Chapter 22, Problem 22.22QP

Interpretation Introduction

Interpretation: The complex ions of chromium (III) 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 chromium (III).

Answer to Problem 22.22QP

Solution

The names of all the complex ions of chromium (III) are rightfully stated.

Explanation of Solution

Explanation

The given complex ions of chromium (III) are $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]\text{Cl}$, $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]{\text{Cl}}_3$ and $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$.

A complex $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]\text{Cl}$ is considered in which the cation is ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]}^{+1}$ which consist of chromium as a central metal ion surrounded by $4$ water and two chlorine ligands. The oxidation state of $\text{Cr}$ is considered to be $x$ and the oxidation state of $\text{Cl}$ is $-1$ and of ${\text{H}}_{\text{2}}\text{O}$ is zero because it is a neutral ligand. The charge on complex ion is $+1$. The oxidation number of $\text{Cr}$ is calculated by using the formula,

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

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

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

The oxidation number of chromium is $+3$. For ${\text{H}}_{\text{2}}\text{O}$ ligand, term aqua is used. Hence, the name of the cation is Tetraaquadichlorochromium (III). The anion in the complex is chloride. Therefore, the name of the complex by joining the cation and anion is Tetraaquadichlorochromium (III) chloride.

A complex $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]{\text{Cl}}_3$ is considered in which the cation is ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]}^{+3}$ which consist of chromium as a central metal ion surrounded by $6$ water ligands. The oxidation state of $\text{Cr}$ is considered to be $x$ and of ${\text{H}}_{\text{2}}\text{O}$ is zero because it is a neutral ligand. The charge on complex ion is $+1$. The oxidation number of $\text{Cr}$ is calculated by using the formula,

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

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

$\begin{array}{c}3=1\left(x\right)+6\left(0\right)+\\ 3=x+0\\ x=3\end{array}$

The oxidation number of chromium is $+3$. For ${\text{H}}_{\text{2}}\text{O}$ ligand, term aqua is used. Hence, the name of the cation is Hexaaquachromium (III). The anion in the complex is chloride. Therefore, the name of the complex by joining the cation and anion is Hexaaquachromium (III) chloride.

A complex $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$ is considered in which the cation is ${\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{Cl}\right]}^{+2}$ which consist of chromium as a central metal ion surrounded by $5$ water and chlorine ligands. The oxidation state of $\text{Cr}$ is considered to be $x$ and the oxidation state of $\text{Cl}$ is $-1$ and of ${\text{H}}_{\text{2}}\text{O}$ is zero because it is a neutral ligand. The charge on complex ion is $+1$. The oxidation number of $\text{Cr}$ is calculated by using the formula,

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

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

$\begin{array}{c}2=1\left(x\right)+5\left(0\right)+1\left(-1\right)\\ 2=x+0-1\\ x=2+1\\ x=3\end{array}$

The oxidation number of chromium is $+3$. For ${\text{H}}_{\text{2}}\text{O}$ ligand, term aqua is used. Hence, the name of the cation is Pentaaquachlorochromium (III). The anion in the complex is chloride. Therefore, the name of the complex by joining the cation and anion is Pentaaquachlorochromium (III) chloride.

Conclusion

The names of the given complex ions are,

• The name of $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]\text{Cl}$ is Tetraaquadichlorochromium (III) chloride.
• The name of $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_6\right]{\text{Cl}}_3$ is Hexaaquachromium (III) chloride.
• The name of $\left[\text{Cr}{\left({\text{H}}_{\text{2}}\text{O}\right)}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$ is Pentaaquachlorochromium (III) chloride.