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91Ó°ÊÓ

In a square planar complex which \(d\) orbital is at lowest energy? Which \(d\) orbital is at highest energy? (Assume the ligands lie along the \(x\) and \(y\) axes.)

Short Answer

Expert verified
In a square planar complex, the \(d_{z^2}\) orbital is at the lowest energy, and the \(d_{x^2-y^2}\) orbital is at the highest energy.

Step by step solution

01

Understand the geometry of the square planar complex

In a square planar complex, the metal is at the center of a square plane with ligands at the four corners. The ligands' positions along the axes affect the splitting of the metal's d orbitals due to ligand field theory.
02

Recall the order of d orbital splitting

For square planar complexes, the typical order of energy levels for the d orbitals, from lowest to highest energy, is: \(d_{z^2} < d_{xy} < d_{xz} = d_{yz} < d_{x^2-y^2}\). This order is influenced by the ligands' positions at 90 degrees relative to each other along the x and y axes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

d orbital splitting
The concept of *d orbital splitting* is central to understanding the behavior of square planar complexes. In transition metals, d orbitals undergo energy level changes when they form complexes with ligands. When a metal ion is surrounded by ligands, such as in a square planar configuration, the electrostatic interactions and arrangements between the ligands and central metal ion lead to the splitting of the degenerate d orbitals (those with the same energy). This splitting occurs because some orbitals experience more repulsion from ligands positioned directly along their axes, leading to higher energy levels, while others experience less, leading to lower energy levels. This is described by ligand field theory, which provides a detailed explanation of how these interactions occur.
ligand field theory
*Ligand field theory* (LFT) plays a key role in understanding the electronic structure of coordination compounds, especially those with complex shapes such as square planar geometries. LFT is an adaptation and extension of crystal field theory that takes into account the covalent nature of the metal-ligand bond. It suggests that ligands are not only point charges that repel electrons in the metal d orbitals but also have interactions based on orbital overlap. In a square planar complex, LFT predicts how the ligand arrangement affects d orbital energy levels:
  • Ligands positioned directly on the x- and y-axes cause significant repulsion for the dx2-y2 orbital.
  • This orbital ends up at the highest energy level because its lobes face the ligands directly.
  • Other orbitals, such as dz2, are affected less directly, resulting in a lower energy state.
By understanding these principles, one can predict the square planar complexes' electronic properties and reactivity.
coordination chemistry
*Coordination chemistry* is a fascinating field that explores how metal ions interact with surrounding molecules or ions, known as ligands, to form coordination compounds. In square planar complexes, coordination chemistry is particularly interesting due to the geometrical arrangement and distribution of ligands around a central metal:
  • The metal is usually a transition metal, which provides the d orbitals that participate in coordinate bonding.
  • Ligands are arranged in a square plane, creating unique electronic environments differing from other geometries such as tetrahedral or octahedral.
These interactions are critical in various applications, including catalysis, materials science, and bioinorganic chemistry. Understanding how ligands affect d orbital energies in square planar complexes allows chemists to tailor these compounds for specific purposes, enhancing their functionality and effectiveness in chemical processes.

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Most popular questions from this chapter

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) (c) \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\) (b) VOCI_ \((d)\left[C u(e n)_{2}\right]^{2+}\)

Nickel and palladium both form complexes of the general formula \(\mathrm{M}\left(\mathrm{PR}_{3}\right)_{2} \mathrm{Cl}_{2}\). (The ligand \(\mathrm{PR}_{3}\) is a phosphine such as \(\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) triphenylphosphine. It is a Lewis base. The nickel( II) compound is paramagnetic whereas the palladium(II) compound is diamagnetic. (a) Explain the magnetic properties of these compounds. (b) How many isomers of each compound are expected?

The complex \(\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) has five unpaired electrons, whereas \(\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{4-}\) has only one. Using the ligand field model, depict the electron configuration for each ion. What can you conclude about the effects of the different ligands on the magnitude of \(\Delta_{0} ?\)

For the high-spin coordination compound \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2},\) identify the following. (a) the oxidation number of manganese (b) the coordination number for manganese (c) the coordination geometry for manganese (d) the number of unpaired electrons per metal atom (e) whether the complex is diamagnetic or paramagnetic (f) the number of geometric isomers

In this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{t}\), for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: \(\mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) \quad K_{\mathrm{f}}=10^{8}\) \(\mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) \(K_{f}=10^{18}\) The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta_{r} G^{\circ}=-R T \ln K\) and \(\Delta_{r} G^{\circ}=\) \(\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} .\) We know from experiment that \(\Delta_{t} H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) and \(\Delta_{i} H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} .\) Is the difference in \(\Delta_{r} H^{\circ}\) suffi- cient to account for the \(10^{10}\) difference in \(K_{f} ?\) Comment on the role of entropy in the second reaction.

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