Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
An octahedral complex of a ion has with a weak ligand and with a strong ligand. The pairing energy of the ion is . Decide the spin state for each ligand, and find the number of unpaired electrons and the CFSE for the strong ligand.
In an octahedral field the set lies at and the set at . A d electron count of 4 to 7 gives high spin if and low spin if . In a tetrahedral field the set lies at and the set at , with . in an octahedron.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The weak ligand. is below . The fourth electron goes to the set. The arrangement is , high spin, with 4 unpaired electrons. Its CFSE is .
Step 2. The strong ligand. is above . The fourth electron pairs up in . The arrangement is , low spin, with 2 unpaired electrons.
Step 3. Its CFSE. .
Answer: the weak ligand gives 4 unpaired electrons and the strong ligand gives 2, with a CFSE of .
Watch out. The low spin complex is not stabilised by the splitting alone. Two electrons have paired, which costs each time.
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