JEE Chemistry 2023 Question with Solution

Given: The species are

Find: The number of paramagnetic species.

1. Determine the electronic configuration and geometry for each species.

In a strong field ligand, it forms a square planar complex. No unpaired electrons, so it is diamagnetic.

In a strong field ligand, it forms a tetrahedral complex. No unpaired electrons, so it is diamagnetic.

In a weak field ligand, it forms a tetrahedral complex. It has 2 unpaired electrons, so it is paramagnetic.

In a strong field ligand, it forms a low-spin octahedral complex. It has 1 unpaired electron, so it is paramagnetic.

It has 1 unpaired electron, so it is paramagnetic.

In a weak field ligand, it forms a high-spin octahedral complex. It has 4 unpaired electrons, so it is paramagnetic.

2. Count the paramagnetic species.

Paramagnetic species are

Therefore, the number of paramagnetic species is $4$.

Given: A set of coordination compounds with different ligands. Find: Which of them contain unpaired electrons.

Use the idea that strong field ligands such as $CN^-$ and $CO$ tend to pair electrons, while weak field ligands such as $Cl^-$ and $H_2O$ often give high-spin configurations.

• $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic.
• $[Ni(CO)_4]$ is diamagnetic.
• $[NiCl_4]^{2-}$ is tetrahedral and paramagnetic.
• $[Fe(CN)_6]^{3-}$ is low spin but still has one unpaired electron, so it is paramagnetic.
• $[Cu(NH_3)_4]^{2+}$ has a $3d^9$ configuration, so one electron remains unpaired.
• $[Fe(H_2O)_6]^{2+}$ is high spin and paramagnetic.

Thus, there are $4$ paramagnetic species.