The complex ion NiCl \(_{4}^{2-}\) has two unpaired electrons, whereas \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) is diamagnetic. Propose structures for these two complex ions.

First, we need to find out the electron configuration of the central metal ion, Ni²⁺. The Nickel (Ni) atom has 28 electrons, but since it loses 2 electrons to form Ni²⁺, it now has 26 electrons. The electron configuration for Nickel (Ni) is [Ar] 3d⁸4s², therefore for Ni²⁺ it becomes [Ar] 3d⁸ (since 2 electrons are removed from 4s shell).

The two ligands, Cl⁻ and CN⁻, differ in their ligand field strength: Cl⁻ is a weak field ligand and CN⁻ is a strong field ligand. Weak field ligands tend to not create a large energy gap between the d-orbitals, while strong field ligands create a large split between the d-orbitals.

Since Cl⁻ is a weak field ligand, the crystal field splitting diagram for a tetrahedral complex would look like this: 1. There would be no splitting in the 4s and 4p orbitals. 2. The d-orbitals will be split into a lower energy set (dxy, dxz, and dyz) containing three orbitals and a higher energy set (dz² and dx²-y²) with two orbitals. 3. The d-electrons from Ni²⁺ ([Ar] 3d⁸) will fill up the orbitals according to Hund's rule (first fill the lower energy orbitals, then the higher energy orbitals with a maximum of two unpaired electrons). In this scenario, the Ni²⁺ ion has 8 d-electrons: 6 will fill the lower energy orbitals, and 2 will fill the higher energy orbitals, leaving 2 unpaired electrons. This aligns with the fact that NiCl₄²⁻ has 2 unpaired electrons.

Since CN⁻ is a strong field ligand, the crystal field splitting diagram for a tetrahedral complex would look like this: 1. There would be no splitting in the 4s and 4p orbitals. 2. The d-orbitals will be split into a lower energy set (dxy, dxz, and dyz) containing three orbitals and a higher energy set (dz² and dx²-y²) with two orbitals. 3. The d-electrons from Ni²⁺ ([Ar] 3d⁸) will fill the orbitals according to the Aufbau principle (starting from the lowest available energy orbitals). In this scenario, the Ni²⁺ ion has 8 d-electrons: all 8 electrons will fill the lower energy orbitals, leaving no unpaired electrons. This aligns with the fact that Ni(CN)₄²⁻ is diamagnetic.

Now that we know how the d-electrons are filled in each complex ion, we can propose structures for both. NiCl₄²⁻: - Central atom: Ni²⁺ - Ligands: 4 Cl⁻ ions - Geometry: Tetrahedral - Bonding: Ni²⁺ will form covalent bonds with the Cl⁻ ions through the filled d-orbitals. Ni(CN)₄²⁻: - Central atom: Ni²⁺ - Ligands: 4 CN⁻ ions - Geometry: Tetrahedral - Bonding: Ni²⁺ will form covalent bonds with the CN⁻ ions through the filled d-orbitals. Thus, the structures of NiCl₄²⁻ and Ni(CN)₄²⁻ are both tetrahedral, with the difference being the ligands and how the d-electrons are filled in each case. NiCl₄²⁻ has two unpaired electrons, while Ni(CN)₄²⁻ is diamagnetic with no unpaired electrons.