Suppose we have 2-bromo butane. The carbon connecting to the bromine will have the following valence electron configuration:
$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} $$
This hybridizes to an sp3 configuration which yields a trigonal pyramidal configuration.
Suppose in an SN1 reaction the bromine comes off. Then the connecting carbon will have a valence electron configuration of
$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert\;]} $$
which yields an sp2 or trigonal planar configuration.
Suppose we instead have an SN2 reaction such as with NaOH. Then the intermediate would be a carbanion that has a carbon that has 5 bonds.
$$ \underset{2s}{[\uparrow ]} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3s}{[\uparrow]} $$
would be the wrong configuration but I'm having trouble understanding what the right one would be.
A trigonal bipyramidal molecular geometry seems to me like it'd fit the shape of what the carbanion would be but the numbers don't seem to add up to me.
In a trigonal bipyramidal geometry there needs to be two $sp$ orbitals two point up and down.
I learned from How are the hybrid orbitals of sulfur hexafluoride shaped? that higher orbitals are usually only slightly involved so the resulting bonds would only have a small d characteristic. For example, instead of a $sp^2d$ hybrid something like a $sp^{2.75}d^{0.25}$ characteristic is more likely.
Would the rest of the bonds have something like a $sp^{2.67}d^{0.33}$ characteristic?
According to molecular orbital theory what would the shape of that carbanion be and how would the groups attached to it be arranged?
