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Chapter 23: Problem 15

Discuss the molecular orbital diagram of a tetrahedral complex with the help of a suitable example

Short Answer

Expert verified
Molecular orbitals of a tetrahedral complex are formed by the combination of atomic orbitals of the atoms involved in the formation of the molecule. The tetrahedral complex of methane is a clear example, where carbon hybridizes its atomic orbitals to form four bonds with the hydrogen atoms.

Step by step solution

01

Concept of Molecular Orbital

Molecular orbitals are formed by the combination of atomic orbitals of different atoms involved in the formation of a molecule. These orbitals describe the behavior of electrons in molecules in terms of their wave-like properties.
02

Formation of Tetrahedral Complex

In a tetrahedral molecule, there are four bonding molecular orbitals and these are formed by the constructive overlap of the atomic orbitals, producing regions of high electron probability or electron clouds around the molecule.
03

Example of Tetrahedral Complex

To give an example, let's discuss the tetrahedral complex of methane (CH4). In methane, carbon forms four bonds with four hydrogen atoms. To form these bonds, the carbon atom hybridizes its 2s and three 2p atomic orbitals, forming four bonding orbitals to overlap with the 1s orbitals of hydrogen. The resulting molecular orbital diagram shows four molecular orbitals produced by the overlap of the atomic orbitals.

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Most popular questions from this chapter

Discuss the main points of Werner theory. What is the significance?

Explain the following: (a) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is paramagnetic and coloured. (b) \(\left(\mathrm{Ni}(\mathrm{CN})_{4}\right)^{2-}\) is square planar, but \(\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]^{2-}\) is tetrahedral. (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is more stable than \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), while \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is more stable than \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) Octahedral complexes are less stable than the square planar complexes. (e) \(\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}\) is a low-spin complex, but \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is high-spin complex.

Discuss the salient features of ligand field theory. How does it differ from crystal field theory?

Discuss the crystal field splitting in a square planar complex.

Discuss in brief the crystal field theory and represent the splitting of ' \(d\) ' orbitals in octahedral and tetrahedral fields.
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