Lone Pair Placement

(Polar molecules, Non-polar molecules, etc.)

Moderators: Chem_Mod, Chem_Admin

Sabah Islam 1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

Lone Pair Placement

Postby Sabah Islam 1G » Sun Nov 12, 2017 11:27 pm

How do you determine where exactly to put the lone pairs on a molecule that has an electron density greater than 4? For instance, on a seesaw, T-shaped, square pyramidal, or square planar molecule?

Miranda 1J
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

Re: Lone Pair Placement

Postby Miranda 1J » Sun Nov 12, 2017 11:31 pm

You should always put lone pairs as far away from each other as possible. I don't think it matters where you put the first pair as long as the second one is the furthest away it can be.

Rachel Formaker 1E
Posts: 86
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

Re: Lone Pair Placement

Postby Rachel Formaker 1E » Sun Nov 12, 2017 11:32 pm

In general, since lone pairs have the greatest repulsion, put them wherever they will be farthest from each other or from other atoms. This will minimize the repulsion in the molecule and result in the most stable structure.

Nancy Dinh 2J
Posts: 59
Joined: Fri Sep 29, 2017 7:07 am

Re: Lone Pair Placement

Postby Nancy Dinh 2J » Sun Nov 12, 2017 11:43 pm

Starting with the base model where both bonds and lone pairs are included, you place the lone pairs where it will affect the least number of bonded pairs at the lowest angle. For example, let's say you first have a trigonal bipyramidal model and are trying where to place the lone pairs. You would place the lone pair on the one that mainly affects the two bonded pairs at 90 degrees (one of the equatorial atoms) and get a see-saw. You wouldn't place the lone pairs at one of the axial atoms because at the lowest angle (90 degrees) it will affect 3 bonded pairs (the horizontal).


Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 1 guest