Problem 4 - Chemistry An Atoms First Approach

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.

Step-by-Step Solution

Answer

Possible metal ions that have two more unpaired electrons in a high-spin octahedral complex than in a low-spin octahedral complex include Manganese (III) ion [Mn(III)] and Iron (II) ion [Fe(II)]. These metal ions have a d5 electron configuration, which satisfies the requirement of having a two-electron difference between high-spin and low-spin configurations.

Step 1: d4 high-spin configuration

In a high-spin d4 configuration, the electrons fill the t2g and eg orbitals as follows: \(t_{2_{g}}^{\uparrow\downarrow \uparrow\uparrow}\) \(e_{g}^{\uparrow\downarrow}\). There are two unpaired electrons.

Step 2: d4 low-spin configuration

In a low-spin d4 configuration, the electrons fill the t2g orbitals mostly due to the large splitting caused by strong-field ligands: \(t_{2_{g}}^{\uparrow\downarrow \uparrow\downarrow}\) \(e_{g}^{\downarrow}\). There is one unpaired electron. As there is only a one-electron difference between high-spin and low-spin d4 configurations, it doesn't satisfy the requirement. #Step 3: Examine d5 configuration#

Step 3: d5 high-spin configuration

In a high-spin d5 configuration, the electrons fill the t2g and eg orbitals as follows: \(t_{2_{g}}^{\uparrow\downarrow \uparrow\uparrow}\) \(e_{g}^{\uparrow\downarrow\uparrow}\). There are three unpaired electrons.

Step 4: d5 low-spin configuration

In a low-spin d5 configuration, the electrons fill the t2g and eg orbitals mostly due to the large splitting caused by strong-field ligands: \(t_{2_{g}}^{\uparrow\downarrow \uparrow\downarrow\uparrow}\) \(e_{g}^{\downarrow}\). There is one unpaired electron. As there is only a two-electron difference between high-spin and low-spin d5 configurations, this satisfies the requirement. #Step 4: Examine d6 and d7 configurations# We can deduce that the d6 and d7 configurations will also not satisfy the requirement as they will have fewer unpaired electrons in high-spin complexes compared to d5 high-spin complexes. Therefore, the d5 configuration is the only one that fulfills the condition. #Step 5: Identify possible metal ions# Now that we have identified the d5 electron configuration, we can name some possible metal ions for which this statement is true. These metal ions include: 1. Manganese (III) ion [Mn(III)] - has d4 configuration in its ground state, and d5 when it loses one more electron 2. Iron (II) ion [Fe(II)] - has d6 configuration in its ground state, and d5 when it loses one electron These metal ions, when part of an octahedral complex, would show a difference of two unpaired electrons between high-spin and low-spin configurations.