Question

Account for the following:
(i) Sulphur in vapour form exhibit paramagnetic behaviour.
(ii) $Sn{Cl}_4$ is more covalent than $Sn{Cl}_2$.
(iii) $H_{3}P{O}_2$ is stronger reducing agent than $H_{3}P{O}_3$.

Answer 1

(i) It is because sulphur in vapour form have unpaired electrons.

(ii) It is due to higher oxidation state (+4) of $Sn$ in $SnC{l}_4$ and it also has high polarising power.

(iii) It is because of two $P-H$ bonds in $H_{3}P{O}_2$ whereas in H_3PO_3 only one $P-H$ bond.