Question

The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

Short answer

The complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains an Mn(II) ion with an electron configuration of \([Ar] 3d^5\). In the octahedral complex, the \(d\) orbitals split into two sets: \(\mathrm{t}_{2\mathrm{g}}\) (lower energy) and \(\mathrm{e}_{\mathrm{g}}\) (higher energy). Due to the weak field ligand ammonia, the splitting energy is small, and electrons occupy the higher energy \(\mathrm{e}_{\mathrm{g}}\) orbitals before pairing up in the lower energy \(\mathrm{t}_{2\mathrm{g}}\) orbitals, resulting in a high-spin configuration: t2g: ↓ ↓ ↓ --- --- --- dxy dyz dxz eg: ↓ ↓ --- --- dx^2-y^2 d(z^2) Thus, the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is a high-spin complex.

Step-by-step solution

Determine the oxidation state of Mn

In the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), the overall charge is +2. Ammonia (\(\mathrm{NH}_{3}\)) is a neutral ligand, which means that the oxidation state of the Mn atom is also +2.

Find the electron configuration of Mn

The atomic number of Mn is 25. In the neutral state, its electron configuration is \([Ar] 3d^5 4s^2\). In the +2 oxidation state, Mn loses two electrons, leaving the electron configuration as \([Ar] 3d^5\).

Apply crystal field theory to split the d orbitals

In an octahedral complex (\(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\)), the crystal field theory predicts that the \(d\) orbitals will split into two sets: \(\mathrm{t}_{2\mathrm{g}}\) (lower energy) and \(\mathrm{e}_{\mathrm{g}}\) (higher energy).

Determine the high-spin or low-spin configuration

Since ammonia is a weak field ligand, the splitting energy (\(\Delta\)) between \(\mathrm{t}_{2\mathrm{g}}\) and \(\mathrm{e}_{\mathrm{g}}\) orbitals will be relatively small. In this case, the Mn(II) ion will adopt a high-spin configuration, where electrons will occupy the higher energy \(\mathrm{e}_{\mathrm{g}}\) orbitals before pairing up in the lower energy \(\mathrm{t}_{2\mathrm{g}}\) orbitals.

Place electrons in the orbitals

The energy-level diagram for the d orbitals in the Mn(II) ion is shown below: t2g: ↓ ↓ ↓ --- --- --- dxy dyz dxz eg: ↓ ↓ --- --- dx^2-y^2 d(z^2) As we can see, there are five unpaired electrons and the ion adopts a high-spin configuration. Therefore, the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is a high-spin complex.

Key concepts

High-spin complexes

High-spin complexes occur when the energy difference, known as the crystal field splitting energy (\(\Delta\)), between different sets of d orbitals in a metal center is small. This situation allows electrons to spread out into higher energy orbitals rather than pairing up in lower ones.

In the case of the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex, the presence of ammonia (\(\mathrm{NH}_{3}\)), a weak field ligand, results in a small crystal field splitting. The electrons prefer to remain unpaired and occupy different orbitals (\(\mathrm{t}_{2\mathrm{g}}\)and \(\mathrm{e}_{\mathrm{g}}\)). This ensures a maximum number of unpaired electrons and leads to a high-spin situation.

Overall, high-spin complexes are generally more colorful and less stable than their low-spin counterparts due to increased electron exchange energy.

Electron configuration

Electron configuration refers to the distribution of electrons in the orbitals of an atom or ion. For transition metals, these configurations primarily involve the d orbitals.

Manganese (Mn), for instance, has an atomic number of 25, giving it the electron configuration \([Ar] 3d^5 4s^2\). When Mn is in a +2 oxidation state, it loses two electrons, both of which are removed from the 4s orbital, resulting in the electron configuration \([Ar] 3d^5\).

This particular electron configuration (\(3d^5\)) aligns with the half-filled d-orbital rule, which often results in increased stability for the ion.

Oxidation states

The oxidation state of a metal within a complex indicates the charge of the metal, after accounting for the electrons lost and the impact of the ligands. In coordination chemistry, calculating the oxidation state is crucial for understanding the electronic structure of the complex.

In the complex\(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), manganese has an oxidation state of +2. This is deduced because ammonia is neutral, contributing no charge. Thus, the +2 charge of the entire complex is solely due to the Mn center.

Having a precise understanding of oxidation states helps in predicting the behavior of the metal center, such as spin states and reactivity.

Ligands in coordination chemistry

Ligands are ions or molecules that donate electron pairs to a central metal atom to form a complex. Their nature is critical in determining the properties of the complex.

In\(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), ammonia (\(\mathrm{NH}_{3}\)) acts as a ligand. It is a lone pair donor and creates coordinate covalent bonds with the metal center. Ammonia is known as a weak field ligand, often leading to high-spin configurations in metal complexes because it causes a small splitting of d orbitals.

Understanding the role of different ligands allows chemists to tailor the design of complexes for specific applications, from catalysis to magnetic materials.