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Chapter 23: Problem 6

How does crystal field theory differ from valence bond theory?

Short Answer

Expert verified
Both Crystal Field Theory and Valence Bond Theory are models that explain the structure and properties of complex compounds. Crystal Field Theory considers the electrostatic interaction between a transition metal and ligands. It includes sigma and pi bonds and acknowledges the repulsion between electrons. Valence Bond Theory explains that bonds form due to the overlap of atomic orbitals, permitting electron pairing between involved atoms, but it primarily focuses on sigma bond and tends to ignore electron-electron repulsion.

Step by step solution

01

Understanding Crystal Field Theory

Crystal Field Theory (CFT) is a model that describes the bonding, levels of energy and the properties of coordination complexes. The key idea behind CFT is that the interaction between a transition metal and ligands arises from the attraction between the positively charged metal cation and negative charge on the non-bonding electrons of the ligand.
02

Understanding Valence Bond Theory

Valence Bond Theory (VBT) on the other hand, explains that bonds form from the overlap of atomic orbitals, which allows the pairing of electrons from the involved atoms. The overlapping orbitals give rise to sigma or pi chemical bonds.
03

Highlighting the Differences

There are several key differences between CFT and VBT. Firstly, CFT considers both sigma and pi bonds as potential sources of attraction between the metal ion and the ligand. In contrast, VBT primarily deals with sigma bond formation. Another crucial difference is that CFT takes into account the repulsion between electrons, whereas VBT largely ignores these electron-electron repulsive effects.

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Most popular questions from this chapter

Draw molecular orbital diagram of a square planar complex showing only \(\pi\) -bonding.

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

Discuss Sidgwick's concept of effective atomic number with the help of suitable example.

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