
What is the shape of $Fe{{(CO)}_{5}}$the molecule and which of the following d-orbitals involved in hybridization?
(A) Tetrahedral, ${{d}_{{{x}^{2}}-{{y}^{2}}}}$
(B) Trigonal bipyramidal, ${{d}_{{{x}^{2}}-{{y}^{2}}}}$
(C) Trigonal bipyramidal, ${{d}_{{{x}^{2}}}}$
(D) Square pyramidal
Answer
131.7k+ views
Hint: Finding the hybridization of the central metal atom will tell the structure of the complex. Hybridization is the process of combining and mixing of atomic orbitals from the same atom to form an entirely new hybrid orbital having the same energy.
Complete step by step solution:
-In simple words, the concept of intermixing of two atomic orbitals with the same energy levels giving rise to the degenerate type of orbitals.
-In hybridization, the atomic orbitals of only similar energy can take parts such as the mixing of two ‘s’ orbitals or two ‘p’ orbitals or an ‘s’ orbital with a ‘p’ orbital or ‘s’ orbital with a ‘d’ orbital.
-The atomic orbitals taking part in hybridization can be both, full-filled or half-filled orbitals provided they must have similar energy.
-Finding the hybridization of $Fe{{(CO)}_{5}}$complex,
Electronic configuration of Fe in the ground state- $[Ar]3{{d}^{6}}4{{s}^{2}}$
Electronic configuration of Fe in an excited state- $[Ar]3{{d}^{6}}4{{s}^{2}}$, as the iron in the complex, is in zero oxidation state.
According to Valence Bond theory,

From spectrochemical series, CO being a strong field ligand thus will cause the pairing of electrons.

where X is the orbitals occupied by the ligand CO.
Therefore, hybridization will be $ds{{p}^{3}}$.
-According to Crystal field theory,

-When the hybridization is $ds{{p}^{3}}$, the shape will be trigonal bipyramidal and the d-orbital involved in hybridization will be $d{{z}^{2}}$.

So, the correct option is (C).
Note: When the coordination number is 2, the geometry of the complex is linear. When the coordination number is 3, the geometries can be trigonal planar geometry, trigonal pyramid geometry, or T-shaped. When the coordination number is 4, the geometries can be tetrahedral or square planar. When the coordination number is 5, the geometries can be square pyramidal or trigonal bipyramidal. When the coordination number is 6, the geometries can be a hexagonal planar, trigonal prism, or octahedral.
Complete step by step solution:
-In simple words, the concept of intermixing of two atomic orbitals with the same energy levels giving rise to the degenerate type of orbitals.
-In hybridization, the atomic orbitals of only similar energy can take parts such as the mixing of two ‘s’ orbitals or two ‘p’ orbitals or an ‘s’ orbital with a ‘p’ orbital or ‘s’ orbital with a ‘d’ orbital.
-The atomic orbitals taking part in hybridization can be both, full-filled or half-filled orbitals provided they must have similar energy.
-Finding the hybridization of $Fe{{(CO)}_{5}}$complex,
Electronic configuration of Fe in the ground state- $[Ar]3{{d}^{6}}4{{s}^{2}}$
Electronic configuration of Fe in an excited state- $[Ar]3{{d}^{6}}4{{s}^{2}}$, as the iron in the complex, is in zero oxidation state.
According to Valence Bond theory,

From spectrochemical series, CO being a strong field ligand thus will cause the pairing of electrons.

where X is the orbitals occupied by the ligand CO.
Therefore, hybridization will be $ds{{p}^{3}}$.
-According to Crystal field theory,

-When the hybridization is $ds{{p}^{3}}$, the shape will be trigonal bipyramidal and the d-orbital involved in hybridization will be $d{{z}^{2}}$.

So, the correct option is (C).
Note: When the coordination number is 2, the geometry of the complex is linear. When the coordination number is 3, the geometries can be trigonal planar geometry, trigonal pyramid geometry, or T-shaped. When the coordination number is 4, the geometries can be tetrahedral or square planar. When the coordination number is 5, the geometries can be square pyramidal or trigonal bipyramidal. When the coordination number is 6, the geometries can be a hexagonal planar, trigonal prism, or octahedral.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

Degree of Dissociation and Its Formula With Solved Example for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Which among the following is the softest metal A Platinum class 11 chemistry JEE_Main

Calculate CFSE of the following complex FeCN64 A 04Delta class 11 chemistry JEE_Main

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Thermodynamics Class 11 Notes: CBSE Chapter 5
