(a)Account for the following
(i) Reducing character decreases from `SO_(2)` to `TeO_(2).`
(ii) `HClO_(3)` is a stronger acid than HClO.
(iii) Xenon forms compounds with fluorine and oxygen only.
(b) Complete the following equations :
(i) `4NaCl+MnO_(2)+4H_(2)SO_(4)to`
(ii) `6XeF_(4)+12H_(2)Oto`

Explanation

(a) (i) Since +6 oxidiation state of S is more stable than +4, therefore, `SO_(2)` act as a reducing agent. Since the stability of +6 oxidatin decreases from S to Te, therfore, reducing character decreases from `SO_(2)` to `TeI_(2).`
(ii) `overset(+5)(HClO_(3)),overset(+1)(HClO)`
Greater the number of oxygen atoms in anion (formed by the removal of Hydrogen ) greater will be the dispersal of negative charge through p `pi-d pi` back bonding and hence greater will be the stability of anions, greater will be the acidity. Hence `HClO_(3)` is stronger acid than HClO.
(iii) Fluorine and oxygen are the most electronegative elements and hence are very reactive. Therefore, Xenon forms compounds with flurine and oxygen only.
(b) (i) `4NaCl+MnO_(2)+4H_(2)SO_(4)to MnCl_(2)+4NaHSO_(4)+2H_(2)O+Cl_(2)`
(ii) `6XeF_(4)+12H_(2)O to4Xe+2XeO_(3)+24HF+3O_(2)`