## axial vs equatorial lone pair

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Annie Lieu-1H
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am
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### axial vs equatorial lone pair

Can someone explain the difference between axial lone pairs and equatorial lone pairs ?How can you tell what type of lone pair a given molecule has? And based on these pairs/shape, that's how we determine bond angles right?

Thank you!

Ishan Saha 1L
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

### Re: axial vs equatorial lone pair

If I understood it correctly, I believe that the lone pairs will always be equatorial first, before going to the axial positions (so seeing lone pairs in the axial position is quite uncommon). This is because lone pairs have more repulsion (they don't want to have a smaller bond angle in relation to the the other bonds like the axial ones do) and will push the other bonds further away from it (and thus closer to each other). So if a molecule had 5 bonding regions with three bonds and 2 lone pairs, then both lone pairs would be on the equatorial plane, giving the molecule a T-shape.

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: axial vs equatorial lone pair

Yes, If you are taking out an electron from a trigonal bipyramidal shape, then take it from the equatorial position so that the lone-pair electron can have room for itself and push the other equatorial bonds closer to each other.

If it is however, an octahedral shape, then remember that there are 6 electron regions around the central atom so when you take a bond out for the first time, it does not matter from which one you take it from because octahedral means the outer electron regions are symmetrically placed around the central atom. But when you take out a second bond to create a lone pair, you have to take out the lone pair from the opposite side of the first lone pair because lone pairs like to have their own room/space remember Lavelle's demonstration. Thus, with two lone pairs on a 6 electron region densities around the central atom, you will end up with a square planar (Think of XeF4).

Hope that helps!