A) It is a octahedral complex
B) \[{{t}_{2g}}\] Orbital contains the \[{{e}^{-}}\] of metal
C) For iron, overlapping between empty orbitals and ligand orbitals takes place
D) All of the above
Correct Answer: D
Solution :
For the complex \[{{[Fe{{(CN)}_{6}}]}^{4-}}\] Oxidation state of \[Fe=+II\] Coordination number = 6 Electronic configuration of \[F{{e}^{2+}}\]\[F{{e}^{2+}}=1{{s}^{2}},2{{s}^{2}},3{{s}^{2}},3{{p}^{6}},3{{d}^{6}},4{{s}^{0}},4{{p}^{0}}\] Since, \[-CN\] is ligand of strong field. Hence, it paired up electrons of d-orbital. So, Hybridization \[={{d}^{2}}s{{p}^{3}}\] Magnetic moment = zero Number of unpaired electrons = zero Geometry = octahedral.You need to login to perform this action.
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