Question

The complex ion \(\mathrm{PdCl}_{4}{ }^{2-}\) is diamagnetic. Propose a structure for \(\mathrm{PdCl}_{4}^{2-}\)

Short answer

The proposed structure for the \(\mathrm{PdCl}_{4}^{2-}\) complex ion is square-planar with the Pd(II) ion as the central atom and four Cl^- ions as ligands. This structure is diamagnetic with no unpaired electrons present.

Step-by-step solution

Identify the oxidation state

We have \(\mathrm{PdCl}_{4}^{2-}\) as the complex ion. Each chlorine atom has a -1 charge, and there are four chlorine atoms. Therefore, the central Pd ion has a charge of +2. So, Pd is in the oxidation state of +2.

Write the electron configuration of Pd and Pd(II)

Write the electron configuration of the central atom, Pd (atomic number 46): Pd: \([Kr] 4d^{10} 5s^0\) Now, consider the Pd(II) configuration after losing two electrons: Pd(II): \([Kr] 4d^8\)

Determine the coordination number and geometry

The coordination number is the number of ligands (Cl^- in this case) attached to the central atom (Pd). Since we have four Cl^- ions acting as ligands, the coordination number is 4. For a coordination number of 4, there are two possible geometries: tetrahedral and square planar.

Determine whether the complex is tetrahedral or square planar

To determine the geometry for the \(\mathrm{Pd^{2+}}\) ion, we need to check if the ion follows the 18-electron rule. For \(\mathrm{Pd^{2+}}\), we have: \([Kr] 4d^8\) By adding the four lone pairs from the Cl^- ions, we get the total number of valence electrons as follows: \(8 + 4(2) = 16\) Since the 18-electron rule is not satisfied, it is unlikely to form a tetrahedral complex. Therefore, the geometry is square-planar in which, the Pd(II) ion has a filled d-orbital, resulting in a diamagnetic complex. Conclusion:

Proposed structure

The proposed structure for the \(\mathrm{PdCl}_{4}^{2-}\) complex ion is square-planar with the Pd(II) ion as the central atom and four Cl^- ions as ligands. This structure is diamagnetic with no unpaired electrons present.