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 energy-level diagram for the \(d\) orbitals of the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) has all five electrons in the \(t_{2g}\) orbitals and no electrons in the \(e_g\) orbitals. The complex ion is a low-spin complex due to the presence of a strong field ligand (\(NH_3\)).

Step-by-step solution

Determine the electron configuration of the Mn2+ ion

First, we need to find the electron configuration of Mn in its 2+ oxidation state. Manganese, being the 25th element in the periodic table, has the electron configuration \([Ar] 3d^5 4s^2\). Losing two electrons will give Mn2+ the electron configuration \([Ar] 3d^5\). Thus, there are five 3d electrons to consider when placing them in the energy-level diagram.

Understand the energy levels and placement of d-orbitals in octahedral complexes

In octahedral complexes, the five \(d\) orbitals are split into two different energy levels due to the crystal field. The three orbitals \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) have the same energy level, while the other two orbitals, \(d_{z^2}\) and \(d_{x^2-y^2}\), have a higher energy level. The energy-level diagram would look like this in simple terms: ``` _______ (eg) (eg) | d_{z^2}, d_{x^2-y^2} _______ (t2g) (t2g) | d_{xy}, d_{xz}, d_{yz} ```

Identify the ligand and its effect on the crystal field.

The ligand in our complex ion is \(NH_3\), which is a strong field ligand. In the presence of a strong field ligand, the energy gap between the \(t_{2g}\) and \(e_g\) orbitals is large, which results in a low-spin complex where electrons preferentially fill the lower-energy \(t_{2g}\) orbitals before occupying the higher-energy \(e_g\) orbitals.

Place electrons in the energy-level diagram

Since we determined that our complex ion is a low-spin complex, we follow the rules of Hund to put the electrons in the \(d\) orbitals. ``` _______ (eg) (eg) | Empty _______ (t2g) (t2g) | 1↑↓, 1↑↓, 1↑0 ```

Conclusion

The energy-level diagram for the \(d\) orbitals of the complex ion \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) has all five electrons in the \(t_{2g}\) orbitals, and no electrons in the \(e_g\) orbitals. The complex ion is a low-spin complex due to the presence of a strong field ligand (\(NH_3\)).