Question

The compound (A) reacts with sulphur to form a compound in which hybridization state of sulphur atom is____________.

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

Answer 1

Answer: (B)

$s{p}^3{d}^2$

The liquid (B) is used for acid etching of glass. This liqiud is therefore hexafluorosilicic acid $\left(HF\right)$.

But $HF$ can not form addition compound with $KF$.

The most reactive gas $\left(F\right)$ is $F_2$ and the monoatomic colorless gas $\left(C\right)$ would be noble gas $Xe$. Both combine in ratio $1:1$ to form $\left(A\right)$.

Therefore $A$ is

$Xe+F_2\to Xe{F}_2$

The reactions can be given as

$Xe{F}_2\left(A\right)+H_2\to HF\left(B\right)+Xe\left(C\right)$

$Xe{F}_2+H_{2}O\to HF\left(B\right)+Xe\left(C\right)+O_2\left(D\right)$

$Xe{F}_2$ reacts with sulphur :

$Xe{F}_2+S\to S{F}_6+Xe$

The $S$ in $S{F}_6$ is $s{p}^3{d}^2$ hybridized. To account for the hexavalency in $S{F}_6$ one electron each from $3s$ and $3p$ orbitals is promoted to $3d$ orbitals These six orbitals get hybridised to form six $s{p}^3{d}^2$ hybrid orbitals. Each of these $s{p}^3{d}^2$ hybrid orbitals overlaps with $2p$ orbital of fluorine to form $S-F$ bond. Thus, SF6 molecule has octahedral structure.