Chapter 2, Problem 2.85QP

(a)

Interpretation Introduction

Interpretation: The chemical formulas of certain compounds are given. The chemical names of the given compounds are to be determined.

Concept introduction: According to IUPAC nomenclature, if the compound is made of metal and non metal, then the name of the compound begins from name of the metal followed by the oxidation state of metal and the name of the non metal. The metal name remains same as in the periodic table, but the name of non metals end with “ide”. The oxidation state is always composed in roman words within the bracket.

To determine: The name of given chemical formula.

Answer to Problem 2.85QP

Solution

The chemical compound ${\text{Cr}}_2{\text{Te}}_3$ is named as chromium $\left(\text{III}\right)$ telluride.

Explanation of Solution

Explanation

The given chemical formula is ${\text{Cr}}_2{\text{Te}}_3$.

Chromium is a metal present in the d block of periodic table, whereas tellurium is a non metal present in ${16}^{\text{th}}$ group of the periodic table. Two chromium cations and three tellurium anions combine to form an electrically balanced compound.

Tellurium consists of $2-$ charge. The compound contains three tellurium atom thus it has $6-$ charge.

The oxidation state of the elements that can exhibit more than one oxidation state is mentioned after the name of the element.

The oxidation state $(x)$ of chromium is calculated by the formula,

$\left(\text{Sum}\text{of}\text{the}\text{oxidation}\text{state}\text{of}\text{the}\text{constituent}\text{atom}\right)=\left(\text{Total}\text{charge}\text{on}\text{the}\text{complex}\right)$

$\begin{array}{c}2x+\left(-6\right)=0\\ 2x=+6\\ x=\frac{6}{2}\\ x=3\end{array}$

Therefore, the chemical compound ${\text{Cr}}_2{\text{Te}}_3$ is named as chromium $\left(\text{III}\right)$ telluride.

(b)

Interpretation Introduction

To determine: The name of given chemical formula.

Answer to Problem 2.85QP

Solution

The chemical compound ${\text{V}}_2{\left({\text{SO}}_4\right)}_3$ is named as vanadium $\left(\text{III}\right)$ sulphate.

Explanation of Solution

Explanation

The chemical formula is ${\text{V}}_2{\left({\text{SO}}_4\right)}_3$.

Vanadium is a metal present in the d block of periodic table whereas sulphate is an anion. Two vanadium cations and three sulphate anions combine to form an electrically balance compound. The sulphate ion consists of $2-$ charge.

The compound contains three sulphate anions thus it has $6-$ charge. The sulphate is neither metal nor non metal. It is an anion.

Thus, its naming never ends with “ide”.

The oxidation state of the elements that can exhibit more than one oxidation state is mentioned after the name of the element.

The oxidation state $(x)$ of vanadium is calculated as,

$\left(\text{Sum}\text{of}\text{the}\text{oxidation}\text{state}\text{of}\text{the}\text{constituent}\text{atom}\right)=\left(\text{Total}\text{charge}\text{on}\text{the}\text{complex}\right)$

$\begin{array}{c}2x+\left(-6\right)=0\\ 2x=+6\\ x=\frac{6}{2}\\ x=3\end{array}$

Therefore, the chemical compound ${\text{V}}_2{\left({\text{SO}}_4\right)}_3$ is named as vanadium $\left(\text{III}\right)$ sulphate.

(c)

Interpretation Introduction

To determine: The name of given chemical formula.

Answer to Problem 2.85QP

Solution

The chemical compound ${\text{Fe}}_2{\text{CrO}}_4$ is named as iron $\left(\text{I}\right)$ chromate.

Explanation of Solution

Explanation

The chemical formula is ${\text{Fe}}_2{\text{CrO}}_4$.

Iron $\left(\text{Fe}\right)$ is a metal present in the d block of the periodic table whereas chromate is an anion. Two iron cations and one chromate anion $\left({\text{CrO}}_4{}^{2-}\right)$ combine to make electrically balance compound. The chromate ion consists of $2-$ charge. It never ends with “ide”. When numerous polyatomic ions are present then the names of polyatomic ions will end in "ate" or "ite".

The oxidation state of the elements that can exhibit more than one oxidation state is mentioned after the name of the element.

The oxidation state $(x)$ of iron is calculated as,

$\left(\text{Sum}\text{of}\text{the}\text{oxidation}\text{state}\text{of}\text{the}\text{constituent}\text{atom}\right)=\left(\text{Total}\text{charge}\text{on}\text{the}\text{complex}\right)$

$\begin{array}{c}2x-2=0\\ 2x=2\\ x=\frac{2}{2}\\ x=1\end{array}$

Therefore, the chemical compound ${\text{Fe}}_2{\text{CrO}}_4$ is named as iron $\left(\text{I}\right)$ chromate.

(d)

Interpretation Introduction

To determine: The name of given chemical formula.

Answer to Problem 2.85QP

Solution

The chemical compound $\text{MnO}$ is named as manganese $\left(\text{II}\right)$ oxide.

Explanation of Solution

Explanation

The chemical formula is $\text{MnO}$.

Manganese is a metal present in the d block of the periodic table whereas oxygen is non metal present in ${16}^{\text{th}}$ group of the periodic table. One manganese cation and one oxygen anion combine to make electrically balance compound. Both cation and anion consist of $2+$ and $2-$ anion respectively.

Oxygen naming ends in “ide” because it is a non metal.

The oxidation state of the elements that can exhibit more than one oxidation state is mentioned after the name of the element.

The oxidation state $(x)$ of manganese is calculated as,

$\left(\text{Sum}\text{of}\text{the}\text{oxidation}\text{state}\text{of}\text{the}\text{constituent}\text{atom}\right)=\left(\text{total}\text{charge}\text{on}\text{the}\text{complex}\right)$

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

Therefore, the chemical compound $\text{MnO}$ is named as manganese $\left(\text{II}\right)$ oxide.

Conclusion

1. a. The given compound ${\text{Cr}}_2{\text{Te}}_3$ is named as chromium $\left(\text{III}\right)$ telluride.
2. b. The given compound ${\text{V}}_2{\left({\text{SO}}_4\right)}_3$ is named as vanadium $\left(\text{III}\right)$ sulphate.
3. c. The given compound ${\text{Fe}}_2{\text{CrO}}_4$ is named as iron $\left(\text{I}\right)$ chromate.
4. d. The given compound $\text{MnO}$ is named as manganese $\left(\text{II}\right)$ oxide