Chapter 11, Problem 72E

Interpretation Introduction

(a)

Interpretation:

The systematic name of ${\text{MgSO}}_4\cdot 7{\text{H}}_{\text{2}}\text{O}$ 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 72E

The systematic name of ${\text{MgSO}}_4\cdot 7{\text{H}}_{\text{2}}\text{O}$ is magnesium sulphate heptahydrate.

Explanation of Solution

In the chemical formula, ${\text{MgSO}}_4\cdot 7{\text{H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is magnesium sulphate. The number of water molecules present is seven. The prefix used to indicate the number of water molecules is hepta. Therefore, the systematic name of ${\text{MgSO}}_4\cdot 7{\text{H}}_{\text{2}}\text{O}$ is magnesium sulphate heptahydrate.

Conclusion

The systematic name of ${\text{MgSO}}_4\cdot 7{\text{H}}_{\text{2}}\text{O}$ is stated above.

Interpretation Introduction

(b)

Interpretation:

The systematic name of ${\text{MnSO}}_{\text{4}}\cdot {\text{H}}_{\text{2}}\text{O}$ 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 72E

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

Explanation of Solution

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

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 one.

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

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The systematic name of $\text{Co}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$ 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 72E

The systematic name of $\text{Co}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$ is cobalt acetate $\left(\text{II}\right)$ tetrahydrate.

Explanation of Solution

In the chemical formula, $\text{Co}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$, the name of anyhdrous compound is cobalt acetate. The number of water molecules present is four. The prefix used to indicate the number of water molecules is tetra.

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

$C={\left(O_N\right)}_{\text{Co}}+x{\left(O_N\right)}_{{\text{CH}}_3{\text{COO}}^{-}}$…(1)

Where,

• $C$ is the overall charge

• ${\left(O_N\right)}_{{\text{CH}}_3{\text{COO}}^{-}}$ is the oxidation number of acetate molecule.

• $x$ is the number of acetate molecules.

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

Substitute $C$ as zero, $x$ as $2$ and ${\left(O_N\right)}_{{\text{CH}}_3{\text{COO}}^{-}}$ as $-1$ in equation (1).

$\begin{array}{rll}0& =& {\left(O_N\right)}_{\text{Co}}+2\left(-1\right)\\ {\left(O_N\right)}_{\text{Cr}}& =& +2\end{array}$

Therefore, the systematic name of $\text{Co}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$ is cobalt acetate $\left(\text{II}\right)$ tetrahydrate.

Conclusion

The systematic name of $\text{Co}{\left({\text{C}}_2{\text{H}}_3{\text{O}}_2\right)}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$ is stated above.

Interpretation Introduction

(d)

Interpretation:

The systematic name of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ 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 72E

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

Explanation of Solution

In the chemical formula, $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$. the name of anyhdrous compound is chromium nitrate. 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 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 $3$ ${\left(O_N\right)}_{{\text{NO}}_3^{-}}$ 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 $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ is chromium nitrate $\left(\text{III}\right)$ trihydrate.

Conclusion

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