Question
Sulphur reacts with chlorine in 1 : 2 ratio and forms $$X.$$ Hydrolysis of $$X$$ gives a sulphur compound $$Y.$$ What is the hybridisation state of central atom in the compound.
A.
$$s{p^3}$$
B.
$$s{p^2}$$
C.
$$sp$$
D.
$$ds{p^2}$$
Answer :
$$s{p^3}$$
Solution :
Hybridisation of $${H_2}S{O_3} = \frac{1}{2}\left( {6 + 2 + 0} \right) = 4 = s{p^3}$$
Hybridisation of $${H_2}S{O_3} = \frac{1}{2}\left( {6 + 2 + 0} \right) = 4 = s{p^3}$$