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The magnetic moment of [fe(nh3)6](no3)2
Accepted Answer
To determine the magnetic moment of [Fe(NH3)6](NO3)2, we need to consider the electronic configuration of the Fe(II) ion.
Fe(II) has 3d6 configuration, which is a high spin complex due to the strong field ligand NH3. This results in 4 unpaired electrons.
Using the spin-only formula: μ = √n(n+2) BM, where n is the number of unpaired electrons, we get:
μ = √4(4+2) BM = √24 BM ≈ 4.89 BM.
Therefore, the magnetic moment of [Fe(NH3)6](NO3)2 is approximately 4.89 Bohr magnetons.
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