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Chapter 21: Problem 60

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

Short Answer

Expert verified
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

01

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.
02

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): \([Kr] 4d^8\)
03

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.
04

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:
05

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.

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Most popular questions from this chapter

Figure \(21.17\) shows that the cis isomer of \(\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}^{+}\) is optically active while the trans isomer is not optically active. Is the same true for \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+} ?\) Explain.

Consider the following data: $$\begin{aligned} \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+} & & \mathscr{E}^{\circ}=1.82 \mathrm{~V} \\ \mathrm{Co}^{2+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+} & K &=1.5 \times 10^{12} \\ \mathrm{Co}^{3+}+3 \mathrm{en} \longrightarrow \mathrm{Co}(\mathrm{en}){ }^{3+} & K &=2.0 \times 10^{47} \end{aligned}$$ where en \(=\) ethylenediamine. a. Calculate \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{Co}(\mathrm{en})_{3}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{en})_{3}^{2+}$$ b. Based on your answer to part a, which is the stronger oxidizing agent, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ? c. Use the crystal field model to rationalize the result in part b.

Give formulas for the following. a. potassium tetrachlorocobaltate(II) b. aquatricarbonylplatinum(II) bromide c. sodium dicyanobis(oxalato)ferrate(III) d. triamminechloroethylenediaminechromium(III) iodide

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that serve as the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}{ }^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood decreases. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (Hint: \(\mathrm{CO}_{2}\) reacts with water to produce carbonic acid.) c. When a person has suffered a cardiac arrest, an injection of a sodium bicarbonate solution is given. Why is this step necessary?

BAL is a chelating agent used in treating heavy metal poisoning. It acts as a bidentate ligand. What type of linkage isomers are possible when BAL coordinates to a metal ion?
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