Chapter 3, Problem 3.82P

Interpretation Introduction

(a)

Interpretation:

The formula of copper (I) sulfite 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 3.82P

  $\begin{array}{ll}\text{Copper(I)}\text{sulfite}\hfill & {\text{Cu}}_{\text{2}}{\text{SO}}_3\hfill \end{array}$

Explanation of Solution

The given name is copper (I) sulfite.

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

  $\begin{array}{l}\begin{array}{l}{\text{Cu}}^{1+}\hfill \end{array}\text{=}+1\\ {\text{SO}}_3{}^{2-}=-2\end{array}$

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

  $1:2$

Therefore, the formula of the compound will be:

  ${\text{Cu}}_{\text{2}}{\text{SO}}_3$

Interpretation Introduction

(b)

Interpretation:

The formula of aluminum 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 3.82P

  $\text{Aluminium}\text{nitrate}\text{-}{\text{Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}$

Explanation of Solution

The given name is aluminum nitrate.

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

  $\begin{array}{l}\begin{array}{l}{\text{Al}}^{3+}\hfill \end{array}\text{=}+3\\ {\text{NO}}_3{}^{-}=-1\end{array}$

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

  $3:1$

Therefore, the formula of the compound will be:

  $\text{Al}{\left({\text{NO}}_3\right)}_3$

Interpretation Introduction

(c)

Interpretation:

The formula of tin (II) acetate 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 3.82P

  $\text{Tin(II)}\text{acetate}\text{-}{\text{Sn(CH}}_{\text{3}}{\text{COO)}}_{\text{2}}$

Explanation of Solution

The given name is tin (II) acetate

Here, tin is cation and acetate is anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Sn}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{CH}}_{\text{3}}{\text{COO}}^{-}=-1\end{array}$

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

  $2:1$

Therefore, the formula of the compound will be:

  $\text{Sn}{\left({\text{CH}}_{\text{3}}\text{COO}\right)}_2$

Interpretation Introduction

(d)

Interpretation:

The formula of lead (IV) carbonate 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 3.82P

  $\text{lead }\left(\text{IV}\right)\text{ carbonate}-{\text{Pb(CO}}_{\text{3}}{\text{)}}_{\text{2}}$

Explanation of Solution

The given name is lead (IV) carbonate.

Here, lead is cation and carbonate is anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Pb}}^{4+}\hfill \end{array}\text{=}+4\\ {\text{CO}}_3^{2-}=-2\end{array}$

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

  $4:2=2:1$

Therefore, the formula of the compound will be:

  $\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_2$

Interpretation Introduction

(e)

Interpretation:

The formula of zinc 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 3.82P

  $\text{Zinc}\text{hydrogen}\text{phosphate-}\text{Zn}\left({\text{HPO}}_{\text{4}}\right)$

Explanation of Solution

The given name is zinc hydrogen phosphate.

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

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

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

  $2:2$

Therefore, the formula of the compound will be:

  $\text{Zn}\left({\text{HPO}}_{\text{4}}\right)$

Interpretation Introduction

(f)

Interpretation:

The formula of manganese dihydrogen 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 3.82P

  $\text{Manganese}\text{}\text{dihydrogen}\text{phosphate-}{\text{Mn(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Explanation of Solution

The given name is manganese dihydrogen phosphate.

Here, manganese is cation and dihydrogen phosphate is anion. The cation and anion with their charges are represented as follows:

  $\begin{array}{l}\begin{array}{l}{\text{Mn}}^{2+}\hfill \end{array}\text{=}+2\\ {\text{H}}_2{\text{PO}}_4{}^{-}=-1\end{array}$

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

  $2:1$

Therefore, the formula of the compound will be:

  ${\text{Mn(H}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{)}}_{\text{2}}$

Interpretation Introduction

(g)

Interpretation:

The formula of ammonium 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 3.82P

  $\text{Ammonium}\text{cyanide }-{\text{ NH}}_{\text{4}}\text{CN}$

Explanation of Solution

The given name is ammonium cyanide.

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

  $\begin{array}{l}\begin{array}{l}{\text{NH}}_4{}^{+}\hfill \end{array}\text{=}+1\\ {\text{CN}}^{-}=-1\end{array}$

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

  $1:1$

Therefore, the formula of the compound will be:

  ${\text{NH}}_{\text{4}}\text{CN}$

Interpretation Introduction

(h)

Answer to Problem 3.82P

  $\text{Iron(II)}{\text{nitrate-Fe(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Explanation of Solution

The name of the compound is iron (II) nitrate.

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

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

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

  $2:1$

Therefore, the formula of the compound will be:

  $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$