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. Which of the following could the metal ion be? A) Cr³⁺ B) Mn²⁺ C) Co²⁺ D) Ti²⁺ E) Cu²⁺

Question

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In high-spin octahedral complexes, the pairing energy required to pair up two electrons into one orbital is greater than the energy difference between the eg and t2g orbitals. As such, electrons are placed singly in each eg orbital before any pairing occurs in the t2g orbitals. This results in the maximum number of unpaired electrons possible for that metal ion.

In low-spin octahedral complexes, the opposite is true. The energy required to pair up two electrons is less than the energy difference between the eg and t2g orbitals. Therefore, electrons are paired up before being placed singly in the eg orbitals, resulting in a lower number of unpaired electrons.

Looking at the answer choices, Co is the only metal ion that can accommodate two more unpaired electrons in high-spin rather than low-spin octahedral complexes. Its d-electron configuration is d7, with three electrons in the t2g orbitals and two in each of the eg orbitals. In low-spin octahedral complexes, these two electrons are paired up in one of the eg orbitals, resulting in only one unpaired electron. In high-spin octahedral complexes, however, each eg orbital has only one electron, resulting in a total of four unpaired electrons.

A Metal Ion in a High-Spin Octahedral Complex Has Two