Chapter 3, Problem 134AE

(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.

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

Answer to Problem 134AE

Answer

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

Explanation of Solution

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

Potassium oxide $\left({\text{K}}_2\text{O}\right)$.

Explanation of Solution

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

Rubidium fluoride $\left(\text{RbF}\right)$.

Explanation of Solution

Explanation

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

Magnesium sulphide $\left(\text{MgS}\right)$.

Explanation of Solution

Explanation

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

Barium iodide $\left({\text{BaI}}_2\right)$.

Explanation of Solution

Explanation

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 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.

Conclusion

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.

(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 134AE

Answer

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

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

Explanation of Solution

Explanation

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

Cesium phosphide $\left({\text{Cs}}_3\text{P}\right)$.

Explanation of Solution

Explanation

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 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.

Conclusion

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.

(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.

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

Answer to Problem 134AE

Answer

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

Explanation of Solution

Explanation

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 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.

Conclusion

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.