Chapter 11, Problem 71E

Interpretation Introduction

(a)

Interpretation:

The systematic name of ${\text{MgCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

A crystalline compound with fixed number of water molecules are known as hydrates. The water molecules are also known as water of crystallization or water of hydration. Each water molecule is attached to a single unit. According to IUPAC nomenclature, the naming of hydrates is done by first naming the anhydrous compound followed by the number of water molecules and hydrate. Also, the oxidation number of metal is written beside the name of anhydrous compound within parenthesis.

Answer to Problem 71E

The systematic name of ${\text{MgCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is magnesium chloride hexahydrate.

Explanation of Solution

In the chemical formula, ${\text{MgCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is magnesium chloride. The number of water molecules present is six. The prefix used to indicate the number of water molecules is hexa. Therefore, the systematic name of ${\text{MgCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is magnesium chloride hexahydrate.

Conclusion

The systematic name of ${\text{MgCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is stated above.

Interpretation Introduction

(b)

Interpretation:

The systematic name of ${\text{MnSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

A crystalline compound with fixed number of water molecules are known as hydrates. The water molecules are also known as water of crystallization or water of hydration. Each water molecule is attached to a single unit. According to IUPAC nomenclature, the naming of hydrates is done by first naming the anhydrous compound followed by the number of water molecules and hydrate. Also the oxidation number of metal is written beside the name of anhydrous compound within parenthesis.

Answer to Problem 71E

The systematic name of ${\text{MnSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$ is manganese sulphate $\left(\text{II}\right)$ heptahydrate.

Explanation of Solution

In the chemical formula, ${\text{MnSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is manganese sulphate. The number of water molecules present is seven. The prefix used to indicate the number of water molecules is hepta.

Water molecules do not contribute in the oxidation number of metal. The oxidation number of manganese is calculated as shown below.

$C={\left(O_N\right)}_{\text{Mn}}+{\left(O_N\right)}_{{\text{SO}}_{{}_4}^{2-}}$…(1)

Where,

• $C$ is the overall charge.

• ${\left(O_N\right)}_{{\text{SO}}_{{}_4}^{2-}}$ is the oxidation number of sulphate molecule.

• ${\left(O_N\right)}_{\text{Mn}}$ is the oxidation number of manganese.

Substitute $C$ as zero and ${\left(O_N\right)}_{{\text{SO}}_{{}_4}^{2-}}$ as $-2$ in equation (1).

$\begin{array}{rll}0& =& {\left(O_N\right)}_{\text{Mn}}+\left(-2\right)\\ {\left(O_N\right)}_{\text{Mn}}& =& +2\end{array}$

Therefore, the systematic name of ${\text{MnSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$ is manganese sulphate $\left(\text{II}\right)$ heptahydrate.

Conclusion

The systematic name of ${\text{MnSO}}_{\text{4}}\cdot 7{\text{H}}_{\text{2}}\text{O}$ is stated above.

Interpretation Introduction

(c)

Interpretation:

The systematic name of $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

A crystalline compound with fixed number of water molecules are known as hydrates. The water molecules are also known as water of crystallization or water of hydration. Each water molecule is attached to a single unit. According to IUPAC nomenclature, the naming of hydrates is done by first naming the anhydrous compound followed by the number of water molecules and hydrate. Also the oxidation number of metal is written beside the name of anhydrous compound within parenthesis.

Answer to Problem 71E

The systematic name of $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ is cobalt cyanide $\left(\text{III}\right)$ trihydrate.

Explanation of Solution

In the chemical formula, $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is cobalt cyanide. The number of water molecules present is three. The prefix used to indicate the number of water molecules is tri.

Water molecules do not contribute in the oxidation number of metal. The oxidation number of cobalt is calculated as shown below.

$C={\left(O_N\right)}_{\text{Co}}+x{\left(O_N\right)}_{{\text{CN}}^{-}}$…(1)

Where,

• $C$ is the overall charge.

• ${\left(O_N\right)}_{{\text{CN}}^{-}}$ is the oxidation number of cyanide molecule.

• $x$ is the number of cyanide molecules.

• ${\left(O_N\right)}_{\text{Co}}$ is the oxidation number of cobalt.

Substitute $C$ as zero $x$ as $3$ and ${\left(O_N\right)}_{{\text{CN}}^{-}}$ as $-1$ in equation (1).

$\begin{array}{rll}0& =& {\left(O_N\right)}_{\text{Co}}+3\left(-1\right)\\ {\left(O_N\right)}_{\text{Cr}}& =& +3\end{array}$

Therefore, the systematic name of $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ is cobalt cyanide $\left(\text{III}\right)$ trihydrate.

Conclusion

The systematic name of $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ is stated above.

Interpretation Introduction

(d)

Interpretation:

The systematic name of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{9H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

A crystalline compound with fixed number of water molecules are known as hydrates. The water molecules are also known as water of crystallization or water of hydration. Each water molecule is attached to a single unit. According to IUPAC nomenclature, the naming of hydrates is done by first naming the anhydrous compound followed by the number of water molecules and hydrate. Also the oxidation number of metal is written beside the name of anhydrous compound within parenthesis.

Answer to Problem 71E

The systematic name of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{9H}}_{\text{2}}\text{O}$ is chromium nitrate $\left(\text{III}\right)$ nonahydrate.

Explanation of Solution

In the chemical formula, $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{9H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is chromium nitrate. The number of water molecules present is nine. The prefix used to indicate the number of water molecules is nona.

Water molecules do not contribute in the oxidation number of metal. The oxidation number of chromium is calculated as shown below.

$C={\left(O_N\right)}_{\text{Cr}}+x{\left(O_N\right)}_{{\text{NO}}_3^{-}}$…(1)

Where,

• $C$ is the overall charge

• ${\left(O_N\right)}_{{\text{NO}}_3^{-}}$ is the oxidation number of nitrate molecule.

• $x$ is the number of nitrate molecules.

• ${\left(O_N\right)}_{\text{Cr}}$ is the oxidation number of chromium.

Substitute $C$ as zero $x$ as $3$ and ${\left(O_N\right)}_{{\text{NO}}_3^{-}}$ as $-1$ in equation (1).

$\begin{array}{rll}0& =& {\left(O_N\right)}_{\text{Cr}}+3\left(-1\right)\\ {\left(O_N\right)}_{\text{Cr}}& =& +3\end{array}$

Therefore, the systematic name of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{9H}}_{\text{2}}\text{O}$ is chromium nitrate $\left(\text{III}\right)$ nonahydrate.

Conclusion

the systematic name of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{9H}}_{\text{2}}\text{O}$ is stated above.