WBJEE 2024 Physics & Chemistry Paper

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: $[\mathrm{CoF}_6]^{3-}$Here, the central ion is $\mathrm{Co}^{3+}$.Cobalt's atomic number is 27 , $\text{so}^{-} \mathrm{Co}^{3+}$ has 24 electrons.The electron configuration of $\mathrm{Co}^{3+}$ is given by removing 3 electrons from Co : $[\mathrm{Ar}]\,3d^6$.Fluoride $(\mathrm{F}^{-})$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: $t_{2g}^4 e_g^2$.Since there are unpaired electrons, the ion $[\mathrm{CoF}_6]^{3-}$ is paramagnetic. Option B: $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$The central ion here is also $\mathrm{Co}^{3+}$.Similarly, $\mathrm{Co}^{3+}: [\mathrm{Ar}]\,3d^6$.Ammonia $(\mathrm{NH}_3)$ 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: $t_{2g}^6 e_g^0$.All electrons are paired.Therefore, the ion $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$ is diamagnetic. Option C: $[\mathrm{Fe}(\mathrm{OH}_2)_6]^{2+}$The central ion is $\mathrm{Fe}^{2+}$.Iron's atomic number is 26 , $\text{so}^{2+} \mathrm{Fe}^{2+}$ has 24 electrons.The electron configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}]\,3d^6$.Water $(\mathrm{H}_2\mathrm{O})$ is a weak field ligand, similarly not causing a large split.In a weak field, the electron configuration remains: $t_{2g}^4 e_g^2$ with unpaired electrons.Therefore, the ion $[\mathrm{Fe}(\mathrm{OH}_2)_6]^{2+}$ is paramagnetic. Option D: $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$For $\mathrm{Fe}^{2+}$ in $[\mathrm{Fe}(\mathrm{CN})_6]^4$.The electron configuration of $\mathrm{Fe}^{2+}$ remains $[\mathrm{Ar}]\,3d^6$.Cyanide $(\mathrm{CN}^{-})$is a very strong field ligand, causing a large split leading to pairing.In a strong field situation: $t_{2g}^6 e_g^0$ with paired electrons.Therefore, the ion $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ is diamagnetic.Therefore, the diamagnetic ions among the given options are:Option B: $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$Option D: $[\mathrm{Fe}(\mathrm{CN})_6]^4$