Question

In which of the following cases hydrolysis takes place through $S_{N}2and{S}_{N}1$ mechanism, respectively?

• A. $P_4{O}_{10},SiC{l}_4$
• B. $NC{l}_3,N{F}_3$
• C. $SiC{l}_4,Si{F}_4$
• D. $S{F}_4,Te{F}_6$

Answer 1

The correct option is D $S{F}_4,Te{F}_6$

The maximum hybridization state possible for group 16 elements is $s{p}^3{d}^2$. $S{F}_4$ undergoes hydrolysis through $S_{N}2$ mechanism as in the transition state, it is possible to accommodate the incoming nucleophile. Sulphur has a total of 6 valence electrons, out of which only 4 are taking part in bonding. The hybridization is $s{p}^{3}d$. It has an oxidation state of +4 while, a maximum of +6 is possible. Here, due to empty d orbital, a nucleophile can attack the S atom. However, the hybridization of $Te$ in $Te{F}_6$ is $s{p}^3{d}^2$. Since, no more d -orbitals are available for the attack of nucleophile, it will undergo $S_{N}1$ mechanism. Thus, $Te{F}_6$ undergoes hydrolysis through $S_{N}1$ mechanism.