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Chapter 24: Problem 10

Discuss the crystal field splitting in a square planar complex.

Short Answer

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The crystal field splitting in a square planar complex occurs such that the two \( d \) orbitals, \( d_{z^2} \) and \( d_{x^2鈭抷^2} \), have higher energy due to direct repulsion from ligands and the three \( d \) orbitals \( d_{xy}, d_{xz}, d_{yz} \) have lower energy. This splitting determines the color, structure, and magnetism of the complex.

Step by step solution

01

Explain Crystal Field Splitting

Crystal Field Splitting refers to the energy difference that arises when d-state degenerate electron orbitals are split into different energy levels upon ligand binding in transition metal complexes. In the absence of any interacting fields, these orbitals are usually of equal energy.
02

Define a Square Planar Complex

A square planar complex refers to a molecular geometry where four ligands are coordinated to the central metal ion in a square plane. Transition metals with 8 d electrons, which results in a d8 configuration, commonly form these types of complexes.
03

Describe Crystal Field Splitting in a Square Planar Complex

In a square planar complex, the crystal field splitting of \( d \) orbitals results in an energy diagram where two \( d \) orbitals 鈥� \( d_{z^2} \) and \( d_{x^2鈭抷^2} \) 鈥� are of higher energy and are oriented along the axes directly at the ligands. These orbitals experience the most repulsion and thus get the higher energy. The other three \( d \) orbitals namely, \( d_{xy}, d_{xz}, d_{yz} \), are of equal energy and are lower. They are orientated between the axes and hence experience less repulsion and get the minimum energy.

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Most popular questions from this chapter

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.

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

How does ligand field theory describe the origin of charge transfer spectra in coordination complexes.

Discuss the molecular orbital treatment for an octahedral complex with the help of a suitable example.

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