Explain the following :
(i) ICI is more reactive than `I_(2)`
(ii) `H_(3)PO_(3)` undergoes disproportionation.
(iii) `K_(a_(2))` is less than `K_(a_(1))` for `H_(2)SO_(4)` in water.

Why is ICI more reactive than I_2 ?

Why is K_(a_(2)) ltlt K_(a_(1)) for H_(2) SO_(4) in water ?

Account for the following : (i) Ammonia is a stronger base than phosphine. (ii) H_(3)PO_(3) is a diprotic acid.

For a 0.072 M NaSO_(4) solution,select the incorrect option(s): (K_(a_(1)) and K_(a_(2)) of H_(2)SO_(4)=oo & 1.2xx10^(-2))

(a) Give reasons : (i) H_(3)PO_(3) undergoes disproportionation reaction but H_(3)PO_(4) does not . (ii) When Cl_(2) reacts with excess of F_(2), ClF_(3) is formed and not FCl_(3) . (iii) Dioxygen is a gas while Sulphur is a solid at room temperature . (b) Draw the structure of the following : (i) XeF_(4) " " (ii) HClO_(3)

In a solution containing 0.1M HCl & 0.1M H_(3)PO_(4) (having dissociation constants K_(a_(1)),K_(a_(2)),K_(a_(3))) assuming alphaltlt1 for H_(3)PO_(4) ,

(a) Write the balanced chemical equations for obtaining XeO_(3) " and " XeOF_(4) "from "XeF_(6). (b) Account for the followingk : (i) H_(2)S is less acidic then H_(2) Te. (ii) H_(3)PO_(2) has reducing nature. (iii) SO_(2) is an air pollutant.