should be mentioned that there are also complexes with vacant coordination sites □, like anything in between" are they just defective complex compounds? Because they don't seem to be present in any of the reactions. $\endgroup$ – Gaurang Tandon Apr 28 '18 at 1:59
  • $\begingroup$ " if you go to higher concentrations, larger $n$ will be preferred." But, according to my logic, when $\ce{[NH3]}>\pu{1M}$, then $\ce{[NH3]}^n$ would increase with increase in $n$, and thus $K_n$ would decrease. So, the tendency for a complex with higher $n$ to form would decrease. Why is this incorrect? $\endgroup$ – Gaurang Tandon Apr 28 '18 at 2:01
  • $\begingroup$ @GaurangTandon There is nothing defective about complexes with empty coordination sites, they exist in equilibrium with the other complexes. I have just not written down all the equations. Your logic is wrong, because with an increasing concentration of one of the reactants, the equilibrium constant won't change; more product will be formed instead. $\endgroup$ – Martin - マーチン♦ Apr 28 '18 at 6:35

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