Question

Draw the \(d\) -orbital splitting diagrams for the octahedral complex ions of each of the following. a. \(\mathrm{Zn}^{2+}\) b. \(\mathrm{Co}^{2+}\) (high and low spin) c. \(\mathrm{Ti}^{3+}\)

Short answer

The d-orbital splitting diagrams for the octahedral complex ions are as follows: a. Zn²⁺: No splitting occurs as all d-orbitals are completely filled (d¹⁰ configuration). b. Co²⁺: - High spin: 4 orbitals with one electron and 1 orbital with two electrons. - Low spin: 3 orbitals with one electron and 1 orbital with three electrons. c. Ti³⁺: One electron in the lowest energy d-orbital and four empty d-orbitals.

Step-by-step solution

a. Zn²⁺ d-orbital Splitting Diagram

Step 1: Determine the electron configuration of Zn²⁺. Zinc has an atomic number of 30, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2\] In the Zn²⁺ ion, it loses two electrons from the 4s orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\] Step 2: Draw the d-orbital splitting diagram. Zn²⁺ ion is a d¹⁰ metal, meaning that all five d-orbitals are completely filled. Since there is no interaction between the d-electrons and ligands, there is no splitting in the Zn²⁺ d-orbitals.

b. Co²⁺ (High and Low Spin) d-Orbital Splitting Diagram

Step 1: Determine the electron configuration of Co²⁺. Cobalt has an atomic number of 27, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\] In the Co²⁺ ion, it loses two electrons, one from the 4s and one from the 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\] Step 2: Draw the d-orbital splitting diagram for high spin. In a high-spin complex, the crystal field splitting energy (Δ) is small, and the electrons will fill all five d-orbitals before any of the orbitals are doubly occupied. The diagram will show 4 orbitals with one electron and 1 orbital with two electrons. Step 3: Draw the d-orbital splitting diagram for low spin. In a low-spin complex, the crystal field splitting energy (Δ) is large, and the electrons will first completely fill the lower energy d-orbitals before moving to the higher energy orbitals. The diagram will show 3 orbitals with one electron and 1 orbital with three electrons.

c. Ti³⁺ d-Orbital Splitting Diagram

Step 1: Determine the electron configuration of Ti³⁺. Titanium has an atomic number of 22, so its ground state electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2\] In the Ti³⁺ ion, it loses two electrons from the 4s and one from the 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\] Step 2: Draw the d-orbital splitting diagram for Ti³⁺. The Ti³⁺ ion is a d¹ metal, meaning it has only one unpaired electron in its d-orbitals. The octahedral splitting diagram will show one electron in the lowest energy d-orbital, and the other four orbitals will be empty.