Which of the following options are correct for [Fe(CN).l3-complex?
(A) d2sp3 hybridisation
(B) sp3d2 hybridisation
(C) Paramagnetic
(D) Diamagnetic
Solution
Let the complex be [Fe(CN)6]3−.
The oxidation state of Fe can be calculated as follows:
Let the oxidation state of Fe be x.
CN− is a ligand which has −1 charge.
x + 6(−1) = 3−
x = +3
So, the oxidation state of Fe is 3+
It has the electronic configuration as [Ar] 3d 5.
The hybridisation scheme is as shown in diagram.
Six pairs of electrons, one from each CN− molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is paramagnetic because of the presence of 1 unpaired electron1. This complex uses inner orbital (3d) in hybridisation (d2sp3).
Hence, the correct answers are options (A) and (C).
Let the complex be [Fe(CN)6]3−.
The oxidation state of Fe can be calculated as follows:
Let the oxidation state of Fe be x.
CN− is a ligand which has −1 charge.
x + 6(−1) = 3−
x = +3
So, the oxidation state of Fe is 3+
It has the electronic configuration as [Ar] 3d 5.
The hybridisation scheme is as shown in diagram.
Six pairs of electrons, one from each CN− molecule, occupy the six hybrid orbitals. Thus, the complex has octahedral geometry and is paramagnetic because of the presence of 1 unpaired electron1. This complex uses inner orbital (3d) in hybridisation (d2sp3).
Hence, the correct answers are options (A) and (C).