Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 76

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} ?

Short Answer

Expert verified
The isomers of Be(tfa)_2 exhibit structural isomerism, while the isomers of Cu(tfa)_2 exhibit geometric isomerism, specifically cis-trans isomerism. In Be(tfa)_2, the isomers have different arrangements of ligands around the central Be atom. In Cu(tfa)_2, the square planar isomers differ in the positions of their tfa ligands as either trans or cis to each other.

Step by step solution

01

Draw tfa ligand

To draw the isomers, first, we need to understand the structure of the tfa (1,1,1-trifluoroacetylacetone) ligand: \[ O=C-C(OH)-C(=O)-CF_3\]
02

Draw Tetrahedral Isomers of Be(tfa)_2

Now let's draw the two isomers of Be(tfa)_2. Since beryllium in Be(tfa)_2 has a tetrahedral geometry, it forms complexes with tfa^(-) having a central Be atom coordinated with four donor atoms from two tfa^- ligands. Isomer 1: Both oxygens are bonded to Be, forming a chelate ring. Isomer 2: Both carbonyls on the same side of the beryllium center, one is bonded to Be through O, and the other through the C=O group.
03

Identify Isomerism in Be(tfa)_2

Both isomers have different ligands positions or spatial arrangements of the atoms around the central beryllium atom. This type of isomerism is referred to as structural isomerism or coordination isomerism. For part b, we will draw the isomers of Cu(tfa)_2 and identify the isomerism exhibited.
04

Draw Square Planar Isomers of Cu(tfa)_2

Now let's draw the two isomers of Cu(tfa)_2. Since copper in Cu(tfa)_2 has a square planar geometry, it forms complexes with tfa^(-) having a central Cu atom coordinated with four donor atoms from two tfa^- ligands in a square planar arrangement. Isomer 1: The two tfa ligands are trans to each other. Isomer 2: The two tfa ligands are cis to each other.
05

Identify Isomerism in Cu(tfa)_2

Both isomers have different ligands positions or spatial arrangements of the atoms around the central copper atom. This type of isomerism is referred to as geometric isomerism or stereochemical isomerism. In this specific case, we have cis-trans isomerism.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The compound cisplatin, \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2},\) has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: \(\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)\) Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

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

Which of the following statement(s) is(are) true? a. The coordination number of a metal ion in an octahedral complex ion is \(8 .\) b. All tetrahedral complex ions are low-spin. c. The formula for triaquatriamminechromium(III) sulfate is \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]_{2}\left(\mathrm{SO}_{4}\right)_ 3}\) d. The electron configuration of \(\mathrm{Hf}^{2+}\) is \([\mathrm{Xe}] 4 f^{12} 6 s^{2}\) e. Hemoglobin contains \(\mathrm{Fe}^{3+}\).

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?

Figure \(20-17\) shows that the \(c i s\) isomer of \(\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}^{+}\) is optically active while the \(t r a n s\) isomer is not optically active. Is the same true for \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+} ?\) Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free