CHEMISTRY COMMUNITY

How do you determine the bond angles for a seesaw shaped molecule?

The bond angles are less than 120 and 90, as it would be a trigonal bipyramidal without the "top" electron, so the lone pair repulsion pushes down and makes the angles slightly less than they were originally with the trigonal bipyramidal shape.

Yes, the bond angles are approximately 90, 120, and 180 degrees.

The bond angles are less than 120 degrees on the equatorial and less than 90 degrees on the axial.

Visually, it's similar to a trigonal bipyramidal but you would have to remove one bonding pair. So it would be 90, 180, 120

It is hard to determine the exact angle of the seesaw shape because every element differs in size and as a result, we can never determine the exact angles unless you remember the bond angles. However, in general you can consider each angles of the trigonal bipyramidal decreased as seesaw’s bond angle.

For seesaw I just remember that its 4 bonds with one lone pair.

Another important concept to remember is that VSEPR cannot provide us the exact bond angle, just approximate if it will be greater or less than our key angles (90, 180, 109.5)