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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 non-observed geometric isomer for the square-planar complex \(\left[\mathrm{Pt}(\mathrm{en})\mathrm{Cl}_{2}\right]\) is the trans isomer, as a valid square-planar geometry cannot be formed with the bidentate ligand (en) occupying two adjacent positions and the Cl ligands opposite each other.

Step by step solution

01

1. Recalling the structure of square-planar complexes

Square-planar complexes have a central metal atom that is bonded to four ligands in a square plane. In this case, the central metal atom is platinum (Pt), with two chloro (Cl) and one ethylenediamine (en) ligands. Ethylenediamine (en) is a bidentate ligand, meaning it has two binding sites, and is commonly represented as \(\mathrm{NH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}\).
02

2. Understanding cis and trans isomers

In square-planar complexes, there are two possible geometric isomers: cis and trans. In cis isomers, two identical ligands are adjacent to each other, and in trans isomers, they are opposite each other. Since this platinum complex has one bidentate ligand, we will analyze the position of the two Cl ligands relative to each other in the possible geometries.
03

3. Analyzing the possible geometric isomers of the complex

In the platinum complex, ethylenediamine (en) has two nitrogen atoms that must bind to the platinum. This means that the two binding sites of en cannot be opposite each other, and must be adjacent. If we are to consider the cis isomer, the bidentate ligand would occupy two adjacent positions, and both Cl ligands would also be adjacent to each other, resulting in a valid square-planar geometry. If we consider the trans isomer, the bidentate ligand would still occupy two adjacent positions, but now, the two Cl ligands have to be opposite each other. In this case, we cannot create a valid square-planar geometry with this ligand arrangement, as the bidentate ligand (en) will remain adjacent.
04

4. Identifying the non-observed geometric isomer

Since the trans isomer cannot form a valid square-planar geometry for the given complex \(\left[\mathrm{Pt}(\mathrm{en})\mathrm{Cl}_{2}\right]\), the non-observed isomer is the trans isomer.

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Most popular questions from this chapter

Among the period 4 transition metals \((\mathrm{Sc}-\mathrm{Zn}),\) which elements do not form ions where there are partially filled 3d orbitals?

Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) (b) square planar \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{3}\right]^{2+}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\)

Sketch the structure of the complex in each of the following compounds and give the full compound name: (a) \(\operatorname{cis}-\left[\operatorname{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{Ru}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{5}\right]\) (c) \(\operatorname{trans} \mathrm{NH}_{4}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) (d) \(\operatorname{cis}-\left[\operatorname{Ru}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]\)

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