Chapter 9, Problem 9C.11E

(a)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{N}}{\text{O}}_{\text{2}}\right)\right]{\text{Br}}_{\text{2}}$ and $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{O}}\text{NO}\right)\right]{\text{Br}}_{\text{2}}$ 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.

Linkage isomers:

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

Example:

${\text{SCN}}^{\text{-}}{\text{/NCS}}^{\text{-}}$.  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.

Answer to Problem 9C.11E

The type of isomer represented by the pair $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{N}}{\text{O}}_{\text{2}}\right)\right]{\text{Br}}_{\text{2}}$ and $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{O}}\text{NO}\right)\right]{\text{Br}}_{\text{2}}$ is linkage isomer.

Explanation of Solution

In the isomer $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{N}}{\text{O}}_{\text{2}}\right)\right]{\text{Br}}_{\text{2}}$ the ${\text{NO}}_{\text{2}}$ ligand is bind to the metal atom through the nitrogen atom while in another complex $\left[\text{Co}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left(\underset{\_}{\text{O}}\text{NO}\right)\right]{\text{Br}}_{\text{2}}$ the ligand is bind to the metal through the oxygen atom.  Thus it is a linkage isomer.

(b)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\left({\text{SO}}_{\text{4}}\right)\right]{\left(\text{OH}\right)}_{\text{2}}$ and $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]\left({\text{SO}}_{\text{4}}\right)$ has to be given.

Concept Introduction:

Ionization Isomers:

Ionization isomers occur when a ligand is bound to the metal center exchanges places with an anion or neutral molecule that was previously outside the coordination complex.

Answer to Problem 9C.11E

The type of isomer represented by the pair $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}\left({\text{SO}}_{\text{4}}\right)\right]{\left(\text{OH}\right)}_{\text{2}}$ and $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]\left({\text{SO}}_{\text{4}}\right)$ is ionization isomer.

Explanation of Solution

In this pair of isomers the hydroxide ion which is bound with central atom is exchanged its place with the sulphate ligand which present outside the nucleus of the complex.  Thus these isomers are ionization isomers.

(c)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair $\left[\text{CoCl}{\left(\text{NC}\underset{\_}{\text{S}}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ and $\left[\text{CoCl}{\left(\underset{\_}{\text{N}}\text{CS}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 9C.11E

The type of isomer represented by the pair $\left[\text{CoCl}{\left(\text{NC}\underset{\_}{\text{S}}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ and $\left[\text{CoCl}{\left(\underset{\_}{\text{N}}\text{CS}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ is linkage isomerism.

Explanation of Solution

In the isomer $\left[\text{CoCl}{\left(\text{NC}\underset{\_}{\text{S}}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ the $\text{NCS}$ ligand is bind to the metal atom through the sulphur atom while in another complex $\left[\text{CoCl}{\left(\underset{\_}{\text{N}}\text{CS}\right)}_{\text{4}}{\left({\text{NH}}_{\text{3}}\right)}_4\right]\text{Cl}$ the ligand is bind to the metal through the nitrogen atom.

(d)

Interpretation Introduction

Interpretation:

The type of isomer represented by the pair $\left[\text{CrCl}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\right]\text{Br}$ and $\left[\text{CrBr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\right]\text{Cl}$ has to be given

Concept Introduction:

Refer to part (b).

Answer to Problem 9C.11E

The type of isomer represented by the pair $\left[\text{CrCl}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\right]\text{Br}$ and $\left[\text{CrBr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\right]\text{Cl}$ is ionization isomer.

Explanation of Solution

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