EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
7th Edition
ISBN: 9781319075125
Author: ATKINS
Publisher: MPS (CC)
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Chapter 9, Problem 9C.12E

(a)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair [Pt(OH2)4][Pt(Cl)6] and [Pt Cl2(OH2)4][Pt(Cl)4] has to be given.

Concept Introduction:

Isomers:

The compounds with same molecular formula but differ in the arrangement of the atom are said to be isomers.

Coordination Isomers:

Coordination isomerism arises in compounds containing complex anionic and cationic parts.  Hence, there are two complex compounds bound together, one with a negative charge and the other with a positive charge.  In coordination isomers, the anion and cation complexes of a coordination compound exchange one or more ligands.

(a)

Expert Solution
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Answer to Problem 9C.12E

The type of isomer represented by the pair [Pt(OH2)4][Pt(Cl)6] and [Pt Cl2(OH2)4] [Pt(Cl)4] is coordination isomer.

Explanation of Solution

In this pair of isomers the chloride ligand of the anionic part exchanges its place between another parts of the coordination sphere thus it is a coordination isomer.

(b)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair [Cr(en)3][Co(Ox)3] and [Cr(ox)3][Co(en)3] has to be given.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 9C.12E

The type of isomer represented by the pair [Cr(en)3][Co(Ox)3] and [Cr(ox)3][Co(en)3] is coordination isomer.

Explanation of Solution

In this pair of isomers the chloride ligand exchanges their place between another parts of the coordination sphere thus it is a coordination isomer.

(c)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair [Fe(CN_)(OH)5]3- and [Fe(C_N)(OH)5]3- has to be given.

Concept Introduction:

Linkage isomers:

Linkage isomerism occurs with ambidentate ligands that are capable of coordinating in more than one way.

Example:

SCN-/NCS-.  The only difference is the atoms bind to the central ion.  The ligand must have more than one donor atom but bind to ion in only one place.

(c)

Expert Solution
Check Mark

Answer to Problem 9C.12E

The type of isomer represented by the pair [Fe(CN_)(OH)5]3- and [Fe(C_N)(OH)5]3- is linkage isomerism.

Explanation of Solution

In the isomer [Fe(CN_)(OH)5]3- the CN ligand is bind to the metal atom through the nitrogen atom while in another complex the ligand is bind to the metal through the oxygen atom.

(d)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair [CoBr2(NH3)4]Br.H2O and [CoBr(NH3)4(H2O)]Br2 has to be given

Concept Introduction:

Hydrate Isomers:

Hydrate isomers results from replacement of a coordinated group by a solvent molecule in this case it is a water molecule.

(d)

Expert Solution
Check Mark

Answer to Problem 9C.12E

The type of isomer represented by the pair [CoBr2(NH3)4]Br.H2O and [CoBr(NH3)4(H2O)]Br2 is hydrate isomer.

Explanation of Solution

In this pair of isomers the aqua ion which is bound with central atom is exchanged its place with the bromide ligand which present outside the nucleus.  Thus these isomers are hydrate isomers.

