Question

Choose the most likely pattern for the crystal field diagram for the complex trans–[Ni(NH₃)₂(CN)₄]^2– where CN^– produces a much stronger crystal field than does NH₃.?

From the answer the pattern show it is a square planer. Could anyone explain it? why is not octahedral?

Answer 1

[Ni(NH₃)₂(CN)₄]^2- shape as square planar can be considered as a special case of octahedral due to jahn-tellar distortion. Here the oxidation state of Ni is 2+, so the electronic configuration will be [Ar] 3d⁸. Now since 6 ligands are there they approach the metal cation along the cartesian coordinate axis. . Thus the eg orbitals, viz 3d_x2-y2 and 3d_z2 will have higher energy than the t2g orbitals, viz dxy, dyz, and dzx. It is bcoz the eg orbitals have lobes in the cartesian cordiantes axis and thus gets more influenced by the incoming ligands, than the t2g orbitals, since their lobes lie in between the cartesian co-ordinates and are thus less affected. Thus the first six electrons will go in the t2g orbitals.

Now, let's assume that the rest 2 electrons are initially uniformly distributed as 1 electron each in 3d_x2-y2 and 3d_z2 orbitals. Since the CN are very strong field ligands, they come very close towards the nucleus of the metal ion.

The single electron in the dx2-y2 orbital is being repelled by four ligands, whilst the electron in the dz' orbital is only being repelled by two ligands. Thus the energy of the dx2-y2 increases relative to that of dz'· If the ligand field is sufficiently strong, the difference in energy between these two orbitals becomes larger than the energy needed to pair the electrons. Under these conditions, a more stable arrangement arises when both the eg electrons pair up and occupy the lower energy dz2 orbital. This leaves the dx2-y2 orbital empty. Thus four ligands can now approach along the +x, -x, +y and -y directions without any difficulty, as the dx2-y2 orbital is empty. However, ligands approaching along the +z and -z directions meet very strong repulsive forces from the filled dz2 orbital. It's like 4 lobes from dx2-y2 has zero electrons, where as d2 with only two lobes have 2 electrons. So there is huge harge difference and hence it pushes the ligands approaching in the +z and -z axis to infinity far direction. Thus only four ligands succeed in bonding to the metal. A square planar complex is formed, the attempt to form an octahedral complex being unsuccessful.