(a) In the titration of FeSO_(4) with KMnO_(4) in the acidic medium, why is dil. H_(2)SO_(4) used instead of dil. HCI ? (b) Give reasons : (i) Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal. (ii) Ce^(4+) is used as an oxidising agent in volumetric analysis. (iii) Transition metals form a number of interstitial compounds. (iv) Zn^(2+) salts are white while Cu^(2+) salts are blue.

(a) In the titration of FeSO_(4) with KMnO_(4) in the acidic medium, w

1.	(a) In the titration of FeSO_(4) with KMnO_(4) in the acidic medium, why is dil. H_(2)SO_(4) used instead of dil. HCI ? (b) Give reasons : (i) Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal. (ii) Ce^(4+) is used as an oxidising agent in volumetric analysis. (iii) Transition metals form a number of interstitial compounds. (iv) Zn^(2+) salts are white while Cu^(2+) salts are blue.
Answer» Solution :(a) HCl gets oxidised to `Cl_(2)` by `KMnO_(4)` whereas `H_(2)SO_(4)` does not. (b) (i) METAL oxoanion may be represented as `MO_(x)^(y-)` where y ranges between 1 to 2 and x ranges between 2 to 4. On calculation of oxidation state of the metal in oxoanion, we find that it is the maximum. (ii) `Ce^(4+)` can gain an electron and, therefore, it acts as an oxidising AGENT. `Ce^(4+)+e^(-)toCe^(3+)` (iii) These elements have VOIDS in which small atoms like C, B can fit TIGHTLY to give interstitial compounds. (iv) `Zn^(2+)` does not have vacant d-orbitals or unpaired electrons in d-orbital. `Cu^(2+)` has ONE unpaired electron and vacant d-orbitals.