Chapter 2, Problem 55E

(a)

Interpretation Introduction

Interpretation: The name of the compound $\text{CuI}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. The element which is positioned left in periodic table is named first followed by another element. The second element’s name also ends with suffix “ide”. Hence, compound’s formula can be easily be utilized to find its name.

Answer to Problem 55E

The name of the compound $\text{CuI}$ is Copper (I) iodide.

Explanation of Solution

The given compound, $\text{CuI}$ contains one copper atom and one iodine atom. Since, copper is positioned left to iodine in periodic table, its name is written first.

The oxidation state of the element is written in bracket after naming the element. The oxidation state of copper in the given compound is $+1$ .

Hence, the compound $\text{CuI}$ has the name Copper (I) iodide.

(b)

Interpretation Introduction

Interpretation: The name of the compound ${\text{CuI}}_2$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. The element which is positioned left in periodic table is named first followed by another element. The second element’s name also ends with suffix “ide”. Hence, compound’s formula can be easily be utilized to find its name.

Answer to Problem 55E

The name of the compound ${\text{CuI}}_2$ is copper (II) iodide.

Explanation of Solution

The given compound, ${\text{CuI}}_2$ contains one copper atom and one iodine atom. Since, copper is positioned left to iodine in periodic table, it’s name is written first.

The oxidation state of the element is written in bracket after naming the element. The oxidation state of copper in the given compound is $+2$ .

Hence, the compound ${\text{CuI}}_2$ has the name Copper (II) iodide.

(c)

Interpretation Introduction

Interpretation: The name of the compound ${\text{CoI}}_2$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. The element which is positioned left in periodic table is named first followed by another element. The second element’s name also ends with suffix “ide”. Hence, compound’s formula can be easily be utilized to find its name.

Answer to Problem 55E

The name of the compound ${\text{CoI}}_2$ is Cobalt (II) iodide.

Explanation of Solution

The given compound, ${\text{CoI}}_2$ contains one cobalt atom and one iodine atom. Since, cobalt is positioned left to iodine in periodic table, it’s name is written first.

The oxidation state of the element is written in bracket after naming the element. The oxidation state of cobalt in the given compound is $+2$ .

Hence, the compound ${\text{CoI}}_2$ has the name Cobalt (II) iodide.

(d)

Interpretation Introduction

Interpretation: The name of the compound ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. In case of ionic compounds, cation’s (positive ion) name is always given first followed by anion’s (negative ion) name. The cation carries its elemental name. Anion and cation combine in a way that overall neutral compound gets generated.

Answer to Problem 55E

The name of the compound ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is sodium carbonate.

Explanation of Solution

The given compound is ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

The cation in ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is sodium ion $\left({\text{Na}}^{+}\right)$ and the anion is carbonate ion $\left({\text{CO}}_3^{2-}\right)$ .

The charge on sodium cation is $+1$ and that on carbonate anion is $-2$ . In order to give a neutral compound, $\text{2}{\text{Na}}^{+}$ combines with $1{\text{CO}}_3^{2-}$ .

Hence, the compound ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ has the name sodium carbonate.

(e)

Interpretation Introduction

Interpretation: The name of the compound ${\text{NaHCO}}_3$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. In case of ionic compounds, cation’s (positive ion) name is always given first followed by anion’s (negative ion) name. The cation carries its elemental name. Anion and cation combine in a way that overall neutral compound gets generated.

Answer to Problem 55E

The name of the compound ${\text{NaHCO}}_3$ is sodium bicarbonate.

Explanation of Solution

The given compound is ${\text{NaHCO}}_3$ .

The cation in ${\text{NaHCO}}_3$ is sodium ion $\left({\text{Na}}^{+}\right)$ and the anion is bicarbonate ion $\left({\text{HCO}}_3^{-}\right)$ .

The charge on sodium cation is $+1$ and that on bicarbonate anion is $-1$ . In order to give a neutral compound,   $1{\text{Na}}^{+}$ . combines with $1{\text{HCO}}_3^{-}$ .

Hence, the compound ${\text{NaHCO}}_3$ has the name sodium bicarbonate.

(f)

Interpretation Introduction

Interpretation: The name of the compound ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. The element which is positioned left in periodic table is named first followed by another element. The second element’s name also ends with suffix “ide”. Hence, compound’s formula can be easily be utilized to find its name.

