I'm trying to understand the Biuret test by first understanding how copper dissolves in the aqueous solution. We first have added the $\ce{CuSO_{4}}$ to a substance than the $\ce{OH^{-}}$.

I've read the following page on the reaction, but I still confused: http://www.docbrown.info/page07/transition09Cu.htm

How is it possible to have tetraaquacopper(II) and hexaaquacopper(II) ions in the solution? How could this configuration be explained using electron configuration? $[\ce{Ar}]\mathrm{3d^{10}4s^{1}}$


1 Answer 1


$\ce{Copper(II)}$ complexes are $\ce{d^9}$ type complexes. This means that there are 9 d-electrons for the metal centre. When a complex forms, a number of ligands coordinate to the metal centre and the species adopts certain geometry. The five d-orbitals are no longer the same energy. They split into sets of different energies depending on the geometry and how ligands interact with them.

For an octahedral complex, orbitals split into two sets of degenerate orbitals - a lower $\ce{t_{2g}}$ ($\ce{d_{xy}},\ce{d_{yz}},\ce{d_{xz}}$) which is bonding to non-bonding and a higher $\ce{e_g}$ ($\ce{d_{z^2}}, \ce{d_{x^2-y^2}}$) which is slightly anti-bonding.

Copper is a bit funny as it experiences what is called a Jahn-Teller distortion where there is an elongation of the metal-ligand bonds along the z-axis. This is due to uneven occupancy of the $\ce{e_g}$ orbitals set. Then, the orbitals further split to remove the degeneracy and gain more stability. For the square planar complex of $\ce{tetraaquacopper(II)}$ to form you would need to go even further and remove those two ligands along the z-axis. The stabilization energy gained overall by going from octahedral to square planar configuration is what allows this switch to occur; it wouldn't happen if it were unfavorable. The gain in stability is borderline and that is why I assume that both configurations are possible.

Also, square planar fields tend to prefer stronger field ligands and water is not such ligand and this might also be considered as a reason why you get a mixture of the two types of geometry.

  • $\begingroup$ So,what would be the electron configuration of the tetraaquacopper(II)? $\endgroup$
    – Matt
    Oct 8, 2016 at 2:12
  • $\begingroup$ What does it means : " it wouldn't happen if it were unfavorable"? In my specific case, my copper ion is bonded with four amines. Is this creating a stronger field than water, considering that is linked with nitrogen? $\endgroup$
    – Matt
    Oct 8, 2016 at 2:22
  • 1
    $\begingroup$ The electron configuration for square planar Cu(II) would be (eg)2(a1g)2(b2g)2(b1g)1. Yes, ammonia is a stronger field ligand than water. By saying "it wouldn't happen if it were unfavourable" I mean that everything happens for a reason and as nature is lazy, no system likes having too much energy so it would utilise every possibility to lower its energy, hence the distortion I mentioned and then forming the square planar complex. However, due to the occupancy of an antibonding orbital it really isn't true square planer, two waters would still hang about at the axials. $\endgroup$
    – Vlad
    Oct 8, 2016 at 2:30
  • $\begingroup$ This is at least what I think happens and how I would explain it. $\endgroup$
    – Vlad
    Oct 8, 2016 at 2:31
  • $\begingroup$ @Matt The complex often described as tetraamminecopper(II) is, in fact, tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$. The two aqua-ligands are just further away from the copper centre. $\endgroup$
    – Jan
    Oct 9, 2016 at 20:26

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