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\[F{{e}^{2+}}\] \[F{{e}^{2+}}=1{{s}^{2}},\,\,2{{s}^{2}},\,\,2{{p}^{6}},\,\,3{{s}^{2}},\,\,3{{p}^{6}},\,\,3{{d}^{6}},\,\,4{{s}^{0}},\,\,4{{p}^{0}}\] Since, \[-CN\] is ligand of strong field, Hence, it paired up electrons of \[d-\]orbital. So,\[{{(Fe{{(CN)}_{6}}]}^{4-}}\] Hybridization\[={{d}^{2}}s{{p}^{3}}\] Magnetic moment = zero Number of unpaired electrons = zero Geometry = octahedral.You need to login to perform this action.
You will be redirected in
3 sec