The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
Question
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:
Solution
The metal ion in question is most likely to be Mn3+. Here's why:
In a high-spin complex, the energy gap between the lower and higher energy sets of d-orbitals (t2g and eg) is small enough that it's easier for electrons to occupy the higher energy orbitals rather than pair up in the lower energy ones. In a low-spin complex, the energy gap is large, so electrons will pair up in the lower energy orbitals before occupying the higher energy ones.
Mn3+ has a d4 configuration. In a high-spin complex, all four of these electrons will be unpaired, each occupying a separate d-orbital. In a low-spin complex, two of these electrons will pair up in the lower energy orbitals, leaving only two unpaired. Thus, the difference in the number of unpaired electrons between the high-spin and low-spin complexes is indeed two.
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