Arrange the following in increasing order of the property indicated :
(a) `H_(3)PO_(3), H_(3)PO_(4), H_(3PO_(2) ("Reducing Character")`
(b) `NH_(3), PH_(3), AsH_(3), SbH_(3), BiH_(3) ("Basic Strength")`

Explanation

(i) `F_(2)` is strong oxidising agent than `Cl_(2)`. Electron gain enthalpy of fluorine is less negative `(-333 kJ//mol)` than that of `Cl(-349KJ//Mol)` but still it is strongest oxidising agent because of its low bond dissociation energy `(158.8 KJ//mol)` and high heat of hydration `(515 KJ//mol)` as compared of those of chlorine for which values are 242.6 and 381 `KJ//mol` respectively.
(ii) Conditions to maximize the yield of `H_(2)SO_(4)`
(I) Low temperature (720K)
(II) High pressure (2 Bar)
(III) Use of Catalyst - `V_(2)O_(5)` (divanandium pentoxide)
(IV) Excess of oxygen
(iii) Reducing character :
`H_(2)PO_(2)` has 2 P - H bonds, `H_(3)PO_(3)` contains one P-H bond whereas, `H_(3)PO_(4)` contains no P-H bond.
Hence, order of reducing character is as follows -
`H_(3) PO_(2) gt H_(3)PO_(3) gt H_(3)PO_(4)`
(b) Basic strength order - `NH_(3) gt PH_(3) gt A_(5)H_(3) gt Sb H_(3) gt BiH_(3)`
Due to the presence of lone pair all are lewis bases. When we move down the group atomic size increases, electron density on central atom decreases, hence, basic strength decreases.