Chapter 21, Problem 37E

(a)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$.

Answer to Problem 37E

Answer

The name of complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$ is hexaamminecobalt(II) chloride.

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Co(NH}}_3{\text{)}}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$.

Oxidation state of Chlorine ion $\left({\text{Cl}}^{-}\right)$ is $-1$.

Oxidation state of neutral Ammonia $\left({\text{NH}}_3\right)$ is $0$.

It is assumed that the oxidation state of metal ion Cobalt $\left(\text{Co}\right)$ is $x$.

The given complex compound contains six Ammonia molecules and two Chlorine molecules (counter ion). Since, the overall charge on stable compound is zero. Therefore,

$\begin{array}{c}6\left(0\right)+2\left(-1\right)+x=0\\ x=+2\end{array}$

Therefore, the oxidation state of Cobalt is $+2$.

The given complex compound contains six Ammonia molecules. Hence, the name of anion is hexaammine. It also contains counter ion chloride salt.

Therefore, the name of complex ion ${\left[{\text{Co(NH}}_3{\text{)}}_{\text{4}}{\left(\text{Cl}\right)}_{\text{2}}\right]}^{+}$ is hexaamminecobalt(II) chloride.

(b)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound $\left[{\text{Co(H}}_2{\text{O)}}_{\text{6}}\right]{\text{I}}_{\text{3}}$.

Answer to Problem 37E

Answer

The name of complex compound $\left[{\text{Co(H}}_2{\text{O)}}_{\text{6}}\right]{\text{I}}_{\text{3}}$ is hexaaquaCobalt(III) iodide.

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Co(H}}_2{\text{O)}}_{\text{6}}\right]{\text{I}}_{\text{3}}$.

Oxidation state of Iodine ion $\left({\text{I}}^{-}\right)$ is $-1$.

Oxidation state of neutral Water $\left({\text{H}}_2\text{O}\right)$ is $0$.

It is assumed that the oxidation state of metal ion Cobalt $\left(\text{Co}\right)$ is $x$.

The given complex compound contains six water molecules and three Iodine ions (counter ion). Since, the overall charge on stable compound is zero. Therefore,

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

Therefore, the oxidation state of Cobalt is $+3$.

The given complex compound contains six water molecules. Hence, the name of anion is hexaaqua. The compound also contains counter ion Iodide salt.

Therefore, the name of complex ion $\left[{\text{Co(H}}_2{\text{O)}}_{\text{6}}\right]{\text{I}}_{\text{3}}$ is hexaaquacobalt(III) iodide.

(c)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound ${\text{K}}_{\text{2}}\left[{\text{PtCl}}_{\text{4}}\right]$.

Answer to Problem 37E

Answer

The name of complex ion ${\text{K}}_{\text{2}}\left[{\text{PtCl}}_{\text{4}}\right]$ is potassium tetrachloroplatinate(II).

Explanation of Solution

Explanation

The complex compound is ${\text{K}}_{\text{2}}\left[{\text{PtCl}}_{\text{4}}\right]$.

Oxidation state of Chlorine $\left({\text{Cl}}^{-}\right)$ is $-1$.

Oxidation state of Potassium $\left({\text{K}}^{+}\right)$ is $+1$.

It is assumed that the oxidation state of metal ion Platinum $\left(\text{Pt}\right)$ is $x$.

The given complex compound contains two Potassium atoms (counter cation) and four Chlorine atoms. Since, the overall charge on stable compound is zero. Therefore,

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

Therefore, the oxidation state of Platinum is $+2$.

The given complex compound contains four Chlorine atoms. Hence, the name of anion is tetrachloro. It also contains counter cation Potassium. The Latin name of Platinum is Platinate.

Therefore, the name of complex ion ${\text{K}}_{\text{2}}\left[{\text{PtCl}}_{\text{4}}\right]$ is potassium tetrachloroplatinate(II).

(d)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound ${\text{K}}_{\text{4}}\left[{\text{PtCl}}_{\text{6}}\right]$.

Answer to Problem 37E

Answer

The name of complex ion ${\text{K}}_{\text{4}}\left[{\text{PtCl}}_{\text{6}}\right]$ is potassium hexachloroplatinate(II).

Explanation of Solution

Explanation

The complex compound is ${\text{K}}_{\text{4}}\left[{\text{PtCl}}_{\text{6}}\right]$.

