InterviewSolution
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InterviewSolution
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(a) Give reasons for the following: (i) Mn^(3+) is a good oxidising agent. (ii) E^(@)_(M^(2+)//M) values are not regular for first row transition metals (3d series). (iii) Although 'F' is more electronegative than 'O', the highest Mn fluoride is MnF_(4), whereas the highest oxide is Mn_(2)O_(7). (b) Complete the following equations : (i) 2CrO_(4)^(2-)2H^(+)to (ii) KMnO_(4)overset("heat")to |
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Answer» Solution :(a) (i) Electronic configuration of `Mn^(3+)` is `d^(4)` in the outer shell. It acts as oxidising AGENT or it gets reduced to `Mn^(2+)`. In doing so its configuration changes to do which is a half-filled stable configuration. (ii) `E^(@)_(M^(2+)//M)` VALUES are not regular for first row transition METALS (3d SERIES). The values of `E^(@)` for Mn, NI and Zn are more negative than expected from the trend. The stability of the half-filled d sub-shell in `Mn^(2+)` and the completely filled `d^(10)` configuration of `Zn^(2+)` are related to their `E^(@)` values. `E^(@)` for Ni is accounted for by the highest negative `Delta_(hyd)H^(@)`. (iii) The ability of O to stabilise these high oxidation states exceeds that of fluorine. Thus, the highest Mn fluoride is `MnF_(4)` whereas the highest oxide is `Mn_(2)O_(7)`. The ability of oxygen to form multiple bonds to metals explains this. In the covalent oxide `Mn_(2)O_(7)`, each Mn is tetrahedrally surrounded by O.s including a Mn - O - Mn bridge. (b) (i) `2CrO_(4)^(2-)+2H^(+)+Cr_(2)O_(7)^(2-)+H_(2)O` (ii) `2KMnO_(4)overset(Heat)toK_(2)MnO_(4)+MnO_(2)+O_(2)` |
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