Question

Which of the following molecule(s) is/are having a see - saw geometry?

• A. $TeB{r}_4$
• B. $TeC{l}_4$
• C. $Xe{O}_2{F}_2$
• D. $S{F}_4$

Answer 1

Answer: (A), (B), (C), (D)

$S{F}_4$

The simplest way to approach this problem is to use VSEPR or the AXE method:
There is a central atom with 4 ligands and one lone pair. In $Xe{O}_2{F}_2$ the oxygen
atoms and the lone pair are in the equatorial plane with the fluorine atoms taking
up the axial positions:

$Te$ belongs to the same group as $O$ and $S$.
In the above structure, if we replace sulphur with $Te$
and the halogens with either $Br$ or $Cl$, we get
the geometries of $TeB{r}_4$ and $TeC{l}_4$.