Chapter 22, Problem 22.28QP

(a)

Interpretation Introduction

Interpretation: The given coordination compounds are to be named.

Concept introduction: The naming of a complex ion expresses the oxidation state and identity of the central metal ion, the number and names of ligands bound to the inner coordination sphere of the metal ion and the overall charge of the complex ion.

To determine: The name of the given coordination compound.

Answer to Problem 22.28QP

Solution

The name of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$ is ammoniumhexacyanocobaltate (III).

Explanation of Solution

Explanation

The naming of coordination compounds is written in two parts. The cation is named first in both the positively or negatively charged coordination complexes and then the name of anion. The ligands are named in alphabetically order before the name of central metal ion or atom.

The given coordination compound is ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$. The given coordination compound contains complex ion and ammonium as a counter ion. The charge on ammonium ion is $+1$, as there are three ammonium ions the charge is $+3$. So in order to maintain electroneutrality the complex ion has $-3$ charge. The complex ion has negative charge. Therefore, the metal name is cobaltate.

The ligand is cyano and the number of ligand is six so prefix for ligand is hexa and charge on $\text{CN}$ is $-1$. The oxidation number of $\text{Co}$ is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Co }\times \text{oxidation state of Co}\right)+\\ \left(\text{Number of CN }\times \text{oxidation state of CN}\right)\end{array}\right]$

Substitute the values of number of atoms and oxidation state in the above formula to calculate the oxidation number of $\text{Co}$,

$\begin{array}{c}-3=1\left(x\right)+6\left(-1\right)\\ -3=x-6\\ x=6-3\\ x=3\end{array}$

The oxidation state of $\text{Co}$ is $+3$. Hence, the name of the complex ion is ammoniumhexacyanocobaltate (III).

(b)

Interpretation Introduction

To determine: The name of the given coordination compound.

Answer to Problem 22.28QP

Solution

The name of $\left[\text{Co}{\left(\text{en}\right)}_{\text{2}}\text{Cl}\right]{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is chlorobis-(ethylenediamine) cobalt (III) nitrate.

Explanation of Solution

Explanation

The given coordination compound is $\left[\text{Co}{\left(\text{en}\right)}_{\text{2}}\text{Cl}\right]{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$. The given coordination compound contains complex ion and nitrate as counter ions. The charge on nitrate ion is $-1$ and there are two nitrate so the charge is $-2$. In order to maintain electroneutrality the complex ion has $+2$ charge. The complex ion has positive charge so the metal is named same as the element name.

Therefore, the metal name is cobalt. There are two types of ligands one is bidentate ligand ethylenediamine and the other is monodentate ligand chlorine. Prefix for bidentate ligand $\left(\text{en}\right)$ is bis and no prefix for ligand $\text{Cl}$ as it is only one. The ligand $\left(\text{en}\right)$ is neutral in nature. The oxidation number of $\text{Co}$ is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Co }\times \text{oxidation state of Co}\right)+\\ \left(\text{Number of en }\times \text{oxidation state of en}\right)+\\ \left(\text{Number of Cl }\times \text{oxidation state of Cl}\right)\end{array}\right]$

Substitute the values of number of atoms and oxidation state in the above formula to calculate the oxidation number of $\text{Co}$,

$\begin{array}{c}+2=1\left(x\right)+2\left(0\right)+1\left(-1\right)\\ +2=x-1\\ x=2+1\\ x=3\end{array}$

The oxidation state of $\text{Co}$ is $+3$. Hence, the name of the complex ion is chlorobis-(ethylenediamine) cobalt (III) nitrate.

(c)

Interpretation Introduction

To determine: The name of the given coordination compound.

Answer to Problem 22.28QP

Solution

The name of $\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]\text{Cl}$ is tetraaquadihydroxoiron (III) chloride.

Explanation of Solution

Explanation

The given coordination compound is $\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]\text{Cl}$. The given coordination compound contains complex ion and chlorine as counter ion. The charge on chlorine ion is $-1$ so in order to maintain electroneutrality the complex ion has $+1$ charge. The complex ion has positive charge so the name of the central metal atom is same as the element name.

Therefore, the name of metal is iron. The ligands are aqua and hydroxo with prefix tetra and di. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Fe }\times \text{oxidation state of Fe}\right)+\\ \left({\text{Number of H}}_2\text{O }\times {\text{oxidation state of H}}_{\text{2}}\text{O}\right)+\\ \left(\text{Number of OH}\times \text{oxidation state of OH}\right)\end{array}\right]$

Substitute the values of number of atoms and oxidation state in the above formula to calculate the oxidation number of $\text{Fe}$,

$\begin{array}{c}+1=1\left(x\right)+4\left(0\right)+2\left(-1\right)\\ +1=x+0-2\\ x=1+2\\ x=3\end{array}$

The oxidation state of $\text{Fe}$ is $+3$. Hence, the name of the complex ion is tetraaquadihydroxoiron (III) chloride.

Conclusion

The names of the given complex ions are,

1. a. The name of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$ is ammoniumhexacyanocobaltate (III).
2. b. The name of $\left[\text{Co}{\left(\text{en}\right)}_{\text{2}}\text{Cl}\right]{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is chlorobis-(ethylenediamine) cobalt (III) nitrate.
3. c. The name of $\left[\text{Fe}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]\text{Cl}$ is tetraaquadihydroxoiron (III) chloride.