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.

In octahedral complexes, the d orbitals are split into two groups with different energy levels: the lower-energy t2g orbitals (dxy, dxz, and dyz), and the higher-energy eg orbitals (dx^2-y^2 and dz^2). High-spin complexes occur when the energy gap between these orbital sets is small, and electrons prefer to occupy all five d orbitals with parallel spins before pairing up. In contrast, low-spin complexes form when the energy gap is large, and electrons pair up in the lower-energy t2g orbitals before occupying the higher-energy eg orbitals.

We are looking for metal ions that have two more unpaired electrons in a high-spin octahedral complex than in a low-spin octahedral complex. Here are some possibilities: 1. M^3+ metal ions: Electron configuration \(t_{2g}^3\). 2. M^4+ metal ions: Electron configuration \(t_{2g}^2\).

Now, let's find metal ions that meet our criteria. Remember that we are looking for metal ions that have two more unpaired electrons in a high-spin octahedral complex than in a low-spin octahedral complex. 1. For M^3+ metal ions, the high-spin electron configuration is \(t_{2g}^3 e_g^0\), with three unpaired electrons. In the low-spin configuration, it is \(t_{2g}^6 e_g^0\), with only one unpaired electron. Examples of M^3+ metal ions include \(Fe^{3+}\) and \(Co^{3+}\). 2. For M^4+ metal ions, the high-spin electron configuration is \(t_{2g}^2 e_g^0\), with two unpaired electrons. In the low-spin configuration, it is \(t_{2g}^4 e_g^0\), with no unpaired electrons. Examples of M^4+ metal ions include \(Cr^{4+}\) and \(Mn^{4+}\). In conclusion, some possible metal ions that have two more unpaired electrons in high-spin octahedral complexes than in low-spin octahedral complexes are \(Fe^{3+}\), \(Co^{3+}\), \(Cr^{4+}\), and \(Mn^{4+}\).