Chapter 2, Problem 54E

(a)

Interpretation Introduction

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

Concept introduction:Thegeneration of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{NaClO}}_{\text{4}}$ has the name Sodium perchlorate.

Explanation of Solution

The cation in ${\text{NaClO}}_{\text{4}}$ is sodium cation $\left({\text{Na}}^{+}\right)$ and the anion is perchlorate anion $\left({\text{ClO}}_4^{-}\right)$ .

The charge on sodium is $+1$ and that on perchlorate is $-1$ giving an overall neutral compound.

Hence, the compound ${\text{NaClO}}_{\text{4}}$ has the name Sodium perchlorate.

(b)

Interpretation Introduction

Interpretation:The name of the compound ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ has the name Magnesium phosphate.

Explanation of Solution

The cation in ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is magnesium ion $\left({\text{Mg}}^{2+}\right)$ and the anion is phosphate ion $\left({\text{PO}}_4^{3-}\right)$ .

The charge on magnesium cation is $+2$ and that on phosphate anion is $-3$ . In order to give a neutral compound,   $\text{3}{\text{Mg}}^{2+}$ combines with $\text{2}{\text{PO}}_4^{3-}$ .

Hence, the compound ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ has the nameMagnesium phosphate.

(c)

Interpretation Introduction

Interpretation:The name of the compound ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ has the nameAluminium sulphate.

Explanation of Solution

The cation in ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ is aluminium cation $\left({\text{Al}}^{3+}\right)$ and the anion is sulphate anion $\left({\text{SO}}_4^{2-}\right)$ .

The charge on aluminiumcation is $+3$ and that on sulphate anion is $-2$ . In order to give a neutral compound,   $2{\text{Al}}^{3+}$ combines with $3{\text{SO}}_4^{2-}$ .

Hence, the compound ${\text{Al}}_2{\left({\text{SO}}_{\text{4}}\right)}_3$ has the nameAluminium sulphate.

(d)

Interpretation Introduction

Interpretation:The name of the compound ${\text{SF}}_{\text{2}}$ is to be predicted.

Concept introduction:Covalent compounds are generated as result of electron sharing among the elements involved. One important point in naming covalent compounds is the use of prefixes which reveal number of atoms involved. The first elememt never makes use of the prefix “mono” but uses the others.The element positioned on the left in perodic table in relation to the other is name first.

Answer to Problem 54E

The compound, ${\text{SF}}_{\text{2}}$ has the name Sulphur difluoride.

Explanation of Solution

The compound contains one Sulphur atom and two fluorine atoms. Since, Sulphur is position left to fluorine in peridic table, it’s name is written first.

Owing to the two fluorine atoms, the prefix “di” is used in the name.

Hence, the compound ${\text{SF}}_{\text{2}}$ has the name Sulphur difluoride.

(e)

Interpretation Introduction

Interpretation:The name of the compound ${\text{SF}}_6$ is to be predicted.

Concept introduction:Covalent compounds are generated as result of electron sharing among the elements involved. One important point in naming covalent compounds is the use of prefixes which reveal number of atoms involved. The first elememt never makes use of the prefix “mono” but uses the others. The element positioned on the left in perodic table in relation to the other is name first.

Answer to Problem 54E

The compound, ${\text{SF}}_6$ has the name Sulphur hexafluoride.

Explanation of Solution

The compound contains one Sulphur atom and six fluorine atoms.Since, Sulphur is position left to fluorine in peridic table, it’s name is written first.

Owing to the six fluorine atoms, the prefix “hexa” is used in the name.

Hence, the compound ${\text{SF}}_6$ has the name Sulphur hexafluoride.

(f)

Interpretation Introduction

Interpretation:The name of the compound ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ has the name Sodium hydrogen phosphate.

Explanation of Solution

The cation in ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is sodium cation $\left({\text{Na}}^{+}\right)$ and the anion is hydrogen phosphate anion $\left({\text{HPO}}_4^{2-}\right)$ .

The charge on sodium cation is $+1$ and that on hydrogen phosphate anion is $-2$ . In order to give a neutral compound,   $2{\text{Na}}^{+}$ combines with $1{\text{HPO}}_4^{2-}$ .

Hence, thecompound ${\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}$ has the nameSodium hydrogen phosphate.

(g)

Interpretation Introduction

Interpretation:The name of the compound ${\text{NaH}}_2{\text{PO}}_{\text{4}}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{NaH}}_2{\text{PO}}_{\text{4}}$ has the name Sodium dihydrogen phosphate.

