Question

Explain why \(\mathrm{Mn}^{3+}\) forms high-spin and low-spin octahedral complexes but \(\mathrm{Mn}^{4+}\) does not.

Short answer

Answer: Mn^3+ can form high-spin and low-spin complexes because its electron configuration allows for the population of both t_2g and e_g orbitals, depending on the ligand's strength and crystal field splitting. Mn^4+ cannot form distinct high-spin and low-spin complexes because its electron configuration remains the same in both cases, with the three electrons occupying the lower energy t_2g orbitals.

Step-by-step solution

Determine the electron configurations of Mn^3+ and Mn^4+

Manganese is an element with the atomic number 25, belonging to the first row of the transition metals. Its ground-state electron configuration is [Ar] 4s^2 3d^5. Now, let's find out the electron configurations of Mn^3+ and Mn^4+ ions. Mn^3+: When Mn loses 3 electrons, its electron configuration becomes [Ar] 3d^4. Mn^4+: When Mn loses 4 electrons, its electron configuration becomes [Ar] 3d^3.

Understand crystal field theory and crystal field splitting in octahedral complexes.

According to the crystal field theory in coordination chemistry, ligands exert an electrostatic field on a central metal ion, influencing its d-orbitals differently. In an octahedral complex, d-orbitals are split into two groups with different energy levels: the lower group contains three orbitals (d_{xy}, d_{yz}, and d_{xz}) known as t_2g orbitals, and the higher group contains two orbitals (d_{z^2} and d_{x^2 - y^2}) known as e_g orbitals.

Explain high-spin and low-spin complexes

High-spin and low-spin refer to the arrangement of electrons in the d-orbitals when crystal field splitting occurs. For high-spin complexes, electrons fill each of the d-orbitals singly before pairing up (following Hund's rule). Low-spin complexes only occur when the energy difference between the two groups of d-orbitals is large enough (large Δ) that pairing of electrons in the lower energy orbitals becomes favorable. This results in less unpaired electrons and lower overall magnetic moments.

Analyze the electron configurations of Mn^3+ and Mn^4+ in octahedral complexes

Let's analyze the electron configurations of Mn^3+ and Mn^4+ in octahedral complexes based on the crystal field split. Mn^3+ ([Ar] 3d^4): In a high-spin Mn^3+ complex, the 4 electrons will occupy the t_2g and e_g levels as follows: t_2g: ↑ ↑ ↑ e_g: ↑ In a low-spin Mn^3+ complex, the 4 electrons will be paired in the lower energy t_2g orbitals: t_2g: ↑↓ ↑↓ e_g: Mn^4+ ([Ar] 3d^3): In both high-spin and low-spin Mn^4+ complexes, the 3 electrons will be located in the lower energy t_2g orbitals with no electrons in the e_g orbitals: t_2g: ↑ ↑ ↑ e_g:

Conclude why Mn^3+ forms high-spin and low-spin complexes but Mn^4+ does not

The Mn^3+ ion has the flexibility to form both high-spin and low-spin octahedral complexes because its electron configuration allows for the population of both t_2g and e_g orbitals, depending on the ligand's strength and Δ. However, the Mn^4+ ion always has the same configuration for both high-spin and low-spin situations (3 electrons in the t_2g orbitals). It does not change its electron configuration, and hence, it can't form distinct high-spin and low-spin octahedral complexes. In conclusion, the reason Mn^3+ forms both high-spin and low-spin octahedral complexes, while Mn^4+ does not, is due to the electron configurations and the crystal field splitting in octahedral geometry. The flexibility in electron configuration for Mn^3+ allows for the formation of distinct high-spin and low-spin complexes, whereas the electron configuration for Mn^4+ remains the same in both cases.