Question

A metal ion in a high-spin octahedral complex has two more unpaired electrons than the same ion does in a low-spin octahedral complex. Name some possible metal ions for which this would be true.

Short answer

Possible metal ions for which high-spin octahedral complexes have two more unpaired electrons than low-spin octahedral complexes are Cr(III) with a d4 electron configuration and Fe(II) with a d6 electron configuration.

Step-by-step solution

Understanding high-spin and low-spin complexes

High-spin and low-spin complexes refer to the electronic configuration of metal ions in coordination complexes. The difference between them lies in the pairing of electrons in the d orbitals of the metal ion. For high-spin complexes, the electrons preferentially occupy different d orbitals, maximizing unpaired electrons. On the other hand, in low-spin complexes, the electrons are more likely to pair up within the same d orbital, minimizing unpaired electrons. This distinction occurs due to the difference in ligands' crystal field splitting energy (CFSE).

Determining electron configuration

In an octahedral complex, the crystal field splitting causes the energy levels to split into two sets of orbitals: $t_{2g}$ (lower energy) and $e_g$ (higher energy). In low-spin complexes, the CFSE is large, meaning the energy difference between the two sets of orbitals is significant, and it is more favorable for electrons to pair up in the $t_{2g}$ orbitals before occupying the higher energy $e_g$ orbitals. In high-spin complexes, the CFSE is small, and electrons will occupy all the $t_{2g}$ orbitals with parallel spins before pairing up in the same orbital or occupying the $e_g$ orbitals.

Identifying possible metal ions

To find possible metal ions for which high-spin octahedral complexes have two more unpaired electrons than low-spin octahedral complexes, we will examine the d-electron configurations of various metal ions and determine if their electron configurations meet the criteria. 1. d4: In a high-spin octahedral complex, all four electrons would occupy the $t_{2g}$ orbitals (3 unpaired electrons). In a low-spin octahedral complex, two electrons would pair up in the $t_{2g}$ orbitals, leaving two unpaired electrons. 2. d6: In a high-spin octahedral complex, the 6 electrons would occupy all the $t_{2g}$ orbitals and one of the $e_g$ orbitals (4 unpaired electrons). In a low-spin octahedral complex, the $t_{2g}$ orbitals would be fully occupied by paired electrons, leaving no unpaired electrons. From the d4 and d6 configurations, we can see that they meet the criteria outlined in the exercise, having two more unpaired electrons in their high-spin octahedral complexes than in their low-spin counterparts.

Possible metal ions

Considering the electron configurations that meet the given criteria, the following metal ions can be used: 1. Cr(III) (d4 electron configuration) 2. Fe(II) (d6 electron configuration) These metal ions would exhibit the desired difference in unpaired electron count when forming high-spin and low-spin octahedral complexes.