Which species (electronegative, double and triple bonded atoms, ions, lone pair etc.) occupy the equitorial position in the trigonal bipyramidal and octahedral geometry when we find the shape using VSEPR theory?
Which species are given more priority and why?

  • 2
    $\begingroup$ This goes some way to explaining it. $\endgroup$
    – bon
    Mar 6, 2016 at 10:14
  • 2
    $\begingroup$ In trigonal bipyramidal geometry, roughly speaking, the more bulky substituent goes to the equatorial position. In octahedral geometry, there is no such thing as equatorial position at all. $\endgroup$ Mar 6, 2016 at 10:59
  • 2
    $\begingroup$ Less electronegative and double/triple bonded groups tend to occupy the equatorial position, while the more electronegative and single bonded groups tend to occupy the axial position. Ron's explanation accounts for these observations: chemistry.stackexchange.com/questions/18427/… $\endgroup$
    – EJC
    Mar 6, 2016 at 11:07

1 Answer 1


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:

$$\ce{X-\overset{+}{A}\bond{...}\overset{-}{X} <-> \overset{-}{X}\bond{...}\overset{+}{A}-X}$$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.