Question

Which of the following ions could exist as either a low-spin or a high-spin octahedral complex depending on the crystal field splitting of the ligands? A) Mn2* B) Ni2* C) Sc* D) Cu2+ E) Zn*

Answer 1

In octahedral complexes, the crystal field splitting determines whether the complex is low-spin or high-spin. Weak-field ligands result in high-spin complexes, while strong-field ligands result in low-spin complexes.

Step-by-step explanation:

In octahedral complexes, the crystal field splitting determines whether the complex is low-spin or high-spin. The size of the crystal field splitting affects the arrangement of electrons in the d orbitals.

For example, in weak-field ligands, such as in the case of C) Sc^3+, the crystal field splitting is small and the compound is high-spin. This means that the electrons will occupy the d orbitals with their spins parallel, resulting in more unpaired electrons.

On the other hand, in strong-field ligands, like in the case of D) Cu^2+, the crystal field splitting is large and the compound is low-spin. In this case, the electrons will pair up in the d orbitals, resulting in fewer unpaired electrons.

Answer 2

Mn2+ can exist as either a low-spin or high-spin octahedral complex depending on the crystal field splitting of the ligands, due to its d5 electron configuration that allows for different distributions of its electrons either pairing or remaining unpaired in the orbitals. So the correct option is A.

Step-by-step explanation:

The question asks which ions could exist as either a low-spin or high-spin octahedral complex depending on the crystal field splitting (Δo) of the ligands. These different spin states occur due to the distribution of electrons in the metal's d orbitals. The magnitude of Δo compared to the spin-pairing energy (P) determines whether an electron will pair with another in a lower energy orbital or occupy a higher energy orbital unpaired. Ions with d4, d5, d6, or d7 electron configurations can exhibit these two different spin states.

For instance, with a d4 configuration, if Δo is less than P, the complex will be high-spin with unpaired electrons in the higher energy eg orbitals, as seen in the [Cr(H2O)6]2+ ion. However, if Δo is greater than P, the complex will be low-spin with the electrons paired up in the lower energy t2g orbitals, exemplified by the [Mn(CN)6]3- ion. Among the options, Mn2+ and Ni2+ can be either high-spin or low-spin, but Ni2+ with its d8 configuration typically does not show high-spin octahedral complexes. Therefore, Mn2+ is the correct answer.