91影视

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

The two ligands, Cl鈦� and CN鈦�, differ in their ligand field strength: Cl鈦� is a weak field ligand and CN鈦� 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鈦� 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 (dz虏 and dx虏-y虏) with two orbitals. 3. The d-electrons from Ni虏鈦� ([Ar] 3d鈦�) 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 Ni虏鈦� 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 NiCl鈧劼测伝 has 2 unpaired electrons.

Since CN鈦� 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 (dz虏 and dx虏-y虏) with two orbitals. 3. The d-electrons from Ni虏鈦� ([Ar] 3d鈦�) will fill the orbitals according to the Aufbau principle (starting from the lowest available energy orbitals). In this scenario, the Ni虏鈦� 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)鈧劼测伝 is diamagnetic.

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