Why are the bond angles of #"SF"_4# less than predicted by VSEPR theory for a trigonal bipyramidal geometry?

1 Answer
Truong-Son N.
Aug 3, 2017

https://upload.wikimedia.org/

Well, the lone pair is on the equatorial plane, so it repels away the bonding-electron pairs for both the equatorial and the axial atoms.

The #"F"-"S"-"F"# bond angle for the axial fluorines decreases by #bb(6.9^@)#, and the #"F"-"S"-"F"# bond angle for the equatorial fluorines decreases by #bb(18.4^@)# from the ideal bond angles.

Since the lone pair of electrons is on the equatorial plane, the equatorial atoms are crunched together the most. That repulsion is the most direct.

There is also only one lone pair, so no counteracting lone pair decreases the influence of the other (unlike in water, which has an #"H"-"O"-"H"# bond angle only #5^@# smaller).

Related questions
See all questions in VSEPR
Impact of this question
2260 views around the world
You can reuse this answer
Creative Commons License