Question

Compare the extent of \(\pi\) -bonding in the following complexes: \(\left[\mathrm{Fe}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{2-}\) and \(\left[\mathrm{V}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{3-}\)

Short answer

\(\left[\mathrm{Fe}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{2-}\) has a greater extent of \( \pi \)-bonding than \(\left[\mathrm{V}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{3-}\) because Fe^2+ (d6) can accept more electron density via \( \pi \)-bonding than V^3+ (d2).

Step-by-step solution

Identify the central atom oxidation states

In order to analyze the bonding, we first need to establish the oxidation states of the metals. For iron in \(\left[\mathrm{Fe}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{2-}\), we have an overall charge of -2. CN is a strong field ligand and causes pairing up. NO is a neutral ligand. So NO+CN- gives 0 overall. This means Fe must be in +2 oxidation state. For \(\left[\mathrm{V}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{3-}\), we get oxidation state of V as +3.

Examine the electronic configurations

Write down the electronic configurations of the central atoms. Iron in the +2 oxidation state (Fe^2+) has a d6 configuration, and vanadium in the +3 oxidation state (V^3+) has a d2 configuration.

Significance of NO as a Ligand

The NO group can act as a ligand in two different ways: as NO^+ (with no electrons to donate, but can accept electrons into its \( \pi^* \) orbitals), or as NO^- (with two electrons to donate). The CN^- is a strong field ligand which would be expected to pair electrons, allowing the vacant \(d\) orbitals to be available.

Compare the extent of \(\pi\)-bonding

Fe^2+, having a d6 configuration, has more vacant \(d\) orbitals as compared to V^3+ (d2). Therefore, it is capable of accepting more electron density from NO via \( \pi \)-bonding. Thus, \(\left[\mathrm{Fe}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{2-}\) has greater extent of pi-bonding than \(\left[\mathrm{V}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{3-}\).

Key concepts

Oxidation States of Metal Ions

Understanding the oxidation state of metal ions is crucial for analyzing the bonding and properties of coordination complexes. The oxidation state indicates the degree of oxidation of an atom within a compound, represented by a numerical value that contributes to the compound's overall electric charge. In our exercise, we calculate the oxidation states for iron and vanadium in complex ions by balancing the overall charge of the complex with the known charges of the ligands.

For instance, in the iron complex \(\left[\mathrm{Fe}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{2-}\), cyanide (CN-) acts as a strong field ligand inducing electron pairing, while nitrosyl (NO) can function as NO^- with a pair of electrons or as NO^+, which has empty \(\pi^*\) orbitals. Since the overall charge of the complex is -2, and cyanide has a -1 charge for each of the five cyanide ions, a balance of charges determines the iron's oxidation state to be +2.

Similarly, vanadium's oxidation state in the \(\left[\mathrm{V}(\mathrm{NO})(\mathrm{CN})_{5}\right]^{3-}\) complex is discerned by considering the overall charge, which results in vanadium being in the +3 oxidation state. The identification of oxidation states serves as a foundational step to predict bonding interactions, such as \(\pi\)-bonding in coordination complexes.

Electronic Configuration of Transition Metals

The electronic configuration of transition metals plays a significant role in the behavior and chemistry of coordination complexes. These metals can exhibit a wide range of oxidation states that affect their electronic structure. In coordination compounds, the d-orbitals of the central transition metal ion can interact with ligand orbitals, facilitating various types of bonding.

For example, with iron in the +2 oxidation state (Fe^2+), we are looking at a d6 configuration, meaning six electrons fill the d orbitals. On the other hand, vanadium in the +3 oxidation state (V^3+) has a d2 configuration with only two electrons. These configurations give insight into the number of electrons available for bonding as well as the number of vacant orbitals.

The electronic configuration determines how the metal ion can bond with ligands. In the case of Fe^2+ (d6), there are more empty orbitals available for bonding compared to V^3+ (d2), which implies a greater capability for Fe^2+ to form \(\pi\)-bonds, especially with ligands capable of \(\pi\)-backbonding like NO.

Ligand Field Theory

Ligand Field Theory (LFT) provides a more advanced understanding of how ligands interact with the central metal ion in a coordination complex. LFT extends the basic ideas of crystal field theory by considering the metal-ligand bonds' covalent nature, especially relevant for understanding \(\pi\)-bonding.

In LFT, the d-orbitals are split into different energy levels in the presence of a ligand field, created by the approach of ligands. Strong field ligands, such as CN^-, cause a larger splitting in the d-orbital energy levels, leading to a low spin configuration through electron pairing. Weak field ligands result in a smaller split and tend to produce high spin configurations where electrons remain unpaired.

The NO ligand is particularly interesting as it can either donate electrons to the metal or accept electrons from the metal into its vacant \(\pi^*\) orbitals, exhibiting versatile bonding modes. The ability of metals with different d-electron configurations to engage in \(\pi\)-backbonding with ligands like NO is influenced greatly by the ligand field created by all the ligands in the complex, which helps explain the variations in \(\pi\)-bonding extent observed in our exercise.