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

T e B r_{4}

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T e C l_{4}

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X e O_{2} F_{2}

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S F_{4}

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Solution

The correct options are
A $T e B r_{4}$
B $T e C l_{4}$
C $X e O_{2} F_{2}$
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 $X e O_{2} F_{2}$ the oxygen
atoms and the lone pair are in the equatorial plane with the fluorine atoms taking
up the axial positions:

$T e$ belongs to the same group as $O$ and $S$ .
In the above structure, if we replace sulphur with $T e$
and the halogens with either $B r$ or $C l$ , we get
the geometries of $T e B r_{4}$ and $T e C l_{4}$ .

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