InterviewSolution
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InterviewSolution
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(a) How will you prepare : (i) K_(2)MnO_(4) from MnO_(2) ? (ii) Na_(2)Cr_(2)O_(7) from Na_(2)CrO_(4) ? (b) Account for the following: (i) Mn^(2+) is more stable than Fe^(2+) towards oxidation to +3 state. (ii) The enthalpy of atomisation is lowest for Zn is 3d series of the transition elements. (iii) Actinoid elements show wide range of oxidation states. |
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Answer» Solution :(a) (i) `K_(2)MnO_(4)` from `MnO_(2)` `MnO_(2)` is fused with potassium hydroxide and an oxidant agent like `KNO_(3)`. `2MnO_(2)+4KOH+O_(2)to2K_(2)MnO_(4)+2H_(2)O` (ii) `Na_(2)Cr_(2)O_(7)` from `Na_(2)CrO_(4)` Yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate can be crystallised. `2Na_(2)CrO_(4)+2H^(+)to Na_(2)Cr_(2)O_(7)+2Na^(+)+H_(2)O` (b) (i) `Mn^(2+)` is more stable than `Fe^(2+)` towards oxidation to +3 state. `Mn^(2+)` has d-orbital configurations as `3d^(5)` and `Mn^(3+)` has the configuration `3d^(4). 3d^(5)` (being half-filled) is a stable configuration compared to `3d^(4)`. Hence, `Mn^(2+)` is more stable towards oxidation to +3 state. `Fe^(2+)` has d-orbital configuration as `3d^(6)` and `Fe^(3+)` has the more stable configuration `3d^(5)`. Hence, `Fe^(2+)` has a greater tendency towards oxidation to +3 state. (ii) Enthalpy of atomisation depends upon the number of unpaired electrons. Since there are no unpaired electrons in the ATOM of zinc in the 3d series, it has the lowest enthalpy of atomisation. (iii) Actinoid elements SHOW a wide range of oxidation STATES. This is due to the fact that 5f, 6d and 7s levels are of comparable energies. Electrons from any of these levels could be lost giving rise to the POSSIBILITY of greater number of oxidation states. |
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