Problem 47 - 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 \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion has Mn in the +2 oxidation state, with an electron configuration of [Ar] 3d^5. In this strong-field complex, the \(d\) orbitals split into t2g orbitals (lower energy) and eg orbitals (higher energy). The electron configuration is t2g^5 eg^0, with all electrons paired and no unpaired electrons. Therefore, the information about five unpaired electrons is incorrect. The complex is classified as a low-spin complex, and the energy-level diagram depicts three t2g orbitals filled with five paired electrons and two empty eg orbitals.

Step 1: Determine the oxidation state of Mn and its electron configuration

First, we must determine the oxidation state of Mn in the complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\). Since the complex ion has an overall charge of +2 and ammonia (NH3) is a neutral ligand, it means that the oxidation state of Mn is +2. The electronic configuration of Mn is [Ar] 3d^5 4s^2. With an oxidation state of +2, the electron configuration of Mn^2+ is [Ar] 3d^5.

Step 2: Determine the ligand field splitting

Ammonia (NH3) is a strong field ligand, which means it can cause a large splitting in the energy levels of the \(d\) orbitals. In a complex involving a strong-field ligand, such as this one, the \(d\) orbitals split into two groups, with three orbitals at a lower energy (t2g) and two orbitals at a higher energy (eg).

Step 3: Fill the orbitals with electrons

Now, we need to fill the t2g and eg orbitals with the five electrons in the Mn^2+ ion. Since NH3 is a strong-field ligand, it will pair the electrons in the lower energy orbitals (t2g) before occupying the higher energy orbitals (eg). The resulting electron configuration for the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex is: t2g^5 eg^0.

Step 4: Determine the number of unpaired electrons and classify the complex

In the t2g^5 eg^0 configuration, all the t2g orbitals are filled, and there are no unpaired electrons. Therefore, the given information about 5 unpaired electrons in the complex ion is incorrect. Since the electrons are all paired, we can classify the complex as a low-spin complex.

Step 5: Sketch the energy-level diagram and place the electrons

To sketch the energy-level diagram for the given complex ion, we will draw five horizontal lines representing the d-orbitals with their respective energy levels. The lower three lines will represent the t2g orbitals, while the upper two lines will represent the eg orbitals. Then, we will place arrows representing the electrons in the orbitals following the aforementioned configuration. The energy-level diagram for the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion can be summarized as follows: - t2g orbitals: 3 orbitals, 5 electrons (paired) - eg orbitals: 2 orbitals, 0 electrons The complex is a low-spin complex.