Answer to Problem 55E

The name of the compound ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ is tetrasulphur tetranitride.

Explanation of Solution

The given compound is ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ .

The compound contains four sulphur atoms and four nitrogen atoms. Since, sulphur is positioned left to nitrogen in periodic table, its name is written first.

Hence, the compound ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ has the name tetrasulphur tetranitride.

(g)

Interpretation Introduction

Interpretation: The name of the compound ${\text{SeBr}}_{\text{4}}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. The element which is positioned left in periodic table is named first followed by another element. The second element’s name also ends with suffix “ide”. Hence, compound’s formula can be easily be utilized to find its name.

Answer to Problem 55E

The name of the compound ${\text{SeBr}}_{\text{4}}$ is Selenium (IV) bromide.

Explanation of Solution

The given compound is ${\text{SeBr}}_{\text{4}}$ .

The compound contains one selenium atom and one bromine atom. Since, selenium is positioned left to bromine in periodic table, its name is written first.

The oxidation state of the element is written in bracket after naming the element. The oxidation state of selenium in the given compound is $+4$ .

Hence, the compound ${\text{SeBr}}_{\text{4}}$ has the name Selenium (IV) bromide.

(h)

Interpretation Introduction

Interpretation: The name of the compound $\text{NaOCl}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. In case of ionic compounds, cation’s (positive ion) name is always given first followed by anion’s (negative ion) name. The cation carries its elemental name. Anion and cation combine in a way that overall neutral compound gets generated.

Answer to Problem 55E

The name of the compound $\text{NaOCl}$ is Sodium hypochlorite.

Explanation of Solution

The given compound is $\text{NaOCl}$ .

The cation in $\text{NaOCl}$ is sodium ion $\left({\text{Na}}^{+}\right)$ and the anion is hypochlorite ion $\left({\text{OCl}}^{-}\right)$ .

The charge on sodium cation is $+1$ and that on hypochlorite anion is $-1$ . In order to give a neutral compound, $1{\text{Na}}^{+}$ combines with $1{\text{OCl}}^{-}$ .

Hence, the compound $\text{NaOCl}$ has the name Sodium hypochlorite.

(i)

Interpretation Introduction

Interpretation: The name of the compound ${\text{BaCrO}}_{\text{4}}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. In case of ionic compounds, cation’s (positive ion) name is always given first followed by anion’s (negative ion) name. The cation carries its elemental name. Anion and cation combine in a way that overall neutral compound gets generated.

Answer to Problem 55E

The name of the compound ${\text{BaCrO}}_{\text{4}}$ is Barium chromate.

Explanation of Solution

The given compound is ${\text{BaCrO}}_{\text{4}}$ .

The cation in ${\text{BaCrO}}_{\text{4}}$ is barium ion $\left({\text{Ba}}^{2+}\right)$ and the anion is chromate ion $\left({\text{CrO}}_4^{2-}\right)$ .

The charge on barium cation is $+2$ and that on chromate anion is $-2$ . In order to give a neutral compound, $1{\text{Ba}}^{2+}$ combines with $1{\text{CrO}}_4^{2-}$ .

Hence, the compound ${\text{BaCrO}}_{\text{4}}$ has the name Barium chromate.

(j)

Interpretation Introduction

Interpretation: The name of the compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is to be stated.

Concept introduction: A chemical compound’s name can be given via elements or ions that are part of that compound. This is done using suitable suffix and prefix. In case of ionic compounds, cation’s (positive ion) name is always given first followed by anion’s (negative ion) name. The cation carries its elemental name. Anion and cation combine in a way that overall neutral compound gets generated.

Answer to Problem 55E

The name of the compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is Ammonium nitrate.

Explanation of Solution

The given compound is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ .

The cation in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is ammonium ion $\left({\text{NH}}_4^{+}\right)$ and the anion is chromate ion $\left({\text{NO}}_3^{-}\right)$ .

The charge on ammonium cation is $+1$ and that on nitrate anion is $-1$ . In order to give a neutral compound, $1{\text{NH}}_4^{+}$ combines with $1{\text{NO}}_3^{-}$ .

Hence, the compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ has the name Ammonium nitrate.