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Chapter 9 Solutions

EBK CHEMICAL PRINCIPLES

Ch. 9 - Prob. 9A.9ECh. 9 - Prob. 9A.10ECh. 9 - Prob. 9A.11ECh. 9 - Prob. 9A.12ECh. 9 - Prob. 9A.13ECh. 9 - Prob. 9A.14ECh. 9 - Prob. 9B.1ASTCh. 9 - Prob. 9B.1BSTCh. 9 - Prob. 9B.2ASTCh. 9 - Prob. 9B.2BSTCh. 9 - Prob. 9B.1ECh. 9 - Prob. 9B.2ECh. 9 - Prob. 9B.3ECh. 9 - Prob. 9B.4ECh. 9 - Prob. 9B.5ECh. 9 - Prob. 9B.6ECh. 9 - Prob. 9B.7ECh. 9 - Prob. 9B.8ECh. 9 - Prob. 9B.9ECh. 9 - Prob. 9B.10ECh. 9 - Prob. 9B.11ECh. 9 - Prob. 9B.12ECh. 9 - Prob. 9B.13ECh. 9 - Prob. 9B.14ECh. 9 - Prob. 9B.15ECh. 9 - Prob. 9B.16ECh. 9 - Prob. 9C.1ASTCh. 9 - Prob. 9C.1BSTCh. 9 - Prob. 9C.2ASTCh. 9 - Prob. 9C.2BSTCh. 9 - Prob. 9C.3ASTCh. 9 - Prob. 9C.3BSTCh. 9 - Prob. 9C.4ASTCh. 9 - Prob. 9C.4BSTCh. 9 - Prob. 9C.1ECh. 9 - Prob. 9C.2ECh. 9 - Prob. 9C.3ECh. 9 - Prob. 9C.4ECh. 9 - Prob. 9C.5ECh. 9 - Prob. 9C.6ECh. 9 - Prob. 9C.7ECh. 9 - Prob. 9C.8ECh. 9 - Prob. 9C.9ECh. 9 - Prob. 9C.10ECh. 9 - Prob. 9C.11ECh. 9 - Prob. 9C.12ECh. 9 - Prob. 9C.13ECh. 9 - Prob. 9C.14ECh. 9 - Prob. 9C.15ECh. 9 - Prob. 9C.16ECh. 9 - Prob. 9C.17ECh. 9 - Prob. 9C.18ECh. 9 - Prob. 9C.19ECh. 9 - Prob. 9C.20ECh. 9 - Prob. 9D.1ASTCh. 9 - Prob. 9D.1BSTCh. 9 - Prob. 9D.2ASTCh. 9 - Prob. 9D.2BSTCh. 9 - Prob. 9D.3ASTCh. 9 - Prob. 9D.3BSTCh. 9 - Prob. 9D.4ASTCh. 9 - Prob. 9D.4BSTCh. 9 - Prob. 9D.1ECh. 9 - Prob. 9D.2ECh. 9 - Prob. 9D.3ECh. 9 - Prob. 9D.4ECh. 9 - Prob. 9D.5ECh. 9 - Prob. 9D.6ECh. 9 - Prob. 9D.7ECh. 9 - Prob. 9D.8ECh. 9 - Prob. 9D.9ECh. 9 - Prob. 9D.10ECh. 9 - Prob. 9D.11ECh. 9 - Prob. 9D.12ECh. 9 - Prob. 9D.13ECh. 9 - Prob. 9D.14ECh. 9 - Prob. 9D.15ECh. 9 - Prob. 9D.16ECh. 9 - Prob. 9D.17ECh. 9 - Prob. 9D.18ECh. 9 - Prob. 9D.19ECh. 9 - Prob. 9D.20ECh. 9 - Prob. 9D.21ECh. 9 - Prob. 9D.22ECh. 9 - Prob. 9D.23ECh. 9 - Prob. 9D.24ECh. 9 - Prob. 9D.25ECh. 9 - Prob. 9D.26ECh. 9 - Prob. 9D.27ECh. 9 - Prob. 9D.28ECh. 9 - Prob. 9D.29ECh. 9 - Prob. 9D.30ECh. 9 - Prob. 9D.31ECh. 9 - Prob. 9D.32ECh. 9 - Prob. 9D.33ECh. 9 - Prob. 9D.34ECh. 9 - Prob. 9.1ECh. 9 - Prob. 9.2ECh. 9 - Prob. 9.3ECh. 9 - Prob. 9.4ECh. 9 - Prob. 9.5ECh. 9 - Prob. 9.6ECh. 9 - Prob. 9.7ECh. 9 - Prob. 9.8ECh. 9 - Prob. 9.9ECh. 9 - Prob. 9.10ECh. 9 - Prob. 9.11ECh. 9 - Prob. 9.12ECh. 9 - Prob. 9.13ECh. 9 - Prob. 9.14ECh. 9 - Prob. 9.15ECh. 9 - Prob. 9.16ECh. 9 - Prob. 9.17ECh. 9 - Prob. 9.18ECh. 9 - Prob. 9.19ECh. 9 - Prob. 9.20ECh. 9 - Prob. 9.21ECh. 9 - Prob. 9.23ECh. 9 - Prob. 9.25E
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