Question

Sulphur trioxide is dissolved in heavy water to form a compound $X$. The state of hybridisation of sulphur in $X$ is:

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

Answer 1

Answer: (B)

$s{p}^3$

Hybridisation is given by the following formula:

$H=\frac{V+X-C+A}{2}$

Here, $V=$ number of valence electrons in central atom, $X=$ number of monovalent atoms around the central atom, $C=$ positive charge on cation and $A=$ negative charge on anion.

If $H=2$, then the compound is $sp$ hybridised.
If $H=3$, then the compound is $s{p}^2$ hybridised.
If $H=4$, then the compound is $s{p}^3$ hybridised.
If $H=5$, then the compound is $s{p}^{3}d$ hybridised.
If $H=6$, then the compound is $s{p}^3{d}^2$ hybridised.
If $H=7$, then the compound is $s{p}^3{d}^3$ hybridised.

Sulphur trioxide dissolves in heavy water to form sulphuric acid.

$S{O}_3+D_{2}O\to D_{2}S{O}_4$

So, its hybridisation $\left(H\right)=\frac{(6+2-0+0)}{2}=4$

Hence, the state of hybridisation is $s{p}^3$.