Why is tetracyanonickelate square planar? Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?



$\ce{CN}$, for being a strong field, would not follow Hund's rule.
the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$
Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be trigonal bipyramidal while it is not.
This is the square planar splitting diagram



Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-

Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.

About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't