Question -
Answer -
(i) [Fe(CN)_6]_4−
(ii) [FeF_6]_3−
(iii) [Co(C_2O_4)3]_3−
(iv) [CoF_6]_3−
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(i) [Fe(CN)6]4−
In the above coordination complex, ironexists in the +II oxidation state.
Fe2+ :Electronic configuration is 3d6
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Orbitalsof Fe2+ ion:
As CN− isa strong field ligand, it causes the pairing of the unpaired 3d electrons.
Sincethere are six ligands around the central metal ion, the most feasiblehybridization is d2sp3.
d2sp3 hybridizedorbitals of Fe2+ are:
6electron pairs from CN− ions occupy thesix hybrid d2sp3orbitals.
Then
Hence, the geometry of the complex isoctahedral and the complex is diamagnetic (as there are no unpaired electrons).
(ii) [FeF6]3−
In this complex, the oxidation state of Fe is+3.
Orbitalsof Fe+3 ion:
Thereare 6 F− ions. Thus, it will undergo d2sp3 or sp3d2 hybridization.As F− is a weak field ligand, it does not cause thepairing of the electrons in the 3d orbital. Hence, the mostfeasible hybridization is sp3d2.
sp3d2 hybridizedorbitals of Fe are:
Hence, the geometry of the complex is foundto be octahedral.
(iii) [Co(C2O4)3]3−
Cobalt exists in the +3 oxidation state inthe given complex.
Orbitals of Co3+ ion:
Oxalateis a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbitalelectrons. As there are 6 ligands, hybridization has to be either sp3d2 or d2sp3 hybridization.
sp3d2 hybridization of Co3+:
The 6electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand)occupy these sp3d2 orbitals.
Hence, the geometry of the complex is foundto be octahedral.
(iv) [CoF6]3−
Cobalt exists in the +3 oxidation state.
Orbitalsof Co3+ ion:
Again,fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons.As a result, the Co3+ ion will undergo sp3d2 hybridization.
sp3d2 hybridizedorbitals of Co3+ ion are:
Hence, the geometryof the complex is octahedral and paramagnetic.