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What is the ground state term for [co(ox)3]^4-
Accepted Answer
The ground state term for [Co(ox)3]^4- is ^4A2g. Here's how to determine it:
1. Determine the electronic configuration of the central metal ion: Cobalt(II) ion (Co^2+) has an electronic configuration of [Ar] 3d^7.
2. Consider the ligand field splitting: Oxalate (ox) is a weak-field ligand, causing a small splitting of the d orbitals.
3. Apply Hund's rules: In a weak field, the electrons will occupy the orbitals individually before pairing up.
4. Determine the total spin (S) and orbital angular momentum (L): For Co^2+, there are three unpaired electrons (S = 3/2). The orbital angular momentum L is determined by the sum of the ml values of the unpaired electrons. Since the d orbitals are split into t2g and eg sets, we consider the t2g orbitals (which have ml values of 2, 1, and 0). The total L = 3.
5. Calculate the term symbol: The term symbol is written as ^2S+1LJ, where J = |L - S|, |L - S + 1|, ..., |L + S|. In this case, S = 3/2 and L = 3. J can be 9/2, 7/2, 5/2, and 3/2. The ground state term is the one with the lowest J value, which is ^4A2g (J = 3/2).
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