Chapter 2, Problem 105AE

(a)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(a)

The formula of the compound is $\left({\text{Ca}}_{\text{3}}{\text{N}}_{\text{2}}\right)$

The name of the binary compound is calcium nitride.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Ca}\text{and}\text{N}$.

The required compound is Calcium nitride $\left({\text{Ca}}_{\text{3}}{\text{N}}_{\text{2}}\right)$.

Calcium belongs to the Group $\text{II}\text{A}$, hence the oxidation state of calcium is $+2$.

Nitrogen is a non-metal of the $5^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =5-8\\ =-3\end{array}$

So, the formula of the given compound is ${\text{Ca}}_{\text{3}}{\text{N}}_{\text{2}}$.

The name of the binary compound.

The name of the binary compound is calcium nitride.

The name of the cation present is Calcium $\left(\text{Ca}\right)$.

The anion present is Nitride $\left({\text{N}}^{3-}\right)$.

Calcium only exists in $+2$ oxidation state.

Hence, the naming of this binary compound is, Calcium nitride.

(b)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(b)

The formula of the compound is $\left({\text{K}}_2\text{O}\right)$.

The name of the binary compound is Potassium oxide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{K}\text{and}\text{O}$

The required compound is Potassium oxide $\left({\text{K}}_2\text{O}\right)$.

Potassium belongs to the Group $\text{I}\text{A}$, hence the oxidation state of Potasium is $+1$.

Oxygen belongs to the $6^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =6-8\\ =-2\end{array}$

So, the formula of the given compound is ${\text{K}}_2\text{O}$.

The name of the binary compound.

The name of the binary compound is Potassium oxide.

The name of the cation present is Potassium $\left(\text{K}\right)$.

The anion present is Oxide $\left({\text{O}}^{2-}\right)$.

Potassium only exists in $+1$ oxidation state.

Hence, the naming of this binary compound is, Potassium oxide.

(c)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(c)

The formula of the compound is $\left(\text{RbF}\right)$.

The name of the binary compound is Rubidium fluoride.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Rb}\text{and}\text{F}$.

The required compound is Rubidium fluoride $\left(\text{RbF}\right)$.

Rubidium belongs to the Group $\text{I}\text{A}$, hence the oxidation state of rubidium is $+1$.

Fluorine belongs to the $7^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =7-8\\ =-1\end{array}$

So, the formula of the given compound is $\text{RbF}$.

The name of the binary compound.

The name of the binary compound is Rubidium fluoride.

The name of the cation present is Rubidium $\left(\text{Rb}\right)$.

The anion present is Fluoride $\left({\text{F}}^{-}\right)$.

Hence, the naming of this binary compound is, Rubidium fluoride.

(d)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(d)

The formula of the compound is $\left(\text{MgS}\right)$.

The name of the binary compound is Magnesium sulphide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Mg}\text{and}\text{S}$.

The required compound is Magnesium sulphide $\left(\text{MgS}\right)$.

Magnesium belongs to the Group $\text{II}\text{A}$, hence the oxidation state of Magnesium is $+2$.

Sulphur belongs to the $6^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =6-8\\ =-2\end{array}$

So, the formula of the given compound is $\text{MgS}$.

The name of the binary compound.

The name of the binary compound is Magnesium sulphide.

The name of the cation present is Magnesium $\left(\text{Mg}\right)$.

The anion present is Sulphide $\left({\text{S}}^{2-}\right)$.

Hence, the naming of this binary compound is, Magnesium sulphide.

(e)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(e)

The formula of the compound is $\left({\text{BaI}}_2\right)$.

The name of the binary compound is Barium iodide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Ba}\text{and}\text{I}$

The required compound is Barium iodide $\left({\text{BaI}}_2\right)$.

Barium belongs to the Group $\text{II}\text{A}$, hence the oxidation state of magnesium is $+2$.

Iodine belongs to the $7^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =7-8\\ =-1\end{array}$

So, the formula of the given compound is ${\text{BaI}}_{\text{2}}$.

The name of the binary compound.

The name of the binary compound is barium iodide.

The name of the cation present is Barium $\left(\text{Ba}\right)$.

The anion present is Iodide $\left({\text{I}}^{-}\right)$.

Hence, the naming of this binary compound is, Barium iodide.

(f)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(f)

The formula of the compound is $\left({\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}\right)$.

The name of the binary compound is Aluminium selenide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Al}\text{and}\text{Se}$.

The required compound is Aluminium selenide $\left({\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}\right)$.

Aluminium belongs to the Group $\text{III}\text{A}$, hence the oxidation state of magnesium is $+3$.

Selenium belongs to the $6^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =6-8\\ =-2\end{array}$

So, the formula of the given compound is ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$.

The name of the binary compound.

The name of the binary compound is Aluminium selenide.

The name of the cation present is Aluminium $\left(\text{Al}\right)$.

The anion present is Selenide $\left({\text{Se}}^{2-}\right)$.

Hence, the naming of this binary compound is, Aluminium selenide.

(g)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(g)

The formula of the compound is $\left({\text{Cs}}_3\text{P}\right)$.

The name of the binary compound is Cesium phosphide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{Cs}\text{and}\text{P}$.

The required compound is Cesium phosphide $\left({\text{Cs}}_3\text{P}\right)$.

Cesium belongs to the Group $\text{I}\text{A}$, hence the oxidation state of magnesium is $+1$.

Phosphorous belongs to the $5^{\text{th}}$ group.

Hence, its oxidation state $\begin{array}{l}=\text{Group}\text{number}-8\\ =5-8\\ =-3\end{array}$

So, the formula of the given compound is ${\text{Cs}}_3\text{P}$.

The name of the binary compound.

The name of the binary compound is Cesium phosphide.

The name of the cation present is Cesium $\left(\text{Cs}\right)$.

The anion present is Phosphide $\left({\text{P}}^{3-}\right)$.

Hence, the naming of this binary compound is, Cesium phosphide.

(h)

Interpretation Introduction

Interpretation:

The formula and the name of the binary compound formed from the given pairs of elements in each case is to be stated.

Concept introduction:

The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as $\text{group number}-8$.  While naming an ionic compound, the cation is named first followed by the naming of the anion.  If the cation exhibits more than one oxidation state then the current oxidation state of the cation is to be mentioned.

Answer to Problem 105AE

(h)

The formula of the compound is $\left({\text{InBr}}_{\text{3}}\right)$.

The name of the binary compound is Indium(III) bromide.

Explanation of Solution

To determine: The formula and the name of the binary compound formed from $\text{In}\text{and}\text{Br}$.

The required compound is Indium(III) bromide $\left({\text{InBr}}_{\text{3}}\right)$.

Indium has an oxidation state of $+3$.

Oxidation state exhibited by Bromine is $-1$.

So, the formula of the given compound is ${\text{InBr}}_{\text{3}}$.

The name of the binary compound.

The name of the binary compound is Indium(III) bromide.

The name of the cation present is Indium $\left(\text{In}\right)$.

The anion present is Bromide $\left({\text{Br}}^{-}\right)$.

Here, indium exhibits the $+3$ oxidation state.

Hence, the naming of this binary compound is, Indium(III) bromide.