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Which theoretical magnetic moment would you calculate for [co(ox)3]4-?
Accepted Answer
To calculate the theoretical magnetic moment of [Co(ox)3]4-, we need to consider the electronic configuration of the cobalt(II) ion. The complex has a coordination number of 6 with oxalate ligands, which are weak field ligands. This means the electrons will likely fill the lower energy orbitals first, resulting in a high spin complex.
Cobalt(II) has an electronic configuration of [Ar] 3d7. In a high spin complex, the 3d orbitals are split into two sets: t2g (lower energy) and eg (higher energy). The 7 electrons will fill these orbitals as follows:
t2g: 3 electrons
eg: 4 electrons
Since we have 3 unpaired electrons in the t2g orbitals, we can use the spin-only formula to calculate the theoretical magnetic moment:
μ = √(n(n+2)) BM
Where n is the number of unpaired electrons.
μ = √(3(3+2)) BM = √15 BM ≈ 3.87 BM
Therefore, the theoretical magnetic moment for [Co(ox)3]4- is approximately 3.87 Bohr Magnetons.
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