Give reasons for the following:
(i) (CH₃)3P = O exists but (CH₃)₃N = O does not.
(ii) Oxygen has less electron gain enthalpy with the negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃.

(i) N atom cannot expand its covalency beyond four due to the absence of vacant d- orbitals, whereas P atom possesses vacant d- orbitals. As a result, (CH₃)₃P = O exists but (CH₃)₃N = O does not.
(ii) Due to the 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 a 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. H₃PO₂ has two P−H bonds while H₃PO₃ has one P−H bond. Thus, H₃PO₂ is a stronger reducing agent than H₃PO₃.