Question

For each of the following complexes, draw a crystal field energylevel diagram, assign the electrons to orbitals, and predict the number of unpaired electrons: (a) \(\left[\mathrm{CrF}_{6}\right]^{3-}\) (b) \(\left[\mathrm{V}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\)

Short answer

(a) 3 unpaired, (b) 2 unpaired, (c) 1 unpaired.

Step-by-step solution

Determine the oxidation state

To determine the oxidation state of the central metal ion in each complex:- (a) For \([\mathrm{CrF}_{6}]^{3-}\), assume each fluoride ion, \( \mathrm{F}^{-}, \) has a charge of \(-1\), so \(x + 6(-1) = -3\). Thus, the oxidation state of chromium is \(+3\).- (b) For \([\mathrm{V}(\mathrm{H}_{2}\mathrm{O})_{6}]^{3+}\), water is a neutral ligand. Therefore, the oxidation state of vanadium is \(+3\).- (c) For \([\mathrm{Fe}( ext{CN})_{6}]^{3-}\), the cyanide ion, \( \text{CN}^{-}\), has a charge of \(-1\). Solving \(x + 6(-1) = -3\), the oxidation state of iron is \(+3\).

Write the electron configuration

Determine the electron configuration of each metal ion in its oxidation state:- (a) \(\mathrm{Cr}^{3+}\) has an electronic configuration of \([\mathrm{Ar}]\, 3d^3\) (since \(\mathrm{Cr}\) is \([\mathrm{Ar}]\, 3d^5\, 4s^1\) and loses three electrons).- (b) \(\mathrm{V}^{3+}\) has an electronic configuration of \([\mathrm{Ar}]\, 3d^2\) (since \(\mathrm{V}\) is \([\mathrm{Ar}]\, 3d^3\, 4s^2\) and loses three electrons).- (c) \(\mathrm{Fe}^{3+}\) has an electronic configuration of \([\mathrm{Ar}]\, 3d^5\) (since \(\mathrm{Fe}\) is \([\mathrm{Ar}]\, 3d^6\, 4s^2\) and loses three electrons).

Apply the crystal field splitting

Differentiate between strong- and weak-field ligands to apply the appropriate crystal field splitting:- (a) \(\mathrm{F}^{-}\) is a weak-field ligand, so use weak-field (high-spin) splitting for \(\mathrm{Cr}^{3+}\): three electrons populate the \(t_{2g}\) and \(e_g\) orbitals, giving a configuration of \(t_{2g}^3\, e_g^0\).- (b) \(\mathrm{H}_2\mathrm{O}\) is a weak-field ligand, hence \(\mathrm{V}^{3+}\) will also have high-spin configuration: two electrons, \(t_{2g}^2\, e_g^0\).- (c) \(\mathrm{CN}^{-}\) is a strong-field ligand, causing low-spin splitting for \(\mathrm{Fe}^{3+}\). Thus, five electrons fully occupy the \(t_{2g}\) set, \(t_{2g}^5\, e_g^0\).

Predict unpaired electrons

Based on the electron configurations obtained in crystal field splitting:- (a) \(\left[\mathrm{CrF}_{6}\right]^{3-}\) has 3 unpaired electrons (in \(t_{2g}^3\)).- (b) \(\left[\mathrm{V}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{3+}\) has 2 unpaired electrons (in \(t_{2g}^2\)).- (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) has 1 unpaired electron (in \(t_{2g}^5\)).

Key concepts

Understanding Oxidation States in Complexes

The oxidation state of a metal within a complex tells us about the charge on the metal after all ligands in the complex have contributed their charges. This is important for understanding electron distribution and the overall charge of the complex. For example, in \([\mathrm{CrF}_{6}]^{3-}\), each fluoride ion (\(\mathrm{F}^{-}\)) contributes a charge of \(-1\), leading to the expression \(x + 6(-1) = -3\), where \(x\) represents the oxidation state of chromium. Solving this gives a \(+3\) oxidation state. Similarly, in \[\mathrm{V}(\mathrm{H}_{2} \mathrm{O})_{6}]^{3+}\], water is neutral, leading to a vanadium oxidation state of \(+3\). Lastly, in \[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}\], cyanide (\(\mathrm{CN}^{-}\)) has a charge of \(-1\). Solving \(x + 6(-1) = -3\) reveals that iron also has an oxidation state of \(+3\). Determining oxidation states helps predict the electron configuration and magnetic properties of complexes.

Exploring Electron Configurations

Electron configuration describes the arrangement of electrons around an atom's nucleus, organized in different orbitals. This arrangement changes when a neutral atom forms a cation by losing electrons. For \([\mathrm{Cr}^{3+}\), the configuration changes from \[\mathrm{Cr}] = [\mathrm{Ar}]\, 3d^5\, 4s^1\] to \[\mathrm{Cr}^{3+} = [\mathrm{Ar}]\, 3d^3\], as it loses three electrons. Vanadium undergoes a similar change: \[\mathrm{V}] = [\mathrm{Ar}]\, 3d^3\, 4s^2\] to \[\mathrm{V}^{3+} = [\mathrm{Ar}]\, 3d^2\]. Iron, too, changes from \[\mathrm{Fe}] = [\mathrm{Ar}]\, 3d^6\, 4s^2\] to \[\mathrm{Fe}^{3+} = [\mathrm{Ar}]\, 3d^5\]. Understanding electron configurations assists in predicting how metal ions and their complexes behave chemically and magnetically.

Impact of Ligand Field Strength

Ligand field strength refers to the power of a ligand to split the d-orbitals of a metal ion into different energy levels, significantly affecting the electronic configuration and magnetic properties of a complex. Ligands are categorized based on whether they cause strong or weak field splitting. \(\mathrm{F}^{-}\) and \(\mathrm{H}_2\mathrm{O}\) are weak-field ligands, resulting in a high-spin configuration for \(\mathrm{Cr}^{3+}\) and \(\mathrm{V}^{3+}\): multiple unpaired electrons occupy the higher energy \(e_g\) orbitals. Conversely, \(\mathrm{CN}^{-}\) is a strong-field ligand, causing \(\mathrm{Fe}^{3+}\) to adopt a low-spin configuration with most electrons paired in the lower energy \(t_{2g}\) orbitals. Understanding ligand field strength is essential for predicting the number of unpaired electrons and hence the magnetic behavior of the complexes, such as paramagnetism or diamagnetism.
To summarize, ligand field strength differences can lead to diverse electronic structures and properties among metal complexes.