Final answer:
The [Fe(NH3)6]2+ complex is expected to have four unpaired electrons due to ammonia being a weak field ligand, which results in a high-spin complex. This means the complex will be paramagnetic.
Step-by-step explanation:
In the complex ion [Fe(NH3)6]2+, we can determine the number of unpaired electrons by considering the strength of the ligand field and its effect on the energy levels of the electrons within the d orbitals. Ammonia (NH3) is generally considered a weak field ligand, which does not cause a large crystal field splitting. Therefore, similar to the case of [Fe(H2O)6]2+, where water as a weak field ligand results in a small crystal field splitting (Aoct<P), the electrons prefer to occupy higher energy orbitals (known as eg orbitals) rather than pair up. This leads to a high-spin complex.
For the iron(II) center in [Fe(NH3)6]2+, which has six d electrons, this means that there will be four unpaired electrons as it is more energetically favorable for the electrons to enter the higher orbitals before pairing occurs. Cementing this conclusion, when comparing to other iron complexes, like the low-spin [Fe(CN)6]4- where all electrons are paired due to strong field ligands, the difference in ligand field strength dictates the number of unpaired electrons. Thus, we would expect the [Fe(NH3)6]2+ to have four unpaired electrons and to be paramagnetic.