Question 22.4: Determine which of these ions can have high-spin and low-spi...
Determine which of these ions can have high-spin and low-spin configurations when part of an octahedral complex: (a) Mn^{4+}; (b) Mn^{2+}; (c) Cu^{2+} .
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Collect and Organize Mn and Cu are in groups 7 and 11 of the periodic table, so their atoms have 7 and 11 valence electrons, respectively. In an octahedral field, a set of five d orbitals splits into a low-energy subset of three orbitals and a high energy subset of two orbitals.
Analyze To determine whether high-spin and low-spin states are possible, we need to determine the number of d electrons in each ion. Then we need to distribute them among sets of d orbitals split by an octahedral field to see whether it is possible for the ions to have different spin states. The electron configurations of the atoms of the three elements are [Ar]3d^{5}4s^{2} for Mn and [Ar]3d^{ 10}4s^{1} for Cu. When they form cations, the atoms of these transition metals lose their 4s electrons first and then their 3d electrons.
Therefore the numbers of d electrons in the ions are 3 in Mn^{4+}, 5 in Mn^{2+}, and 9 in Cu^{2+}. It is likely that the ion with the fewest d electrons (Mn^{4+}) and the one with nearly the most d electrons possible (Cu^{2+}) will each have only one spin state.
Solve
a. Mn^{4+}: Following Hund’s rule and putting three electrons into the lowest-energy 3d orbitals available and keeping them as unpaired as possible gives this orbital distribution of electrons (Figure 22.19a). There is only one spin state for Mn^{4+}.
b. Mn^{2+}: There are two options for distributing five electrons among the five 3d orbitals (Figure 22.19b). Thus Mn^{2+} can have a high-spin (on the left) or a low-spin (on the right) configuration in an octahedral field.
c. (Cu^{2+}): The nine 3d electrons fill the lower-energy orbitals and nearly fill the higher-energy ones. Only one arrangement is possible, so (Cu^{2+}) has only one spin state (Figure 22.19c).
Think About It In an octahedral field, metal ions with 4, 5, 6, or 7 d electrons can exist in high-spin and low-spin states. Those ions with 3 or fewer d electrons have only one spin state, in which all the electrons are unpaired and in the lower energy set of orbitals.
Ions with 8 or more d electrons have only one spin state because their lower-energy set of orbitals is filled. The magnitude of the crystal field splitting energy, \Delta_{o}, determines which spin state an ion with 4, 5, 6, or 7 d electrons occupies.
