(B) The central metal atom in $Fe(CO)_5$ is $Fe$. $Fe$ has an atomic number of $26$ and its electronic configuration is $[Ar] 3d^6 4s^2$. In $Fe(CO)_5

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(B) The central metal atom in $Fe(CO)_5$ is $Fe$. $Fe$ has an atomic number of $26$ and its electronic configuration is $[Ar] 3d^6 4s^2$. In $Fe(CO)_5$,$Fe$ is in the $0$ oxidation state. Due to the strong field ligand $CO$,the electrons in $3d$ and $4s$ orbitals pair up,resulting in $dsp^3$ hybridization. This hybridization corresponds to a trigonal bipyramidal geometry.

Class 12 Chemistry Coordination Compounds Hybridisation and Geometry

Medium

The shape of pentacarbonyl iron,$Fe(CO)_5$,is:

A
Square planar
B
Trigonal bipyramidal
C
Triangular
D
None of these

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Class 12

Chemistry Questions

Chapter

Coordination Compounds

Subtopic

Hybridisation and Geometry

What is the hybridization of the complex $[Mn(CN)_6]^{3-}$?

Medium

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Match the coordination number and type of hybridisation with the distribution of hybrid orbitals in space based on Valence Bond Theory.

Coordination number and type of hybridisation Distribution of hybrid orbitals in space

$a. 4, sp^3$ $i. \text{trigonal bipyramidal}$

$b. 4, dsp^2$ $ii. \text{octahedral}$

$c. 5, sp^3d$ $iii. \text{tetrahedral}$

$d. 6, d^2sp^3$ $iv. \text{square planar}$

Select the correct option.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$a. 4, sp^3$	$i. \text{trigonal bipyramidal}$
$b. 4, dsp^2$	$ii. \text{octahedral}$
$c. 5, sp^3d$	$iii. \text{tetrahedral}$
$d. 6, d^2sp^3$	$iv. \text{square planar}$