(Polar molecules, Non-polar molecules, etc.)
5 posts • Page 1 of 1
I know that when there is an unpaired set of electrons, the bonded pairs get pushed away due to the electrostatic repulsion. In a molecule with a square planar VSEPR Structure, will the angles between the remaining 4 bonded atoms remain 90 degrees , or will the repulsion push them closer together?
For the square planar VSEPR model, the electron pairs will be oriented opposite of each other, and they will have equal contributions to the four substituents that are on the "x-axis". At the same time, the four groups are also minimizing the interactions between themselves. So in terms of the model, there will be no excess repulsion that causes the four substituents to be pushed closer together.
In a square planar, the lone pairs are on opposite sides of the central atom, keeping the four bonding electron pairs on one plane. The bonding pairs experience equal repulsion from both lone pairs, so 4 bonds split on a 2-D plane is 90 degrees of separation between each bonding pair. However, what I don't get is if a square pyramidal has 90 and 180 degrees between each of its bonding pairs since that would mean the bonding pair opposite of the single lone pair would have the same repulsion as a lone pair, otherwise the bond angles would be pushed slightly smaller by the stronger lone pair.
You're right, in a square pyramidal structure the bond angles are slightly less than 90 degrees according to VSEPR theory. And yes we do have to know about molecules in that form, which have the shorthand abbreviation AX5E.
Who is online
Users browsing this forum: No registered users and 2 guests