Chapter 3, Problem 71P

Interpretation Introduction

(a)

Interpretation:

The chemical formula of the ionic compound magnesium phosphate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\begin{array}{ll}Magnesiumphosphate\hfill & M{g}_3{\left(P{O}_4\right)}_2\hfill \end{array}$

Explanation of Solution

The name of the ionic compound is magnesium phosphate.

Here, magnesium forms cation and phosphate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Mg}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{PO}}_4{}^{3-}=-3\end{array}$

The ratio of charge of cation to anion is as follows:

  $3:2$

Therefore, the formula of the compound will be:

  ${\text{Mg}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

Interpretation Introduction

(b)

Interpretation:

The chemical formula of the ionic compound nickel sulfate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Nickel sulfate }-{\text{ NiSO}}_{\text{4}}$

Explanation of Solution

The name of the ionic compound is nickel sulfate.

Here, nickel forms cation and sulfate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Ni}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{SO}}_4{}^{2-}=-2\end{array}$

The ratio of charge of cation to anion is as follows:

  $2:2$

Therefore, the formula of the compound will be:

  ${\text{NiSO}}_{\text{4}}$

Interpretation Introduction

(c)

Interpretation:

The chemical formula of the ionic compound copper (II) hydroxide needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Copper(II)}\text{hydroxide}-{\text{Cu(OH)}}_{\text{2}}$

Explanation of Solution

The name of the ionic compound is copper (II) hydroxide.

Here, copper forms cation and hydroxide forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Cu}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{OH}}^{-}=-1\end{array}$

The ratio of charge of cation to anion is as follows:

  $2:1$

Therefore, the formula of the compound will be:

  $\text{Cu}{\left(\text{OH}\right)}_2$

Interpretation Introduction

(d)

Interpretation:

The chemical formula of the ionic compound potassium hydrogen phosphate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\begin{array}{ll}\text{Potassium hydrogenphosphate}\hfill & {\text{K}}_{\text{2}}{\text{HPO}}_{\text{4}}\hfill \end{array}$

Explanation of Solution

The name of the ionic compound is potassium hydrogen phosphate.

Here, potassium forms cation and hydrogen phosphate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{K}}^{+}\hfill \end{array}\text{=}+1\\ {\text{HPO}}_{\text{4}}{}^{2-}=-2\end{array}$

The ratio of charge of cation to anion is as follows:

  $1:2$

Therefore, the formula of the compound will be:

  ${\text{K}}_{\text{2}}{\text{HPO}}_{\text{4}}$

Interpretation Introduction

(e)

Interpretation:

The chemical formula of the ionic compound gold (II) nitrate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Gold(II)}{\text{nitrate-Au(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Explanation of Solution

The name of the ionic compound is gold (II) nitrate.

Here, gold forms cation and nitrate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Au}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{NO}}_3{}^{-}=-1\end{array}$

The ratio of charge of cation to anion is as follows:

  $2:1$

Therefore, the formula of the compound will be:

  ${\text{Au(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Interpretation Introduction

(f)

Interpretation:

The chemical formula of the ionic compound lithium phosphate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Lithium}\text{phosphate}-{\text{Li}}_{\text{3}}{\text{PO}}_{\text{4}}$

Explanation of Solution

The name of the ionic compound is lithium phosphate.

Here, lithium forms cation and phosphate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Li}}^{+}\hfill \end{array}\text{=}+1\\ {\text{PO}}_4{}^{3-}=-3\end{array}$

The ratio of charge of cation to anion is as follows:

  $1:3$

Therefore, the formula of the compound will be:

  ${\text{Li}}_3{\text{PO}}_4$

Interpretation Introduction

(g)

Interpretation:

The chemical formula of the ionic compound aluminum bicarbonate needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Aluminium}\text{bicarbonate}\text{-}{\text{Al(HCO}}_{\text{3}}{\text{)}}_{\text{3}}$

Explanation of Solution

The name of the ionic compound is aluminum bicarbonate.

Here, aluminum forms cation and bicarbonate forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Al}}^{3+}\hfill \end{array}\text{=}+3\\ {\text{HCO}}_3{}^{-}=-1\end{array}$

The ratio of charge of cation to anion is as follows:

  $3:1$

Therefore, the formula of the compound will be:

  ${\text{Al(HCO}}_{\text{3}}{\text{)}}_{\text{3}}$

Interpretation Introduction

(h)

Interpretation:

The chemical formula of the ionic compound chromous cyanide needs to be determined.

Concept Introduction:

When the number of electrons increases or decreases from the atomic number, ions are formed. Cation is a positively charged ion formed by losing electron/s and anion is a negatively charged ion formed by gaining electron/s. While writing name of the ionic compounds, the name of cation is always written first followed by the name of the anion. In order to form an ionic compound, the cation and anion combine in such a way that the total charge is zero.

Answer to Problem 71P

  $\text{Chromous}\text{cyanide}\text{-}{\text{Cr(CN)}}_{\text{2}}$

Explanation of Solution

The name of the ionic compound is chromous cyanide.

Here, chromium forms cation and cyanide forms anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Cr}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{CN}}^{-}=-1\end{array}$

The ratio of charge of cation to anion is as follows:

  $2:1$

Therefore, the formula of the compound will be:

  ${\text{Cr(CN)}}_{\text{2}}$