Question

Sulphur reacts with chlorine in $1:2$ ratio and forms $X$. Hydrolysis of $X$ gives a sulphur compound $Y$. The hydridisation of central atom in the anion $Y$ is:

• A. $s{p}^3$
• B. $s{p}^2$
• C. $s{p}^{3}d$
• D. $sp$

Answer 1

Answer: (A)

$s{p}^3$

When sulphur reacts with chlorine in $1:2$ ratio the $SC{l}_4$ is obtained which on hydrolysis gives $H_{2}S{O}_4$. Hence, the compound ${}^{\prime }{X}^{\prime }$ is $SC{l}_4$ and ${}^{\prime }{Y}^{\prime }$ is $H_{2}S{O}_4$

$S+2C{l}_2\to \underset{\underset{\left(x\right)}{\text{Sulphur tetrachloride}}}{SC{l}_4}$

$SC{l}_4+4H_{2}O\to \underset{Unstable}{S(OH{)}_4}+4HCl$

$S(OH{)}_4\to \underset{Sulphurousacid}{H_{2}S{O}_4}+H_{2}O$

The hybridisation of sulphur in $S{O}_3^{2-}$ is $s{p}^3$.