1.

(a) How would you account for the following: (i) Highest fluoride of Mn is MnF, whereas the highest oxide is Mn_(2)O_(7). (ii) Transition metals and their compounds show catalytic properties. (b) Complete the following equation: 3MnO_(4)^(2-)+4H^(+)to

Answer»

Solution :(a) (i) The ability of oxygen to form MULTIPLE bonds to METALS explains why the highest oxide of Mn is `Mn_(2)O_(7)` while the highest fluoride is `MnF_(4)`.
(ii) Catalytic properties of transition metals and their compounds is due to their ability to adopt multiple oxidation STATES and to form COMPLEXES.
(b) The completed EQUATION is :
`3MnO_(4)^(2-)+4H^(+)to2MnO_(4)^(-)+MnO_(2)+2H_(2)O`


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