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Chapter 20: Problem 46

Draw all geometrical and linkage isomers of square planar \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\)

Short Answer

Expert verified
There are four isomers for the square planar complex \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{SCN})_{2}\): 1. Cis isomer with both thiocyanate ligands S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Cis isomer with both thiocyanate ligands N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) 3. Trans isomer with both thiocyanate ligands S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 4. Trans isomer with both thiocyanate ligands N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\)

Step by step solution

01

Identify the central atom and ligands

In this square planar complex, the central atom is Platinum (Pt), and there are two types of ligands: ammine (NH3) and thiocyanate (SCN).
02

Draw the square planar complex

The initial complex consists of Platinum (Pt) surrounded by the ligands in a square planar arrangement. There are two ammine ligands and two thiocyanate ligands.
03

Determine the possible geometrical isomers

Geometrical isomers are formed when the ligands are arranged differently around the central atom. In this case, there are two possibilities: 1. The ammine ligands are adjacent to each other (cis isomer). 2. The ammine ligands are opposite each other (trans isomer).
04

Draw the geometrical isomers

For the square planar complex with the given formula, we have two geometrical isomers: 1. Cis isomer: The ammine ligands are adjacent to each other and the thiocyanate ligands are opposite each other in the square plane. ![cis-isomer](https://i.imgur.com/3zwLUR5.png) 2. Trans isomer: The ammine ligands are opposite to each other and the thiocyanate ligands are also opposite each other in the square plane. ![trans-isomer](https://i.imgur.com/HGH1o2X.png)
05

Determine the possible linkage isomers

Linkage isomers arise from different sites of attachment of ligands. In thiocyanate (SCN), there are two possible sites of attachment: sulfur (S) or nitrogen (N). Thus, the thiocyanate ligand can create two linkage isomers: S-bound or N-bound.
06

Draw the linkage isomers for cis and trans isomers

Here are the possible linkage isomers for the cis and trans geometrical isomers: - Cis isomer: 1. Cis - Both thiocyanate ligands are S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Cis - Both thiocyanate ligands are N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) - Trans isomer: 1. Trans - Both thiocyanate ligands are S-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{S}\mathrm{CN})_{2}\) 2. Trans - Both thiocyanate ligands are N-bound: \(\mathrm{Pt}(\mathrm{NH}_{3})_{2}(\mathrm{N}\mathrm{CS})_{2}\) So, there are a total of four isomers for the given square planar complex: two geometrical isomers with each having two linkage isomers.

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Most popular questions from this chapter

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?

A compound related to acetylacetone is 1,1,1 -trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise \(47 .\) Both \(\mathrm{Be}^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa \(^{-}\) having the formula \(\mathrm{M}\) (tfa) \(_{2}\). Two isomers are formed for each metal complex. a. The \(\mathrm{Be}^{2+}\) complexes are tetrahedral. Draw the two isomers of Be(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} ?

Name the following coordination compounds. a. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}\) b. \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{I}_{3}\) c. \(\mathrm{K}_{2}\left[\mathrm{PtCl}_{4}\right]\) d. \(K_{4}\left[P_{t} C l_{6}\right]\) e. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\) f. \(\left[\operatorname{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right]\)

Give formulas for the following. a. hexakis(pyridine) cobalt(III) chloride b. pentaammineiodochromium(III) iodide c. tris(ethylenediamine)nickel(II) bromide d. potassium tetracyanonickelate(II) e. tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

In which of the following is(are) the electron configuration(s) correct for the species indicated? a. Cu \([\mathrm{Ar}] 4 s^{2} 3 d^{9}\) b. \(\mathrm{Fe}^{3+} \quad[\mathrm{Ar}] 3 d^{5}\) c. Co \([\mathrm{Ar}] 4 s^{2} 3 d^{7}\) d. La \([\mathrm{Ar}] 6 s^{2} 4 f^{1}\) e. \(\mathrm{Pt}^{2+} \quad[\mathrm{Xe}] 4 f^{14} 5 d^{8}\)
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