CBSE Class 12 Chemistry 2014 Delhi Set 1 Solved Paper

Question : 29

Total: 30

(a) How do you prepare:
(i) KMnO4 from MnO2
(ii) Na2Cr2O7 from Na2CrO4
(b) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
(iii) Actionoid elements show wide range of oxidation states.
OR
(i) Name the elements of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive E°(M2+∕M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(iv) Name a member of the Lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation
MnO4−+8H++5e−→

Solution:

(i) KMnO4 can be prepared from pyrolusite (MnO2)⋅MnO2 is ignited with KOH in the presence of catalysts agents, such as oxygen from the air or KNO3 or KClO4 to give KMnO4.
2MnO2+4‌KOH+O2───▸KMnO4‌ (green) ‌ +2H2O
(ii) For the preparation of Na2Cr2O7, the yellow solution of sodium chromate (Na2CrO4) is acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7⋅2H2O can be crystallized.
‌2Na2CrO4‌ (yellow) ‌+2H+

───▸

∆

‌Na2Cr2O7‌ (orange) ‌+2Na++H2O
(b) (i) Electronic configuration of Mn2+ is [Ar]3d5
Electronic configuration of Fe2+ is [Ar]3d6
It is known that the half-filled orbitals are more stable.
Therefore, Mn in +2 state has a stable d5 configuration, shows resistance to the oxidation to Mn3+.
Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration and it gets oxidised to Fe3+ easily.
(ii) In all transition metals (except Zn, electronic configuration: 3d104s2 ), there are some unpaired electrons that account for their stronger metallic bonding.
Zn has the least enthalpy of atomization because it lacks these unpaired electrons, which makes its inter-atomic electronic bonding the weakest.
(iii) Actinides exhibit larger oxidation states because of very small energy gap between 5f,6d and 7s sub-shells.
Since, all these sub-shells have similar (n+l) value, therefore all can be involved in bonding resulting in a larger oxidation number for actinoids.
OR
(i) Manganese ([Ar] 3d54s2) shows maximum number of oxidation state as its atoms have five unpaired electrons in 3d orbitals. It shows all the oxidation state from +2 to +7 .
(ii) Cu has positive EO‌

M2+

value, because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(iii) Mn3+ is stronger oxidising agent as the charge from Mn3+ to Mn2+ results in half filled, d5 configuration which has extra stability.
(iv) Europium, (Eu2+) is formed by losing the two 5 s electrons and its electronic configuration becomes [Xe ]4f7 which is quite stable configuration.
(v) MnO4−+8H++8e−→Mn2+ +4H2O

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