The square planar crystal field

    A square planar arrangement of ligands can be formally derived from an octahedral array by removal of two transligands (Figure 1.1). If we remove the ligands lying along the z axis, then the d_z2 orbital is greatly stabilized; the energies of the d_yz and d_xz orbitals are also lowered, although to a smaller extent. The resultant ordering of the metal d orbitals is shown at the left-hand side of Figure 1.2. The fact that square planar d⁸ complexes such as [Ni(CN)₄]^2- are diamagnetic is a consequence of the relatively large energy difference between the d_xy and d_x²_-y² orbitals. Worked example 20.1 shows an experimental means (other than single-crystal X-ray diffraction) by which square planar and tetrahedral d⁸ complexes can be distinguished.

Fig. 1.1 A square planar complex can be derived from an octahedral complex by the removal of two ligands, e.g. those on the z axis; the intermediate stage is a Jahn–Teller distorted (elongated) octahedral complex.

Fig. 1.2 Crystal field splitting diagrams for some common fields referred to a common barycentre; splittings are given with respect to Δ_oct. For tetrahedral splitting,

Worked example 1.1 Square planar and tetrahedral d⁸ complexes d 8 complexes 

 The d⁸ complexes [Ni(CN(₄]^2- and [NiCl₄]^2- are square planar and tetrahedral respe ctively. Will these complexes be paramagnetic or diamagnetic?

Consider the splitting diagrams shown in Figures 20.8 and 20.10. For [Ni(CN)₄]^2- and [NiCl₄]^2- , the eight electrons occupy the d orbitals as follows:

    Thus, [NiCl₄]^2- is paramagnetic while [Ni(CN)₄]^2- is diamagnetic.  Although [NiCl₄]^2-  is tetrahedral and paramagnetic, [PdCl₄]^2- and [PtCl₄]^2- are square planar and diamagnetic.

   This difference is a consequence of the larger crystal field splitting observed for second and third row metal ions compared with their first row congener; Pd(II) and Pt(II) complexes are invariably square planar (but see Box 20.7).