How to predict the hybridisation of a coordinaton complex when 3 bidentate ligads are bound to it and it has 4 eectrons in its valence shell?
Answers
Explanation:
Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for transition metal complexes?
Now, with that disclaimer out of the way, let's examine how that flawed theory works. In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be sp3sp3 and dsp2dsp2 respectively. Since [Ni(CO)4][Ni(CO)X4] is tetrahedral and [Ni(CN)4]2−[Ni(CN)X4]X2− is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated theory, it should have a table telling you what hybridisation each geometry corresponds to. Here is an example from Housecroft's Inorganic Chemistry 4th ed. (p 667). To their credit, there is a subsequent section debunking this theory.

These are generally based on post hocjustifications. Essentially, somebody looked at the geometry and then tried to piece together which orbitals could be combined to form such a geometry. So, I do not think there is a good way to predict them a priori. This formula you have dug up does not work very well, even for main group elements. For transition metals the factors deciding geometries are much more complex, and simply cannot be captured in one line of mathematics.