Chapter 21, Problem 38E

 (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 names 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 $\left[{\text{Cr(NH}}_3{\text{)}}_{\text{6}}\right]{\text{Cl}}_{\text{2}}$

Answer to Problem 38E

Answer

The name of complex compound $\left[{\text{Cr(H}}_2{\text{O)}}_{\text{5}}\text{Br}\right]{\text{Br}}_{\text{2}}$ is pentaaquabromochromium(III) bromide.

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Cr(H}}_2{\text{O)}}_{\text{5}}\text{Br}\right]{\text{Br}}_{\text{2}}$.

Oxidation state of Bromine ion $\left({\text{Br}}^{-}\right)$ 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 Chromium $\left(\text{Cr}\right)$ is $x$.

The given complex compound contains five water molecules and three bromine 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 Chromium is $+3$.

The given complex compound contains five water molecules and one bromine ion. Hence, the name of anion is pentaaquabromo. It also contains counter ion bromide salt.

Therefore, the name of the complex ion $\left[{\text{Cr(H}}_2{\text{O)}}_{\text{5}}\text{Br}\right]{\text{Br}}_{\text{2}}$ is pentaaquabromochromium(III) bromide.

(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 names 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 ${\text{Na}}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$

Answer to Problem 38E

Answer

The name of complex ion ${\text{Na}}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$ is sodium hexacyanocobaltate(III).

Explanation of Solution

Explanation

The complex compound is ${\text{Na}}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$.

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

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

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

The given complex compound contains three Sodium atoms (counter cation) and six Cyanide atoms. Since, the overall charge on stable compound is zero. Therefore,

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

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

The given complex compound contains six Cyanide ligands. Hence, the name of anion is hexacyano. It also contains counter cation Sodium. The Latin name of Cobalt is Cobaltate.

Therefore, the name of complex ion ${\text{Na}}_{\text{3}}\left[\text{Co}{\left(\text{CN}\right)}_{\text{6}}\right]$ is sodium hexacyanocobaltate(III).

(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 names 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.

Todetermine: The name of the complex compound $\left[{\text{Fe(NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2{\text{)}}_{\text{2}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}\right]\text{Cl}$

Answer to Problem 38E

Answer

The name of complex compound   $\left[{\text{Fe(NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2{\text{)}}_{\text{2}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}\right]\text{Cl}$ is bis(ethylene diammine)dinitito-N-iron(III) chloride.

Explanation of Solution

Explanation

The complex compound is $\left[{\text{Fe(NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2{\text{)}}_{\text{2}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}\right]\text{Cl}$.

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

Oxidation state of neutral ligand Ethylenediammine $\left({\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2\right)$ is $0$.

Oxidation state of ligand ${\text{NO}}_2{}^{-}$ is $-1$.

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

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

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

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

The compound contains two ligands and one counter ion.

Since, the ligand Ethylenediammine already contains ‘di’. Therefore, ‘bis’ is used in the prefix of anion instead of ‘di’. Hence, the name of anion is bis(ethylene diammine)dinitito. The compound also contains counter ion Chloride salt.

Therefore, the name of complex ion $\left[{\text{Fe(NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2{\text{)}}_{\text{2}}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}\right]\text{Cl}$ is bis(ethylene diammine)dinitito-N-iron(III) chloride.

(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 names 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 $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$

Answer to Problem 38E

Answer

The name of complex ion $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ is tetraamminediiodoplatinum(IV) tetraiodoplatinate(II).

Explanation of Solution

Explanation

The complex compound is $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$. In this compound, the oxidation state of first platinum should be written by looking at the charge of the other coordination sphere, that is, ${\text{I}}_{\text{4}}$. Since, the total charge on ion sphere ${\text{I}}_{\text{4}}$ is $-4$, therefore, the charge on first platinum should be $+4$ to balance the charge.

Similarly, oxidation state of second platinum should be calculated according to the charge on first ion sphere ${\text{I}}_{\text{2}}$. Since, total charge on ion sphere ${\text{I}}_{\text{2}}$ is $-2$, therefore, the charge on second platinum should be $+2$ to balance the charge.

The first coordination sphere contains four Ammonia ligands and two Iodine ligands, whereas second ionsphere contains only two Iodine ligands.

The Latin name of Platinum is Platinate.

Therefore, the name of complex ion $\left[\text{Pt}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{I}}_{\text{2}}\right]\left[{\text{PtI}}_{\text{4}}\right]$ is tetraamminediiodoplatinum(IV) tetraiodoplatinate(II).

Conclusion

Conclusion

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