Square Planar or Tetrahedral

How can you tell whether a complex ion/molecule, with a coordination number of 4, is tetrahedral or square planar in shape?

Other than d8, you get tetrahedral for those without lone pairs, but for those with one lone pair and 4 bonding pairs, that'd constitute square pyramidal, while if you have two lone pairs and 4 bonding pairs, then you'd get square planar, where the lone pairs hover above and below.

Just like VSEPR when applied to main group.

Square planar is favoured electronically by d8 complexes. They find this configuration particularly favourable as they completely fill the "lower" (i.e. not the dx2-y2 orbital) set of square planar orbitals as opposed to populating the t2 set of tetrahedral orbital energies which are higher in energy. This can be overriden, however, with very large ligands which might prefer the larger 109.5 degree tetrahedral angles. If you're confused by my references to the orbital energies, have a look at the molecular orbital diagrams on this page: http://wwwchem.uwimona.edu.jm:1104/courses/CFT.html

Thanks guys, that's really helpful! So it's sort of like a rule of thumb that d8s form square planar molecules?



Also, say you have CoCl4, it has four bonding pairs and it has two lone pairs but it forms tetrahedral?

But then we are talking about Co4+ which is d5, losing two from d and two from s; it uses 4 electrons for bonding, so one is left, that cant be counted as an electron pair. Hence, it would be tetrahedral; to be honest, I am not sure what happens to this one electron; I am sure someone can come up with a better explanation.

CoCl4 has an overall charge of 2- so we are talking about Co2+, it's only lost two from 4s and then it has 3 and a half pairs, which doesn't help. Baffled.

Quoted from an inorganic text:

Cobalt(II) forms numerous complexes, mostly either octahedral or tetrahedral but 5-coordinate and square planar complexes are also known.,

There are more TETRAHEDRAL complexes of Co(II) than for other Tm ions, because for a d7 ion, ligand field stabilization energies disfavor the tetrahedral configuration relative to the octahedral one to a smaller extent than for any other dn(1<n<9; meant to be greater than or equal) config. The only d7 ion of common occurrence is Co(II).

kyri, isn't that only true for 3d, not for 4d and 5d where 4-coordinated complexes are exclusively square planar(apart from PdF2 which I mentioned is not)? just a question, not doubting your explanation.

Yes planar, I simply see no other solution. It must be!

Please why does Pd2 and Pt2 have higher tendencies to form square planar structure than Ni2