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InterviewSolution
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How do you prepare : (i) `k_(2)MnO_(4) from MnO_(2)`? (ii) `Na_(2)Cr_(2)O_(7) from NaCrO_(4)`? (b) Account for the following : (i) `Mn^(2+)` is more stable then `Fe^(2+)` towards oxidation to + 3 state. (ii) The enthelpy of atomization is lowest for Zn in 3d series of the transition elements. (iii) Actinoid elements show wide range of oxidation states. |
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Answer» (i) `K_(2)MnO_(4)` can be prepared from pyrolusite `(MnO_(2))`. The ore is fused with KOH in the prsence of either atmopheric oxygen or an oxidising agrant,such as `KNO_(3) or KClO_(4)` to given `K_(2)MnO_(4)` . `2MnO_(2) + 4KOH+ O_(2) to 2K_(2)MnO_(4)+2H_(2)O` (ii) `Na_(2) Cr_(2)O_(7)` can be prepared from `Na_(2)CrO_(4)` in the following way: For the preparation ltbtgt of solution dichromate is acidified with sulphuric acid to give a solution from which orange sodium dichromate, `Na_(2)Cr_(2)O_(7).2H_(2)O` can be crystallised . Balanced equation for above reactions is as follows : `2Na_(2)underset("Yellow")(CrO_(4))+2H^(+)toNaunderset("Orange")(Cr_(2))O_(7) + 2Na^(+) + H_(2)O` (b) (i) Electronic configuration of `Mn^(2+) is [Ar]^(18) 3d^(5)` Electronic configuration of `Mn^(2+)is [Ar]^(18) 3d^(5)` . lt is known that half-filled and fully-filled orbitals are more stable . Therefore , Mn in `.^(+2)` state has a stable `d^(5)` configuration. Therefore, `Mn^(2+)` shows resistance to oxidation of `Mn^(3+)`. Also, `Fe^(2+) has 3d^(6)` configuration and by losing one electron, its configuration changes to more stable `3d^(5)` configuration. Therefore, `Fe^(2+)` gets oxidised to `Fe^(3+)` easily. The extent of metallic bonding an element undergoes, decides the enthalpy of atomisation. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomisation. Zinc only form metallic bonding but no d-d overlapping Whereas in other metals, both metallic as well as covalent bonding involve, enthalpy of atomisation is lowest for zinc. (iii) Actinides exhibit larger oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells . The energies are calculated on the basis of (n + 1) rule. The (n+ 1) values of the three orbitals are : 5f = 5 + 3 = 8 6d = 6 + 2 = 8 7s = 7 + 0 = 7 Since, all the values are almost same, therefore all orbitals can involved in bonding resulting in larger oxidation number for actionids. |
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