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