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CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

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Question : 11 of 27
Marks: +1, -0
(a) Using valence bond theory, write the hybridization and magnetic character of the complex [Fe(CN)6]4−[\mathrm{Fe}(\mathrm{CN})_6]^{4-}. (Atomic no. of Fe=26\mathrm{Fe}=26 )
(b) Write the electronic configuration of d6d^6 on the basis of crystal field theory when:
(i) Δ0<\Delta_0 < P and
(ii) Δ0>P\Delta_0 > \mathrm{P}
Solution:  
(a) Oxidation number of Fe\mathrm{Fe} is +2 in [Fe(CN)6]4−[\mathrm{Fe}(\mathrm{CN})_6]^{4-}
Fe2+−3d6\mathrm{Fe}^{2+}-3 d^6
Six CN−\mathrm{CN}^{-}ions give their six pairs of electron to the empty orbitals. The orbitals utilized for hybridization is d2sp3d^2 \mathrm{sp}^3.
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