Question
$$S{F_2},S{F_4}$$ and $$S{F_6}$$ have the hybridisation at
sulphur atom respectively as :
A.
$$s{p^2},s{p^3},s{p^2}{d^2}$$
B.
$$s{p^3},s{p^3},s{p^3}{d^2}$$
C.
$$s{p^3},s{p^3}d,s{p^3}{d^2}$$
D.
$$s{p^3},sp{d^2},{d^2}s{p^3}$$
Answer :
$$s{p^3},s{p^3}d,s{p^3}{d^2}$$
Solution :
$$\eqalign{
& {\text{Hybridisation :}} \cr
& 1.\,\,\,\,:\ddot S{F_2} \Rightarrow \frac{1}{2}\left( {6 + 2} \right) = 4 = s{p^3} \cr
& 2.\,\,\,:S{F_4} \Rightarrow \frac{1}{2}\left( {6 + 4} \right) = 5 = s{p^3}d \cr
& 3.\,\,\,S{F_6} \Rightarrow \frac{1}{2}\left( {6 + 6} \right) = 6 = s{p^3}{d^2} \cr} $$