Question

For each of the following complexes, describe the bonding using valence bond theory. Include orbital diagrams for the free metal ion and the metal ion in the complex. Indicate which hybrid orbitals the metal ion uses for bonding, and specify the number of unpaired electrons. (a) \(\left[\mathrm{AuCl}_{4}\right]^{-}\) (square planar) (b) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) (c) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) (high-spin) (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) (low-spin)

Short answer

(a) 0 unpaired e-, (b) 0 unpaired e-, (c) 4 unpaired e-, (d) 0 unpaired e-.

Step-by-step solution

Determine the Oxidation State and Electron Configuration

Calculate the oxidation state of each metal by considering the charges of the ligands and the overall charge of the complex.- (a) For \([\mathrm{AuCl}_{4}]^{-}\), \( \mathrm{Au} \) is in the +3 oxidation state.- (b) For \([\mathrm{Ag(NH}_{3})_{2}]^{+}\), \( \mathrm{Ag} \) is in the +1 oxidation state.- (c) For \([\mathrm{Fe(H}_{2}\mathrm{O})_{6}]^{2+}\), \( \mathrm{Fe} \) is in the +2 oxidation state.- (d) For \([\mathrm{Fe(CN)}_{6}]^{4-}\), \( \mathrm{Fe} \) is in the +2 oxidation state.Write the electron configuration for each free metal ion:- (a) \(\mathrm{Au}^{3+}: [Xe]4f^{14}5d^{8}\)- (b) \(\mathrm{Ag}^{+}: [Kr]4d^{10}\)- (c) \(\mathrm{Fe}^{2+}: [Ar]3d^{6}\)- (d) \(\mathrm{Fe}^{2+}: [Ar]3d^{6}\)

Orbital Diagrams for Free Metal Ions

Draw the orbital diagram for each free metal ion based on their electron configuration:- (a) \(\mathrm{Au}^{3+}: \) \(5d\) orbitals have 8 electrons, distributed as \(\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\).- (b) \(\mathrm{Ag}^{+}: \) \(4d\) orbitals are completely filled with 10 electrons.- (c) \(\mathrm{Fe}^{2+}: \) \(3d\) orbitals have 6 electrons, with one electron in each of the five \(d\) orbitals and one paired.- (d) \(\mathrm{Fe}^{2+}: \) \(3d\) orbitals have paired electrons in low-spin, typically \(\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\uparrow\).

Identify Hybrid Orbitals for Bonding

Determine which orbitals are hybridized in each complex for bonding with ligands:- (a) \([\mathrm{AuCl}_{4}]^{-}: \) Uses \( dsp^{2} \) hybridization to achieve square planar geometry.- (b) \([\mathrm{Ag(NH}_{3})_{2}]^{+}: \) Utilizes \( sp \) hybridization for linear geometry.- (c) \([\mathrm{Fe(H}_{2}\mathrm{O})_{6}]^{2+}: \) Forms \( sp^{3}d^{2} \) hybrid orbitals for an octahedral shape (high-spin).- (d) \([\mathrm{Fe(CN)}_{6}]^{4-}: \) Uses \( d^{2}sp^{3} \) hybrid orbitals for low-spin octahedral shape.

Count Unpaired Electrons

Determine the number of unpaired electrons in the complex by analyzing the filling of orbitals:- (a) \([\mathrm{AuCl}_{4}]^{-}: \) 0 unpaired electrons, as all \(5d\) electrons are paired in square planar.- (b) \([\mathrm{Ag(NH}_{3})_{2}]^{+}: \) 0 unpaired electrons, since the \(4d\) shell is fully paired in \(\mathrm{Ag}^{+}\).- (c) \([\mathrm{Fe(H}_{2}\mathrm{O})_{6}]^{2+}: \) 4 unpaired electrons in \(3d\) orbitals, high-spin configuration.- (d) \([\mathrm{Fe(CN)}_{6}]^{4-}: \) 0 unpaired electrons, all \(3d\) electrons are paired (low spin).

Key concepts

Bonding in Coordination Complexes

Coordination complexes occur when a central metal atom binds with surrounding molecules or ions, called ligands, to form a larger complex. Understanding the bonding in these complexes is crucial to predict their shapes and properties. Valence bond theory helps us here by examining how the metal and ligands interact at an atomic level.
The metal atom typically utilizes empty or partially filled orbitals to form bonds with the ligands. In a complex like \([\mathrm{AuCl}_{4}]^{-}\), which is square planar, the \(\mathrm{Au}\) atom uses \(dsp^{2}\) hybridization. This means it combines one \(d\), one \(s\), and two \(p\) orbitals to create the hybrid orbitals that form bonds with the four chloride ions.
The geometrical arrangement, such as square planar, is determined by both the hybridization and the repulsions between ligands. In this way, valence bond theory provides insights into why complexes have their characteristic shapes.

Orbital Hybridization

In coordination complexes, orbital hybridization is a key concept that describes how atomic orbitals mix to form new, equivalent hybrid orbitals. These hybrid orbitals facilitate the formation of strong covalent bonds between the metal and its ligands. Different complexes will utilize different types of hybridizations depending on their coordination number and geometry.
For instance, in the complex \([\mathrm{Ag(NH}_{3})_{2}]^{+}\), the silver ion undergoes \(sp\) hybridization. This involves the mixing of one \(s\) and one \(p\) orbital to form two linearly arranged hybrid orbitals. These hybrid orbitals accommodate the lone pairs from the ammonia ligands, resulting in a linear structure.
Another example is \([\mathrm{Fe(H}_{2}\mathrm{O})_{6}]^{2+}\), which involves \(sp^{3}d^{2}\) hybridization, resulting in an octahedral geometry. The metal's empty \(3d\), \(4s\), and \(4p\) orbitals combine to form six equivalent hybrid orbitals, allowing it to coordinate with six water molecules simultaneously.

Unpaired Electrons in Metal Complexes

The presence of unpaired electrons in metal complexes affects their magnetic properties and reactivity. Unpaired electrons lead to paramagnetism, which can be detected by substances being attracted to a magnetic field. Valence bond theory aids in illustrating the electron configuration within the complex.
Consider the high-spin \([\mathrm{Fe(H}_{2}O)_{6}]^{2+}\) complex. Here, the iron ion has 4 unpaired electrons in its \(3d\) orbitals. This creates a paramagnetic complex with significant magnetic moment.
In contrast, low-spin complexes such as \([\mathrm{Fe(CN)}_{6}]^{4-}\) typically pair up the electrons within the \(d\) orbitals, sometimes altering their spin state due to the strong field created by ligands like cyanide. As a result, complexes with no unpaired electrons, like this one, are diamagnetic and show no magnetic response.