Question

Give reasons for the following:
(i) ${\left(C{H}_3\right)}_{3}P=O$ exists but ${\left(C{H}_3\right)}_{3}N=O$ does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) $H_{3}P{O}_2$ is a stronger reducing agent than $H_{3}P{O}_3$.

Answer 1

(i) N atom cannot expand its covalency beyond four due to absence of low lying vacant d- orbitals, whereas P atom possesses low lying vacant d- orbitals. As a result, $(C{H}_3{)}_{3}P=O$ exists but $(C{H}_3{)}_{3}N=O$ does not.
(ii) Due to small size and compact nature of the oxygen atom, the incoming electron is not accommodated with ease. As a result, oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) Greater the number of element−hydrogen (E−H) bonds present in a compound, greater is the reducing nature of the compound. $H_{3}P{O}_2$has two P−H bonds while $H_{3}P{O}_3$ has one P−H bond. Thus, $H_{3}P{O}_2$ is a stronger reducing agent than$H_{3}P{O}_3$.