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Chapter 24: Problem 16

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

Short Answer

Expert verified
The molecular orbital diagram of a square planar complex involves horizontal lines corresponding to the energy levels with \(\pi\)-bonding molecular orbital and \(\pi\)-anti-bonding molecular orbital. In a square planar, four \(\pi\)-bonds are formed and shown in the diagram as pairs of electrons occupying the \(\pi\)-bonding molecular orbitals.

Step by step solution

01

Understand the Square Planar Complex

A square planar molecule is a molecule that forms the shape of a square in a single plane. For this exercise, only consider \(\pi\)-bonding MO diagram.
02

Draw Molecular orbitals (MO)

Start by drawing a horizontal line which will denote the energy level. Above and below this line, draw two more lines. These lines represent the atomic orbitals of the central atom. It's important to note that the line in the middle represents a non-bonding atomic orbital, the line above it represents a bonding molecular orbital, and the line below it represents an anti-bonding molecular orbital.
03

Represent \(\pi\)-bonding

In \(\pi\)-bonding, electrons are shared between two atoms in regions above and below a plane. Now, show four \(\pi\)-bonds for a square planar complex. Place one pair of electrons (two dots or lines) in each of the bonding molecular orbitals.
04

Complete the Molecular Orbital Diagram

Label the \(\pi\)-bonding molecular orbital and the \(\pi\)-anti-bonding molecular orbital. Lastly, draw arrows connecting each ligand鈥檚 atomic orbital to the corresponding molecular orbital i.e. \(\pi\)-bonding and \(\pi\)-anti-bonding molecular orbitals.

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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.

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

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

How does crystal field theory differ from valence bond theory?

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