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CBSE Class 12 Chemistry 2023 Delhi Set 2 Solved Paper

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Question : 11 of 12
Marks: +1, -0
SECTION - C
(a) On the basis of crystal field theory write the electronic configuration for d5d^5 ion with a weak ligand for which delta 0<p0 < p. (b) Explain [Fe(CN)6]3−[\mathrm{Fe}(\mathrm{CN})_6]^{3-} is an inner orbital complex whereas [FeF6]3−[\mathrm{FeF}_6]^{3-} is an outer orbital complex.
[Atomic number: Fe=26\mathrm{Fe}=26 ]
It is a magnitude difference in energy between the two sets of d-orbital i.e. t2gt_{2g} and eg electronic configuration of d5d^5 if Δ0<P\Delta_0 < P
thent2g3t_{2g}^3 and eg2e_g^2
Because it follows weak field ligand phenomina when weak field ligands are present in a structure pairing of electrons do not take place.
∵ so
(b) [Fe(CN)6]3−=CN[\mathrm{Fe}(\mathrm{CN})_6]^{3-}=\mathrm{CN} is strong field ligand
Fe=26, [Ar] 3d6 4s2\mathrm{Fe}=26,\,[\mathrm{Ar}]\,3d^6\,4s^2
Fe3+=3d5\mathrm{Fe}^{3+}=3 d^5
Inner d-complex
If strong field ligand is available then
Δ0>P=t2g5eg\Delta_0 > P = t_{2g}^5 e_g
In [Fe(F6)]3−, Fe=[Ar] 3d6 4s2[\mathrm{Fe}(\mathrm{F}_6)]^{3-},\,\mathrm{Fe}=[\mathrm{Ar}]\,3d^6\,4s^2
Fe3+=3d5\mathrm{Fe}^{3+}=3 d^5
with a weak field ligand Δ0<P\Delta_0 < P
∵\because so, there is no pairing of electrons in 3d3d orbitals.
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