Question

The \(\mathrm{CrF}_{6}^{4-}\) ion is known to have four unpaired electrons. Does the \(\mathrm{F}^{-}\) ligand produce a strong or weak field?

Short answer

The \(\mathrm{F}^{-}\) ligand produces a weak field in the \(\mathrm{CrF}_{6}^{4-}\) complex ion, as the presence of four unpaired electrons indicates that the splitting of the d orbitals is not sufficient to pair up the electrons.

Step-by-step solution

Recall the basics about crystal field theory and ligand field strength

According to crystal field theory, when a transition metal ion is surrounded by ligands, the degenerate d orbitals of the metal ion split into two sets with different energies. The energy difference between these two sets depends on the field strength of the ligands. In general, the ligand strength is classified as either strong or weak field. Strong-field ligands cause a larger splitting of d orbitals, resulting in low-spin complexes, whereas weak-field ligands cause a smaller splitting, resulting in high-spin complexes.

Identify the central metal ion and determine its electron configuration

In the given complex ion, \(\mathrm{CrF}_{6}^{4-}\), the central metal ion is \(\mathrm{Cr^{4+}}\). Chromium is in Group 6 of the periodic table and has an atomic number of 24. The electron configuration of \(\mathrm{Cr}\) is: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5\] With a charge of +4, four electrons are removed from the outermost shells, which results in the electron configuration of the \(\mathrm{Cr^{4+}}\) ion as: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^1\]

Calculate the number of unpaired electrons in \(\mathrm{Cr^{4+}}\) ion

The electron configuration of \(\mathrm{Cr^{4+}}\) in its 3d orbital is given as: 3d: ↑ There is 1 unpaired electron in the \(\mathrm{Cr^{4+}}\) ion.

Recall the unpaired electron information for \(\mathrm{CrF}_{6}^{4-}\) complex ion

The given information states that \(\mathrm{CrF}_{6}^{4-}\) ion has four unpaired electrons.

Determine if the fluoride ligand produces a strong or weak field

The presence of four unpaired electrons in \(\mathrm{CrF}_{6}^{4-}\) ion indicates that the splitting of the d orbitals due to the fluoride ligand (\(\mathrm{F}^{-}\)) is not sufficient to pair up the electrons. As a result, the fluorine creates a weak-field effect. To summarize, the \(\mathrm{F}^{-}\) ligand produces a weak field in the \(\mathrm{CrF}_{6}^{4-}\) complex ion.

Key concepts

Ligand Field Strength

Crystal Field Theory (CFT) is a theoretical model that helps explain the behavior of transition metal complexes, such as the one in the exercise \(\mathrm{CrF}_{6}^{4-}\). This model describes how the presence of ligands can affect the energies of the d orbitals of the metal ion at the center of the complex.

Ligands are ions or molecules that can donate electron pairs to form coordinate bonds with the metal ion, essentially creating a field around it. The strength of the field produced by the ligands influences the splitting of the metal ion's d orbitals into two energy levels: the lower-energy t2g set and the higher-energy eg set.

How does ligand field strength affect electron arrangement?

In the presence of a strong-field ligand, the energy gap between these two sets of orbitals is large. This can lead to electrons pairing up in the lower t2g set before occupying the higher eg set, resulting in a low-spin configuration. Conversely, weak-field ligands produce a smaller energy gap, and electrons will occupy all five d orbitals singly before any pairing occurs, leading to a high-spin state.

Identifying Ligand Field Strength

Ligands are arranged in a series known as the spectrochemical series, which ranks them from weak field (causing high spin) to strong field (causing low spin). We can conclude from our exercise that fluoride \(\mathrm{F}^{-}\) is a weak-field ligand as it creates a high-spin state seen in the \(\mathrm{CrF}_{6}^{4-}\) ion with its four unpaired electrons.

Electron Configuration

Understanding electron configuration is key to grasping the concepts of CFT and the properties of transition metal complexes. The electron configuration of an atom or ion tells us how the electrons are distributed among the various orbitals.

For the central ion in our exercise, chromium \(\mathrm{Cr^{4+}}\), which has lost four electrons, it has only one electron in the 3d orbital after the removal of electrons (accounting for its +4 charge). It's important to remember that the electrons are removed from the outermost orbitals first, which for transition metals is typically the 4s orbital and then the 3d orbital.

Electron Configuration Notation

To denote the configuration, we use notations such as \(1s^2 \) to indicate that the 1s orbital is filled with two electrons. The configuration of \(\mathrm{Cr^{4+}}\) shows that all subshells are fully filled except for the 3d subshell, which is of prime interest in crystal field theory. The detailed configuration reveals the electronic structure and helps predict the chemical and magnetic properties of the ion.

Unpaired Electrons

Unpaired electrons in a transition metal complex play a significant role in determining its magnetic and spectral properties. An unpaired electron is one that is alone in an orbital and is not paired with another electron of opposite spin.

Importance of Unpaired Electrons

In the context of CFT, the number of unpaired electrons can be used to deduce whether a ligand field is strong or weak, as observed with the \(\mathrm{CrF}_{6}^{4-}\) ion that has four unpaired electrons, indicative of a weak-field caused by the fluoride ligands. Due to these unpaired electrons, such complexes often exhibit paramagnetism, which is a type of magnetism that only occurs in the presence of an externally applied magnetic field.

Unpaired Electrons and Magnetic Properties

By knowing the number of unpaired electrons, we can predict the magnetic behavior of a complex. Complexes with one or more unpaired electrons are paramagnetic and those without any unpaired electrons are diamagnetic. In the exercise, the presence of four unpaired electrons means the \(\mathrm{CrF}_{6}^{4-}\) complex is paramagnetic. This is in stark contrast with a complex with paired electrons, which would exhibit diamagnetism due to the absence of net magnetic moments in the orbitals.