Explanation of Solution

The cation in ${\text{NaH}}_2{\text{PO}}_{\text{4}}$ is sodium cation $\left({\text{Na}}^{+}\right)$ and the anion is dihydrogen phosphate anion $\left({\text{H}}_2{\text{PO}}_4^{-}\right)$ .

The charge on sodium cation is $+1$ and that on dihydrogen phosphate anion is $-1$ which on combining gives an overall neutral compound.

Hence, the compound ${\text{NaH}}_2{\text{PO}}_{\text{4}}$ has the nameSodium dihydrogen phosphate.

(h)

Interpretation Introduction

Interpretation:The name of the compound ${\text{Li}}_{\text{3}}\text{N}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{Li}}_{\text{3}}\text{N}$ has the name Lithium nitride.

Explanation of Solution

The cation in ${\text{Li}}_{\text{3}}\text{N}$ is lithium cation $\left({\text{Li}}^{+}\right)$ and the anion is nitride anion $\left({\text{N}}_3^{-}\right)$ .

The charge on lithium cation is $+1$ and that on nitride anion is $-3$ . Hence, $3{\text{Li}}^{+}$ combines with $1{\text{N}}_3^{-}$ to ultimately give a neutral compound.

Hence, the compound ${\text{Li}}_{\text{3}}\text{N}$ has the name Lithium nitride.

(i)

Interpretation Introduction

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

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, $\text{NaOH}$ has the name Sodium hydroxide.

Explanation of Solution

The cation in $\text{NaOH}$ is sodium cation $\left({\text{Na}}^{+}\right)$ and the anion is hydroxide anion $\left({\text{OH}}^{-}\right)$ .

The charge on sodium cation is $+1$ and that on hydroxide anion is $-1$ and hence, on combinination gives an overall neutral compound.

Hence, the compound $\text{NaOH}$ has the name Sodium hydroxide.

(j)

Interpretation Introduction

Interpretation:The name of the compound $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ has the name Magnessium hydroxide.

Explanation of Solution

The cation in $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ is magnesium cation $\left({\text{Mg}}^{2+}\right)$ and the anion is hydroxide anion $\left({\text{OH}}^{-}\right)$ .

The charge on magnessium cation is $+2$ and that on hydroxide anion is $-1$ and hence, $\text{1}{\text{Mg}}^{2+}$ combines with $\text{2}{\text{OH}}^{-}$ to give an overall neutral compound.

Hence, the compound $\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$ has the nameMagnessium hydroxide.

(k)

Interpretation Introduction

Interpretation:The name of the compound $\text{Al}{\left(\text{OH}\right)}_3$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, $\text{Al}{\left(\text{OH}\right)}_3$ has the name Aluminiumhydroxide.

Explanation of Solution

The cation in $\text{Al}{\left(\text{OH}\right)}_3$ is aluminium cation $\left({\text{Al}}^{3+}\right)$ and the anion is hydroxide anion $\left({\text{OH}}^{-}\right)$ .

The charge on aluminium cation is $+3$ and that on hydroxide anion is $-1$ and hence, $\text{1}{\text{Al}}^{3+}$ combines with $3{\text{OH}}^{-}$ to give an overall neutral compound.

Hence, the compound $\text{Al}{\left(\text{OH}\right)}_3$ has the name Aluminiumhydroxide.

(l)

Interpretation Introduction

Interpretation:The name of the compound ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be predicted.

Concept introduction:The generation of ionic compounds involve combination of anion with cation. While naming these ionic compounds, some very simple rules are to be kept in mind. The naming of cation is always done first followed by naming anion. Cation is named by its elemental name however in naming an anion the end part of elemental name is substituted by the suffix “ide”.

Answer to Problem 54E

The compound, ${\text{Ag}}_2{\text{CrO}}_4$ has the name Silver chromate.

Explanation of Solution

The cation in ${\text{Ag}}_2{\text{CrO}}_4$ is silver cation $\left({\text{Ag}}^{+}\right)$ and the anion is chromate anion $\left({\text{CrO}}_4^{2-}\right)$ .

The charge on silver cation is $+1$ and that on chromate anion is $-2$ . Hence, to form a neutral compound $\text{2}{\text{Ag}}^{+}$ reacts with $\text{1}{\text{CrO}}_4^{2-}$ .

Hence, the compound ${\text{Ag}}_2{\text{CrO}}_4$ has the name Silver chromate.