How many atom(s) of $S F_{2} (C H_{3})_{2}$ can lie in equatorial plane?

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Solution

In $S F_{2} (C H_{3})_{2}$ , the central $S$ atom undergoes $s p^{3} d^{2}$ hybridization as it contains $1$ lone pair and $4$ bond pairs of electrons. This results in trigonal bipyramidal geometry. The lone pair is present in the equatorial position as it minimizes electron electron repulsion. Two $F$ atoms are present in axial positions. Thus, a maximum of $5$ atoms can lie in the equatorial plane. This includes one $S$ atom, $2 C$ atoms and $2 H$ atoms.

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