(a) Account for the following :
(i) Hydration enthalpy of `F^(-)` ion is more than `Cl^(-)` ion.
(ii) `SO_(2)` is a reducing agent, whereas `TeO_(2)` is an oxidising agent in group-16 oxides.
(b) Write the reaction of `F_(2)` with water. Why does `I_(2)` not react with water ?
(c) Draw the structure of `XeF_(2)`.

Explanation

(a) (i) Due to small size to F, the density of the negative charge is greater on `F^(-)` than on `Cl^(-)`. Therefore hydration enthalpy of `F^(-)` ion is more than that of `Cl^(-)`.
(ii) `SO_(2)` is a reducing agent because S has d-orbitals and can expand its oxidation state from +4 to +6 and thus behave as a reducing agent. Heavier elements like Te do not expand the oxidation state from +4 to +6 due to inert pair effect. It acts as an oxidising agent.
(b) `2F_(2)(g)+2H_(2)O(l)to4H^(+)(aq)+4F^(-)(aq)+O_(2)(g)`
`I_(2)` does not react with water due to lower value of reduction potential (0.54 V).
(c) Structure of `XeF_(2)`