(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).

(a) Account for the following : (i) Hydration enthalpy of F^(-) ion i

1.	(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).
Answer» SOLUTION :(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)`