Give reasons for the following
(i) `(CH_(3))_(3)` P=0 exists but `(CH_(3))_(3)` N=0 does not exists.
(ii) Oxygen has less electron gain enthalypy with negative sign than sulphur
(iii) `H_(3)PO_(2)` is a strong reducing agent than `H_(3)PO_(3)`

Explanation

(i)Due to absence of vacant d robitals N can not form 5 convent bonds
(ii) O has exceptionally smely size. Incoming electrons feels more repulsion than expected and its negative electron gain enthalpy becomes less than expected
(iii) In `H_(3)PPO_(2)` oxidation state of P is '+1' while in `H_(3)PO_(3)` oxidation state of P is '+3' .In `H_(3)PO_(2)` oxidaiton state of P is lower than that in `H_(3)PO_(3)`