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 \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion has an octahedral coordination geometry due to the 6 ammonia ligands. The weak field ligands lead to minimal splitting of the \(d\) orbitals. The 5 \(d\) electrons will singly occupy the orbitals in the following manner: \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{z^2}\), and \(d_{x^2-y^2}\). This results in a high-spin complex with 5 unpaired electrons.

Step-by-step solution

Analyzing the given complex

The given complex is \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) which contains Mn (Manganese) as the central metal ion and 6 NH\(_3\) (Ammonia) ligands surrounding it.

Determine the coordination geometry

Since there are six ligands surrounding the metal ion, the coordination geometry is octahedral.

Apply the crystal field theory and determine the energy-level splitting

In an octahedral field, the \(d\) orbitals split into two sets: - \(t_{2g}\) orbitals (lower energy): \(d_{xy}\), \(d_{yz}\), and \(d_{xz}\) - \(e_{g}\) orbitals (higher energy): \(d_{z^2}\) and \(d_{x^2-y^2}\) Ammonia (NH\(_3\)) is a weak field ligand, which means it will not split the \(d\) orbitals significantly.

Add electrons to the energy levels accordingly

Mn\(^{2+}\) has 5 \(d\) electrons. Since ammonia is a weak field ligand and does not cause significant splitting, the electrons will first occupy the lower energy \(t_{2g}\) orbitals, and then singly occupy the higher energy \(e_{g}\) orbitals, following Hund's rule. The energy level placement will look like this: - \(d_{xy}\): 1 electron - \(d_{yz}\): 1 electron - \(d_{xz}\): 1 electron - \(d_{z^2}\): 1 electron - \(d_{x^2-y^2}\): 1 electron

Determine if the complex is a high-spin or low-spin complex

As the electrons are maximumly singly occupied in the higher energy orbitals, the complex is a high-spin complex. The energy-level diagram for the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion has 5 unpaired electrons in the \(d\) orbitals, making it a high-spin complex.