Solution:
To determine which of the given ions are diamagnetic, we need to look at their electronic configurations and the nature of the ligands involved. A diamagnetic substance is one that has all of its electrons paired.
Let's analyze each option one by one:
Option A: [CoF6]3−
Here, the central ion is Co3+.
Cobalt's atomic number is 27 , so−Co3+ has 24 electrons.
The electron configuration of Co3+ is given by removing 3 electrons from Co : [Ar]3d6.
Fluoride (F−)is a weak field ligand, meaning it does not cause a large split in the d-orbitals.
In a weak field, the electron configuration remains as it is without pairing up, thus: t2g4eg2.
Since there are unpaired electrons, the ion [CoF6]3− is paramagnetic.
Option B: [Co(NH3)6]3+
The central ion here is also Co3+.
Similarly, Co3+:[Ar]3d6.
Ammonia (NH3) is a strong field ligand, causing a large split in the d -orbitals which leads to pairing of electrons.
In a strong field, the electron configuration can be rearranged to pair up electrons: t2g6eg0.
All electrons are paired.
Therefore, the ion [Co(NH3)6]3+ is diamagnetic.
Option C: [Fe(OH2)6]2+
The central ion is Fe2+.
Iron's atomic number is 26 , so2+Fe2+ has 24 electrons.
The electron configuration of Fe2+ is [Ar]3d6.
Water (H2O) is a weak field ligand, similarly not causing a large split.
In a weak field, the electron configuration remains: t2g4eg2 with unpaired electrons.
Therefore, the ion [Fe(OH2)6]2+ is paramagnetic.
Option D: [Fe(CN)6]4−
For Fe2+ in [Fe(CN)6]4.
The electron configuration of Fe2+ remains [Ar]3d6.
Cyanide (CN−)is a very strong field ligand, causing a large split leading to pairing.
In a strong field situation: t2g6eg0 with paired electrons.
Therefore, the ion [Fe(CN)6]4− is diamagnetic.
Therefore, the diamagnetic ions among the given options are:
Option B: [Co(NH3)6]3+
Option D: [Fe(CN)6]4
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