Question

Give reasons for the following :

(i) $(C{H}_3{)}_{3}P=O$ exists but $(C{H}_3{)}_{3}N=O$ 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 cannot form $p\pi -d\pi$ multiple bonds due to absence of d-orbitals. N, therefore, cannot expand its covalency beyond 4 but in $(C{H}_3{)}_{3}N=O$, its covalency is 5. Therefore, this compound does not exist. P has vacant d-orbitals. It can form $p\pi -d\pi$ multiple bonds and hence can expand its covalency beyond 4. Therefore, $(C{H}_3{)}_{3}P=O$ exists in which covalency of phosphorus is 5.

(ii) The size of oxygen atom is smaller than that of sulphur atom. Thus, when an electron is added to isolated neutral gaseous atom, the electron-electron repulsions experienced in smaller 2p subshell of oxygen atom are comparatively larger than in the bigger 3p subshell of S atom. As a result, more energy is released in case of S atom. In other words, oxygen has less electron gain enthalpy than sulphur atom.

(iii) $H_{3}P{O}_2$ has two P-H bonds but $H_{3}P{O}_3$ has only one P-H bond. Due to greater number of P-H bonds, $H_{3}P{O}_2$ is stronger reducing agent than
$H_{3}P{O}_3$.