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Shape of $Fe(CO)_5$ is :
A
Octahedral
B
Square planar
C
Trigonal bipyramidal
D
Square pyramidal
Detailed Solution
In $Fe(CO)_5$, Fe is in the zero oxidation state: $[Ar]\,3d^6\,4s^2$. With strong field CO, the electrons pair up to give $3d^8$, leaving one 3d, one 4s and three 4p orbitals for bonding.
These form five $dsp^3$ hybrid orbitals, which accept lone pairs from five CO ligands.
$dsp^3$ hybridisation gives a trigonal bipyramidal shape (3 equatorial and 2 axial CO groups). The complex is diamagnetic.
So the shape of $Fe(CO)_5$ is trigonal bipyramidal.
These form five $dsp^3$ hybrid orbitals, which accept lone pairs from five CO ligands.
$dsp^3$ hybridisation gives a trigonal bipyramidal shape (3 equatorial and 2 axial CO groups). The complex is diamagnetic.
So the shape of $Fe(CO)_5$ is trigonal bipyramidal.