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Chapter 24: Problem 13

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

Short Answer

Expert verified
In the octahedral complex \([Mo(CN)_6]^{3-}\), Mo's d-orbitals are split into \( e_{g} \) and \( t_{2g} \) sets by ligand's electric field. They overlap with ligand orbitals to form molecular orbitals wherein the electrons are filled following certain rules creating bonding and antibonding orbitals.

Step by step solution

01

Octahedral Complexes and Central Atom

Firstly, identify the central atom and its valence d-orbital energy levels. For the \([Mo(CN)_6]^{3-}\) complex, the central atom is molybdenum (Mo) which is in the d-block of the periodic table, so its valence (outermost) electron shell houses d-orbitals.
02

Ligand Field Splitting

Under the influence of ligands (like CN- in our example), the d-orbital energy levels of the central atom split into two different energy groups due to unequal electric field experienced by d-orbitals. In an octahedral complex, this results in the lower energy \( t_{2g} \) set (includes \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals) and the higher energy \( e_{g} \) set (includes \(d_{z^2}\) and \(d_{x^2 - y^2}\) orbitals).
03

Formation of Molecular Orbitals

Molecular orbitals are formed by the overlap of atomic orbitals of the same or similar energy. Ligands have lone pair of electrons, and in our example CN- ligands have d-orbitals and sigma (σ) bonding orbitals. These orbitals overlap with Mo's \( e_{g} \) and \( t_{2g} \) orbitals to form bonding and anti-bonding molecular orbitals.
04

Electron Filling

Now fill in the electrons starting with the lowest energy orbital following the Pauli Exclusion Principle (each orbital can house at most 2 electrons with opposite spin) and Hund's Rule (every orbital in a subshell is singly occupied before any orbital is doubly occupied). In the case of \([Mo(CN)_6]^{3-}\), Mo has 6 electrons in its 4d orbital. Considering the three extra electrons brought in by the charge, the total becomes 9 which will fill the bonding orbitals of the octahedral complex.

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Most popular questions from this chapter

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

Discuss the factors affecting the magnitude of crystal field splitting.

Discuss the crystal field splitting in a square planar complex.

Discuss the molecular orbital diagram of a tetrahedral complex with the help of a 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.
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