Problem 63 - Chemistry The Central Science

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?

Step-by-Step Solution

Answer

The complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) has a central \(\mathrm{Mn}^{2+}\) ion with 5 electrons in the \(d\) orbitals. The ligand \(\mathrm{NH}_{3}\) is a weak field ligand. In the energy-level diagram, three electrons fill the \(t_{2g}\) set and two electrons fill the \(e_{g}\) set, resulting in five unpaired electrons. The complex is a high-spin complex due to the maximization of unpaired electrons.

Step 1: Electron configuration of the central metal ion

The central metal ion in this complex is \(\mathrm{Mn}\) (Manganese), and we need to find its electron configuration. The atomic number of manganese is 25, which corresponds to an electron configuration of \(\mathrm{1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{5}}\) in its neutral state. However, we are given the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) cation, so we must remove 2 electrons from this configuration. These will be removed from the 4s shell, giving us a new electron configuration of \(\mathrm{1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}}\) for the \(\mathrm{Mn}^{2+}\) ion, with 5 electrons in the \(d\) orbitals.

Step 2: Determine the field of the ligand

The ligand in this complex is \(\mathrm{NH}_{3}\) (Ammonia). Ammonia is considered a weak field ligand, meaning it does not cause significant splitting of the \(d\) orbitals in the energy-level diagram.

Step 3: Sketch the energy-level diagram and place electrons

In an octahedral complex with weak-field ligands, the energy-level diagram for the \(d\) orbitals is split into a lower energy \(t_{2g}\) set consisting of three degenerate orbitals (\(d_{xy}\), \(d_{xz}\), and \(d_{yz}\)) and a higher energy \(e_{g}\) set consisting of two degenerate orbitals (\(d_{x^{2}-y^{2}}\) and \(d_{z^{2}}\)). Since the ligand is weak-field, the energy difference between the \(d\) orbitals is small, and the electrons will fill the orbitals according to Hund's rule (maximizing the number of unpaired electrons with parallel spins). There are 5 electrons in the \(d\) orbitals for the \(\mathrm{Mn}^{2+}\) ion, so they will fill the orbitals in the following manner: - 3 electrons in the \(t_{2g}\) set, one in each orbital - 2 electrons in the \(e_{g}\) set, one in each orbital The energy-level diagram shows five unpaired electrons, as given in the exercise.

Step 4: Determine high-spin or low-spin complex

We filled the \(d\) orbitals by maximizing the number of unpaired electrons, meaning the electrons occupy all orbitals before starting to pair up. This configuration indicates that the complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) is a high-spin complex.