Which species occupy the equatorial positions in VSEPR's trigonal bipyramidal and octahedral geometry
Trigonal planar is a molecular structure that requires five ‘substituents’ around a central atom (counting lone pairs as ‘substituents’). Assuming main-group compounds, that means, we have already left the realm of easily derived molecules and need to resort to 4-electron-3-centre bonds to properly explain the molecule and be in line with the octet rule.

This very good answer by Ron does a good job of explaining what the central atom’s electronic structure should be considered to be and what the 4-electron-3-centre bond is. In a nutshell, assume the central atom to be $\mathrm{sp^2}$ hybridised with an unhybridised $\mathrm{p}$ orbital. This additional $\mathrm{p}$ orbital then interacts with two electronegative substituents forming a 4-electron-3-centre bonds; this bond can be rationalised in a Lewis formalism by looking at these two resonance structures:

$$\ce{X-\overset{+}{A}\bond{...}\overset{-}{X} &lt;-&gt; \overset{-}{X}\bond{...}\overset{+}{A}-X}$$

The bond order of a 4-electron-3-centre bond is approximately ½, in good agreement with a single bond and a non-bond forming equivalent resonance structures for each $\ce{X\bond{...}A}$ fragment. Each outer atom has a partial negative charge while the central atom has a partial positive one.

This already means a few things:


a lone pair can never be in axial position as this position is formed by a $\mathrm{p}$ orbital. If a lone pair were in axial position, no second ligand can be there and the structure is actually better described as trigonal planar.
the more electronegative a substituent is, the more likely it is to occupy the axial positions as these positions require stabilising a partial negative charge. Hence, fluorine is most frequently found here.


This is also somewhat of an example for Bent’s rule — the more electropositive a substituent is, the higher the $\mathrm{s}$ orbital participation for the orbital forming the bond to said substituent. Conversely, the more electronegative the substitutent is, the higher the $\mathrm{p}$ content. Naturally, a pure $\mathrm{p}$ orbital is on one end of that spectrum, clearly preferring electronegative substituents.