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Chapter 23: Problem 74

The square-planar complex $\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_{2}\right]$ only forms in one of two possible geometric isomers. Which isomer is not observed: cis or trans?

Short Answer

Expert verified
The cis isomer is not observed for the square-planar complex \(\left[\mathrm{Pt}(\mathrm{en})\mathrm{Cl}_{2}\right]\) as it would not allow the ethylenediamine ligand to occupy adjacent sites on the metal center due to the placement of chloride ligands. The trans isomer is observed as it allows for a stable square-planar configuration with the platinum center, ethylenediamine, and the two chloride ligands.

Step by step solution

01

Understand the structure of the complex

The complex \(\left[\mathrm{Pt}(\mathrm{en})\mathrm{Cl}_{2}\right]\) contains a platinum(II) metal center (Pt), with ethylenediamine (en) and two chloride ligands (Cl) attached to it. Ethylenediamine is a bidentate ligand, meaning it can attach to the metal center at two points. The square planar configuration implies that all ligands lie in the same plane.
02

Identify cis and trans isomers

In a square-planar complex, the cis isomer has similar ligands adjacent to each other, while the trans isomer has similar ligands opposite each other. For \(\left[\mathrm{Pt}(\mathrm{en})\mathrm{Cl}_{2}\right]\), the cis isomer would have the two chloride ligands next to each other, and the trans isomer would have them diagonally opposite.
03

Determine the stability and existence of the isomers

The ethylenediamine ligand is a chelating compound that can form stable, ring-shaped structures with metal ions. In the complex, the en ligand occupies two adjacent coordination sites on the platinum center due to its bidentate nature. In a cis configuration, the chloride ligands would be on adjacent sites, leaving no space for the en ligand to occupy adjacent sites. Therefore, a cis isomer cannot form. On the other hand, in a trans configuration, the chloride ligands would be opposite each other, leaving two adjacent sites for the en ligand to attach to and form a stable chelate.
04

Identify the unobserved isomer

Based on the above reasoning, the cis isomer cannot form as it would not allow the ethylenediamine ligand to occupy adjacent sites on the metal center. Therefore, the cis isomer is not observed for the given complex. The trans isomer is observed as it allows both the ethylenediamine and the two chloride ligands to form a stable square-planar configuration with the platinum center.

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Most popular questions from this chapter

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2},$, (b) $\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]$, (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\), (d) $[\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4},(\mathbf{e})\left[\mathrm{Cd}\left(\mathrm{H}_{2} ,\mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$.

The complexes \(\left[\mathrm{CrBr}_{6}\right]^{3-}\) and \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) are both known. (a) Draw the \(d\) -orbital energy-level diagram for octahedral \(\mathrm{Cr}(\mathrm{III})\) complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy?

For each of the following molecules or polyatomic ions, draw the Lewis structure and indicate if it can act as a monodentate ligand, a bidentate ligand, or is unlikely to act as a ligand at all: (a) ethylamine, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\), (b) trimethylphosphine, \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\), (c) carbonate, \(\mathrm{CO}_{3}^{2-}\) (d) ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\).

Which periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and \(8 ?(\mathbf{a})\) The number of valence electrons reaches a maximum at group 8. (b) The effective nuclear charge increases on moving left across each period. (c) The radii of the transition-metal elements reach a minimum for group \(8,\) and as the size of the atoms decreases it becomes easier to remove electrons.

The \(E^{\circ}\) values for two low-spin iron complexes in acidic solution are as follows: $$ \begin{aligned} \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{3+}(a q)+\mathrm{e}^{-} \rightleftharpoons \\ \left[\mathrm{Fe}(o-\mathrm{phen})_{3}\right]^{2+}(a q) & E^{\circ}=1.12 \mathrm{~V} \end{aligned} $$ $$ \begin{aligned} \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}(a q)+\mathrm{e}^{-} \rightleftharpoons & \\ &\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}(a q) \quad E^{\circ}=0.36 \mathrm{~V} \end{aligned} $$ (a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, \(\left[\mathrm{Fe}(o \text { -phen })_{3}\right]^{3+}\) or \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},\) is more difficult to reduce? (c) Suggest an explanation for your answer to (b).
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