Chapter 22, Problem 22.27QP

(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 22.27QP

Solution

The name of $\left[\text{Zn}{\left(\text{en}\right)}_{\text{2}}\right]{\text{SO}}_{\text{4}}$ is ethylenediaminezinc (II) sulphate.

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 $\left[\text{Zn}{\left(\text{en}\right)}_{\text{2}}\right]{\text{SO}}_{\text{4}}$. The given coordination compound contain complex ion and sulfate as a counter ion. The charge on sulfate ion is $-2$, so 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 zinc. The ligand is ethylenediamine and the number of ligand is one so prefix is not required. The oxidation number of $\text{Zn}$ is assumed to be $x$ and is calculated by using the formula,

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

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

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

The oxidation state of $\text{Zn}$ is $+2$. Hence, the name of the complex ion is ethylenediaminezinc (II) sulphate.

(b)

Interpretation Introduction

To determine: The name of the given coordination compound.

Answer to Problem 22.27QP

Solution

The name of $\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left({\text{H}}_{\text{2}}\text{O}\right)\right]{\text{Cl}}_{\text{2}}$ is pentaammineaquanickel (II) chloride.

Explanation of Solution

Explanation

The given coordination compound is $\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left({\text{H}}_{\text{2}}\text{O}\right)\right]{\text{Cl}}_{\text{2}}$. The given coordination compound contains complex ion and chlorine as counter ions. The charge on chlorine ion is $-1$ and there are two chlorine 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 nickel. There are two types of ligands one is ammonia and other is water.

There are $5$ ammonia so the prefix for ligand ammonia is penta and no prefix for ligand ${\text{H}}_{\text{2}}\text{O}$ as it is only one. Both the ligands are neutral in nature. The oxidation number of $\text{Ni}$ is assumed to be $x$ and is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Ni }\times \text{oxidation state of Ni}\right)+\\ \left({\text{Number of NH}}_3\times {\text{oxidation state of NH}}_3\right)+\\ \left({\text{Number of H}}_2\text{O }\times {\text{oxidation state of H}}_{\text{2}}\text{O}\right)\end{array}\right]$

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

$\begin{array}{c}+2=1\left(x\right)+5\left(0\right)+1\left(0\right)\\ +2=x+0\\ x=2\end{array}$

The oxidation state of $\text{Ni}$ is $+2$. Hence, the name of the complex ion is pentaammineaquanickel (II) chloride.

(c)

Interpretation Introduction

To determine: The name of the given coordination compound.

Answer to Problem 22.27QP

Solution

The name of ${\text{K}}_{\text{4}}\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]$ is potassiumhexacyanoferrate (II).

Explanation of Solution

Explanation

The given coordination compound is ${\text{K}}_{\text{4}}\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]$. The given coordination compound contains complex ion and potassium as counter ions. The charge on potassium ion is $+1$ and there are four potassium so the charge is $+4$. In order to maintain electroneutrality the complex ion has $-4$ charge. The complex ion has negative charge so the name of the central metal atom is ferrate.

The ligand is cyano and the number of ligands is six, so the prefix used for ligand is hexa. The oxidation number of $\text{Fe}$ is assumed to be $x$ and is calculated by using the formula,

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

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

$\begin{array}{c}+4=1\left(x\right)+6\left(-1\right)\\ +4=x-6\\ x=6-4\\ x=2\end{array}$

The oxidation state of $\text{Fe}$ is $+2$. Hence, the name of the complex ion is potassiumhexacyanoferrate (II).

Conclusion

The names of the given complex ions are,

1. a. The name of $\left[\text{Zn}{\left(\text{en}\right)}_{\text{2}}\right]{\text{SO}}_{\text{4}}$ is ethylenediaminezinc (II) sulphate.
2. b. The name of $\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\left({\text{H}}_{\text{2}}\text{O}\right)\right]{\text{Cl}}_{\text{2}}$ is pentaammineaquanickel (II) chloride.
3. c. The name of ${\text{K}}_{\text{4}}\left[\text{Fe}{\left(\text{CN}\right)}_{\text{6}}\right]$ is potassiumhexacyanoferrate (II).