Frequently Asked Question

To determine the ground state energy term of [Co(ox)3]^4-, we need to follow these steps:
1. Determine the oxidation state of cobalt: In this complex, oxalate (ox) has a -2 charge, and the overall charge is -4. Since there are three oxalate ligands, cobalt must have a +2 oxidation state. Therefore, we are dealing with Co(II).
2. Determine the electronic configuration of Co(II): Co has an atomic number of 27. The electronic configuration of Co is [Ar] 3d7 4s2. Removing two electrons (for Co(II)) gives us [Ar] 3d7.
3. Consider the ligands and their effect on the d-orbitals: Oxalate is a weak field ligand, meaning it does not cause a large splitting of the d-orbitals. This means we can assume a high-spin configuration for Co(II) in this complex.
4. Apply Hund's rule to fill the d-orbitals: In a high-spin configuration, the electrons will fill each of the five d-orbitals individually before pairing up. This gives us 7 electrons distributed with a maximum number of unpaired electrons.
5. Determine the total spin quantum number (S): With 7 unpaired electrons, S = 3.5 (sum of spins of all unpaired electrons, each spin = 1/2)
6. Determine the total orbital angular momentum quantum number (L): This requires considering the shape of the orbitals. For a high-spin d7 complex, the configuration is t2g4 eg3. Since all orbitals are singly occupied, the total orbital angular momentum is zero. This means L = 0.
7. Calculate the total angular momentum quantum number (J): J = L + S, so J = 3.5. The ground state term symbol is written as 2S+1LJ.
Therefore, the ground state energy term of [Co(ox)3]^4- is ^8S3/2.