(a) Out of Ag_(2)SO_(4),CuF_(2),MgF_(2) and CuCl, which compound will be coloured and why ? (b) Explain : (i) CrO_(4)^(2-) is a strong oxidising agent while MnO_(4)^(2-) is not (ii) Zr and Hr have identical sizes. (iii) The lowest oxidation state of manganese is basic while the highest is acidic. (iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series.

(a) Out of Ag_(2)SO_(4),CuF_(2),MgF_(2) and CuCl, which compound will

1.	(a) Out of Ag_(2)SO_(4),CuF_(2),MgF_(2) and CuCl, which compound will be coloured and why ? (b) Explain : (i) CrO_(4)^(2-) is a strong oxidising agent while MnO_(4)^(2-) is not (ii) Zr and Hr have identical sizes. (iii) The lowest oxidation state of manganese is basic while the highest is acidic. (iv) Mn (II) shows maximum paramagnetic character amongst the divalent ions of the first transition series.
Answer» Solution :(a) `CuF_(2)` is coloured due to vacant d-orbital and presence of unpaired electron. (B) (i) Cr in `CrO_(4)^(2-)` has oxidation state +6. It can reduce its oxidation state to +3 (in `Cr^(3+)`). THUS, it can act as oxidising agent. MN in `MnO_(4)^(2-)` has oxidation state +6. But its most stable oxidation state is +7. Therefore, it does not act as oxidising agent. (ii) Zr and Hf have identical sizes due to lanthanoid contraction. (III) Oxides of Mn in lowest oxidation state are ionic and form basic solution, whereas oxides in highest oxidation state are covalent and form acidic solution with water. `Mn^(2+)` has the maximum number of unpaired electrons and, therefore, has the maximum paramagnetic character.