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Chapter 21: Problem 66

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

Short Answer

Expert verified
The structure of the complex ion \(\mathrm{PdCl}_{4}^{2-}\) is square planar, with the Pd(II) ion at the center and the 4 Cl- ions surrounding it. This geometry allows all electrons to be paired up, making the complex diamagnetic.

Step by step solution

01

Determine the electron configuration of Pd

Pd is a palladium ion, which is in the 5th period and 10th group of the periodic table. Its atomic number is 46, so its ground state electron configuration is [Kr] 4d^10 5s^0.
02

Calculate the charge of Pd ion in the complex

The complex ion given is \(\mathrm{PdCl}_{4}^{2-}\). The total charge of the complex is -2. Chlorine ions (Cl-) have a charge of -1 each. Since there are 4 Cl- ions in the complex, Pd must have a charge of +2 in order to ensure that the overall charge of the complex ion is -2.
03

Determine the electron configuration of Pd(II)

We now know that the Pd ion has a charge of +2. In order to achieve the +2 charge, Pd must lose 2 electrons from its valence shell. In this case, Pd(II) will have an electron configuration of [Kr] 4d^8 5s^0.
04

Determine the coordination number and structure

A diamagnetic complex has all electrons in the complex ion paired up. We know that for \(\mathrm{PdCl}_{4}^{2-}\), there are 4 Cl- ligands surrounding the Pd(II) ion. Considering that Cl- is a monodentate ligand (coordinating through a single coordinate covalent bond), the coordination number for Pd(II) is 4.
05

Identify geometry

Coordination complexes with a coordination number of 4 typically exhibit one of two geometries: tetrahedral or square planar. Since \(\mathrm{PdCl}_{4}^{2-}\) is diamagnetic and Pd(II) has an electron configuration of [Kr] 4d^8 5s^0, a square planar geometry would allow all electrons to be paired up in the complex ion.
06

Propose structure

Given the information that we have about the complex ion \(\mathrm{PdCl}_{4}^{2-}\) being diamagnetic and having a coordination number of 4, the proposed structure for this complex ion is a square planar geometry with the Pd(II) ion at the center and the 4 Cl- ions surrounding it.

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Most popular questions from this chapter

A compound related to acetylacetone is 1,1,1-trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise 49.) Both Be \(^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa - having the formula \(\mathrm{M}(\mathrm{tfa})_{2}\) . Two isomers are formed for each metal complex. a. The Be \(^{2+}\) complexes are tetrahedral. Draw the two isomers of \(\mathrm{Be}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Cu}(\mathrm{tfa})_{2} ?\)

Acetylacetone, abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac \(^{-},\) as shown below, where \(\mathrm{M}\) is a transition metal: Which of the following complexes are optically active: cis-$\mathrm{Cr}(\mathrm{acac})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2},\( trans-Cr(acac) \)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2},\( and \)\mathrm{Cr}(\mathrm{acac})_{3} ?$

Amino acids can act as ligands toward transition metal ions. The simplest amino acid is glycine $\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right) .$ Draw a structure of the glycinate anion \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right)\) acting as a bidentate ligand. Draw the structural isomers of the square planar complex \(\mathrm{Cu}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}\right)_{2}\)

Draw geometrical isomers of each of the following complex ions. a. $\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}-$ b. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}^{2+}\) c. \(\operatorname{Ir}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\) d. $\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{I}_{2}^{+}$

Qualitatively draw the crystal field splitting of the \(d\) orbitals in a trigonal planar complex ion. (Let the \(z\) axis be perpendicular to the plane of the complex.)
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