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Which theoretical magnetic moment would be used to calculate the magnetic moment off [co(ox)3]^4-
Accepted Answer
The theoretical magnetic moment for [Co(ox)3]4- can be calculated using the spin-only formula: μ = √(n(n+2)) BM, where n is the number of unpaired electrons. To determine n, you need to know the electronic configuration of the cobalt ion in the complex. Here's how to approach it:
1. Determine the oxidation state of cobalt: In [Co(ox)3]4-, oxalate (ox) has a -2 charge, and there are three of them, totaling -6. Since the overall charge is -4, cobalt must have an oxidation state of +2 (Co²⁺).
2. Determine the electronic configuration of Co²⁺: The electronic configuration of Co is [Ar] 3d⁷ 4s². When cobalt loses two electrons to form Co²⁺, the configuration becomes [Ar] 3d⁷.
3. Consider the crystal field splitting: In an octahedral complex like [Co(ox)3]4-, the d-orbitals split into two sets: the lower energy t2g orbitals and the higher energy eg orbitals. The oxalate ligands are weak field ligands, meaning they don't cause a large splitting. Therefore, the electrons will likely fill the t2g orbitals before pairing up in them.
4. Determine the number of unpaired electrons: With 7 electrons in the d orbitals, the t2g orbitals will be filled with 5 electrons, leaving 2 unpaired electrons in the eg orbitals.
5. Calculate the magnetic moment: Using the spin-only formula with n = 2: μ = √(2(2+2)) BM = √8 BM ≈ 2.83 BM.
Therefore, the theoretical magnetic moment for [Co(ox)3]4- would be approximately 2.83 Bohr magnetons (BM).
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