Oxidation state of Chlorine $\left({\text{Cl}}^{-}\right)$ is $-1$.

Oxidation state of Potassium $\left({\text{K}}^{+}\right)$ is $+1$.

It is assumed that the oxidation state of metal ion Platinum $\left(\text{Pt}\right)$ is $x$.

The given complex compound contains four Potassium atoms (counter cation) and six Chlorine atoms. Since, the overall charge on stable compound is zero. Therefore,

$\begin{array}{c}4\left(+1\right)+6\left(-1\right)+x=0\\ x=+2\end{array}$

Therefore, the oxidation state of Platinum is $+2$.

The given complex compound contains six chlorine atoms. Hence, the name of anion is hexachloro. It also contains counter cation Potassium. The Latin name of Platinum is Platinate.

Therefore, the name of complex ion ${\text{K}}_{\text{4}}\left[{\text{PtCl}}_{\text{6}}\right]$ is potassium hexachloroplatinate(II).

(e)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$.

Answer to Problem 37E

Answer

The name of complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$ is pentaamminechlorocobalt(II) chloride.

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Co(NH}}_3{\text{)}}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$.

Oxidation state of Chlorine ion $\left({\text{Cl}}^{-}\right)$ is $-1$.

Oxidation state of neutral Ammonia $\left({\text{NH}}_3\right)$ is $0$.

It is assumed that the oxidation state of metal ion Cobalt $\left(\text{Co}\right)$ is $x$.

The given complex compound contains five Ammonia molecules and one Chlorine atom. It also contains counter Chloride ions. Since, the overall charge on stable compound is zero. Therefore,

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

Therefore, the oxidation state of Cobalt is $+3$.

The given complex compound contains five Ammonia molecules and one Chlorine atom. Hence, the name of anion is pentaamminechloro. The compound also contains Chloride salt.

Therefore, the name of complex ion $\left[{\text{Co(NH}}_3{\text{)}}_{\text{5}}\text{Cl}\right]{\text{Cl}}_{\text{2}}$ is pentaamminechlorocobalt(II) chloride.

(f)

Interpretation Introduction

Interpretation: The name of the given complex compounds is to be stated.

Concept introduction: Rules followed in the naming of coordination compound:

• In the naming of coordination compound, the name of anion (ligand) comes first. Metal ion is named second, which is the name of element.

• An ‘o’ is added to the root name of anion.

• Greek prefix is used to express the number of ligands.

• Roman numeral is used to denote the oxidation state of metal ion.

• In case of more than one type of ligand present in a compound, they are named alphabetically.

• The suffix ‘-ate’ is added to metal ion if complex ion contains negative charge. Latin names are also used to identify the name of metals.

• If counter ions are present in complex compound, then cation is named first, which is the name of element, whereas anion is named last, which is the name of salt.

To determine: The name of the complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{3}}{\left({\text{NO}}_2\right)}_3\right]$.

Answer to Problem 37E

Answer

The name of complex compound $\left[{\text{Co(NH}}_3{\text{)}}_{\text{3}}{\left({\text{NO}}_2\right)}_3\right]$ is triamminetrinitro-N-cobalt(III).

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Co(NH}}_3{\text{)}}_{\text{3}}{\left({\text{NO}}_2\right)}_3\right]$.

Oxidation state of Nitrite $\left({\text{NO}}_2{}^{-}\right)$ is $-1$.

Oxidation state of neutral Ammonia $\left({\text{NH}}_3\right)$ is $0$.

It is assumed that the oxidation state of metal ion Cobalt $\left(\text{Co}\right)$ is $x$.

The given complex compound contains three Ammonia molecules and three Nitrite molecule. Since, the overall charge on stable compound is zero. Therefore,

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

Therefore, the oxidation state of Cobalt is $+3$.

The given complex compound contains three Ammonia molecules and three Nitrite molecule. Hence, the name of anion is triamminetrinitro.

Therefore, the name of complex ion $\left[{\text{Co(NH}}_3{\text{)}}_{\text{3}}{\left({\text{NO}}_2\right)}_3\right]$ is triamminetrinitro-N-cobalt(III).

Conclusion

Conclusion

In the naming complex ion, anion is named first, whereas metal ion is named second.