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Chapter 20: Problem 75

Draw all the geometrical isomers of \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+}\). Which of these isomers also have an optical isomer? Draw the various isomers.

Short Answer

Expert verified
The \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+}\) complex has three geometrical isomers: cis arrangement, trans arrangement, and mixed arrangement. Out of these three isomers, only the mixed arrangement isomer has an optical isomer.

Step by step solution

01

Draw the bidentate ligand

Let's start by drawing the ethylenediamine (en) ligand. The ethylenediamine is a bidentate ligand, which means it can form two coordinate bonds with the central metal ion, Cr. We will represent the en ligand as a straight line connecting two coordination sites on the central metal ion.
02

Identify the possible positions for the remaining ligands

Since we have four coordination sites remaining on the Cr ion, there are three possible arrangements for the NH3, Br, and Cl ligands: 1) cis arrangement (both the NH3 ligands are adjacent), 2) trans arrangement (both the NH3 ligands are opposite), and 3) mixed arrangement (Br and Cl ligands are adjacent).
03

Draw the geometrical isomers

Now, let's draw the geometrical isomers. Isomer 1 (cis arrangement): Both NH3 ligands are adjacent, and Br and Cl ligands are opposite. Isomer 2 (trans arrangement): Both NH3 ligands are opposite, and Br and Cl ligands are adjacent. Isomer 3 (mixed arrangement): Br and Cl ligands are adjacent, and one NH3 ligand is opposite to Br and the other NH3 ligand is opposite to Cl.
04

Determine optical isomers

Optical isomers are non-superimposable mirror images, which means they cannot be superimposed on each other. Now, let's check each geometrical isomer for the presence of an optical isomer. Isomer 1 (cis arrangement) and Isomer 2 (trans arrangement) are symmetrical and do not have mirror images that are non-superimposable. Hence, they do not have optical isomers. Isomer 3 (mixed arrangement) is not symmetrical, and it has a non-superimposable mirror image. Hence, Isomer 3 has an optical isomer. In conclusion, the \(\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}^{+}\) complex has three geometrical isomers. Out of these three isomers, the mixed arrangement isomer has an optical isomer.

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Most popular questions from this chapter

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

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a. In the absorption spectrum of the complex ion \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) there is a band corresponding to the absorption of a photon of light with an energy of \(1.75 \times 10^{4} \mathrm{cm}^{-1}\). Given 1 \(\mathrm{cm}^{-1}=1.986 \times 10^{-23} \mathrm{J},\) what is the wavelength of this photon? b. The \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle in \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) is predicted to be \(180^{\circ} .\) What is the hybridization of the \(\mathrm{N}\) atom in the NCS - ligand when a Lewis acid-base reaction occurs between \(\mathrm{Cr}^{3+}\) and \(\mathrm{NCS}^{-}\) that would give a \(180^{\circ}\) \(\mathrm{Cr}-\mathrm{N}-\mathrm{C}\) bond angle? \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}\) undergoes substitution by ethylenediamine (en) according to the equation \(\mathrm{Cr}(\mathrm{NCS})_{6}^{3-}+2 \mathrm{en} \longrightarrow \mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}+4 \mathrm{NCS}^{-}\) Does \(\operatorname{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit geometric isomerism? Does \(\mathrm{Cr}(\mathrm{NCS})_{2}(\mathrm{en})_{2}^{+}\) exhibit optical isomerism?

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