Chapter 22, Problem 22.21QP

Interpretation Introduction

Interpretation: The complex ions of platinum (IV) are to be named.

Concept introduction: 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.

To determine: The names of complex ions of platinum (IV).

Answer to Problem 22.21QP

Solution

The names of all the complex ions of platinum (IV) are rightfully stated.

Explanation of Solution

Explanation

The given complex ions of platinum (IV) are, $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_6\right]{\text{Cl}}_{\text{4}}$, $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_3$, $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_4{\text{Cl}}_2\right]{\text{Cl}}_2$, $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_3{\text{Cl}}_3\right]\text{Cl}$ and $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_2{\text{Cl}}_4\right]$.

A complex $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_6\right]{\text{Cl}}_{\text{4}}$ is considered in which the cation is ${\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_6\right]}^{4+}$ which consist of platinum as a central metal ion surrounded by six ammonia ligand. The oxidation state of $\text{Pt}$ is considered to be $x$ and the oxidation state of ${\text{NH}}_{\text{3}}$ is zero because it is a neutral ligand. The charge on complex ion is $+4$. The oxidation number of $\text{Pt}$ is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Pt }\times \text{oxidation state of Pt}\right)+\\ \left({\text{Number of NH}}_3\times {\text{oxidation state of NH}}_3\right)\end{array}\right]$.

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

$\begin{array}{c}4=1\left(x\right)+6\left(0\right)\\ 4=x+0\\ x=4\end{array}$

Hence, the oxidation state of $\text{Pt}$ is $+4$ and the name of the cation is hexaammineplatinum (IV).

The anion in the complex is chloride. Therefore, the name of the complex by joining the cation and anion is hexaammineplatinum (IV) chloride.

A complex $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_3$ is considered in which the cation is ${\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{Cl}\right]}^{3+}$ consists of platinum as a central metal ion surrounded by five ammonia and $1$ chlorine ligand. The oxidation state of $\text{Pt}$ is considered to be $x$ and the oxidation state of $\text{Cl}$ is $-1$ and of ${\text{NH}}_{\text{3}}$ is zero because it is a neutral ligand. The charge on complex ion is $+3$. The oxidation number of $\text{Pt}$ is calculated by using the formula,

$\text{Charge on complex}\text{ion}=\left[\begin{array}{c}\left(\text{Number of Pt }\times \text{oxidation state of Pt}\right)+\\ \left({\text{Number of NH}}_3\times {\text{oxidation state of NH}}_3\right)+\\ \left(\text{Number}\text{of}\text{Cl}\times \text{oxidation}\text{state}\text{of}\text{Cl}\right)\end{array}\right]$.

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

$\begin{array}{c}3=1\left(x\right)+5\left(0\right)+1\left(-1\right)\\ 3=x+0-1\\ x=3+1\\ x=4\end{array}$

The oxidation state of platinum is $+4$ and the name of the cation is pentaaminechloroplatinum (IV). The anion in the complex is chloride. Hence, the name of the complex is pentaamminechloroplatinum (IV) chloride.

A complex $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_4{\text{Cl}}_2\right]{\text{Cl}}_2$ is considered in which the cation is ${\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_4{\text{Cl}}_2\right]}^{2+}$ consists of platinum as a central metal ion surrounded by four ammonia and $2$ chlorine ligands. Similarly oxidation state of platinum is,

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

The oxidation state of platinum is $+4$ and the name of the cation is tetraamminedichloroplatinum (IV). The anion in the complex is chloride. Hence, the name of the complex is tetraamminedichloroplatinum (IV) chloride.

A complex $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_3{\text{Cl}}_3\right]\text{Cl}$ is considered in which the cation is ${\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_3{\text{Cl}}_3\right]}^{+}$ consists of platinum as a central metal ion surrounded by three ammonia and $3$ chlorine ligands. Similarly oxidation state of platinum is,

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

The oxidation state of platinum is $+4$ and the name of the cation is triamminetrichloroplatinum (IV). The anion in the complex is chloride. Hence, the name of the complex is triamminetrichloroplatinum (IV) chloride.

A complex $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_2{\text{Cl}}_4\right]$ is considered which is a neutral complex consists of platinum as a central metal ion surrounded by two ammonia and $4$ chlorine ligands. The oxidation state of platinum is $+4$. Hence, the name of the complex is diamminetetrachloroplatinum (IV).

Conclusion

The names of the given complex ions are,

• The name of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_6\right]{\text{Cl}}_{\text{4}}$ is hexaammineplatinum (IV) chloride.
• The name of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_3$ is pentaamminechloroplatinum (IV) chloride.
• The name of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_4{\text{Cl}}_2\right]{\text{Cl}}_2$ is tetraamminedichloroplatinum (IV) chloride.
• The name of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_3{\text{Cl}}_3\right]\text{Cl}$ is triamminetrichloroplatinum (IV) chloride.
• The name of $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_2{\text{Cl}}_4\right]$ is diamminetetrachloroplatinum (IV).