CBSE Class 12 Chemistry 2023 Delhi Set 3 Solved Paper

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Question : 13
Total: 13
SECTION - D
(a) Write the number of unpaired electrons in Cr3+.
(Atomic number of Cr=24 )
(b) Complete the reaction mentioning all the products formed:
Cr2O72+3H2S+8H+
(c) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) Copper has exceptionally positive E(M2+M) value.
(iii) Eu2+ with electronic configuration [Xe]4f76s2 is a strong reducing agent.
(a) Cr=25,[Ar]3d44s2 Ground state structure
Cr3+=[Ar]3d4
There are four unpaired electrons present in Cr3+.
(b) Cr2O72+3H2S+8H+
2Cr3+
cromiumion
+
7H2O
water
+
3S
sulphur

(c) (i) Electronic configuration-
Mn=25,[Ar]3d54s2Mn2+ =[Ar]3d5donating 2e
Fe=26,[Ar]3d64s2Fe3+ =[Ar]3d5donating 3e
Mn = after loosing 2 electron 3d orbital is half filled and Mn2+ is stable.
Fe=After loosing three electrons 3d orbitals is half filled and more stable. That's why, Fe3+ is stable at +3 oxidation state and Mn2+ is stable in +2 oxidation state.
(ii) E° value of copper is +ve i.e, 0.34V. This is due to presence of high enthalpy of atomization and low enthalpy of hydrogen which make it exceptionally positive.
(iii) Electronic configuration of Eu2+[Xe]4f76s2
It has the tendency to loose two electron and attain a stable half filled configuration. So, it oxidized by loosing 2 electrons and reduce to other species.
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