Chapter 2, Problem 52E

(a)

Interpretation Introduction

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

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. The cation’s name is always written first followed by anion’s name. Cation’s name is same as its elemental name however in anion’s name, the suffix “ide” is substituted at the end part of its element’s name.

Answer to Problem 52E

The compound, ${\text{BaSO}}_{\text{3}}$ has the name barium sulfite.

Explanation of Solution

The cation in ${\text{BaSO}}_{\text{3}}$ is barium cation $\left({\text{Ba}}^{2+}\right)$ and the anion issulfite anion $\left({\text{SO}}_3^{2-}\right)$.

The charge on barium is $+2$ and that onsulfiteis $-2$ giving an overall neutral compound.

Hence, the compound ${\text{BaSO}}_{\text{3}}$ has the namebarium sulfite.

(b)

Interpretation Introduction

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

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. The cation’s name is always written first followed by anion’s name. Cation’s name is same as its elemental name however in anion’s name, the suffix “ide” is substituted at the end part of its element’s name.

Answer to Problem 52E

The compound, ${\text{NaNO}}_{\text{2}}$ has the name sodium nitrite

Explanation of Solution

The cation in ${\text{NaNO}}_{\text{2}}$ is sodium ion $\left({\text{Na}}^{+}\right)$ and the anion is nitrite ion $\left({\text{NO}}_2^{-}\right)$.

The charge on sodium cation is $+1$ and that on nitrite anion is $-1$ and therefore on combination gives a neutral compound.

Hence, the compound ${\text{NaNO}}_{\text{2}}$ has the name sodium nitrite

(c)

Interpretation Introduction

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

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. The cation’s name is always written first followed by anion’s name. Cation’s name is same as its elemental name however in anion’s name, the suffix “ide” is substituted at the end part of its element’s name.

Answer to Problem 52E

The compound, ${\text{KMnO}}_{\text{4}}$ has the name potassium permanganate.

Explanation of Solution

The cation in ${\text{KMnO}}_{\text{4}}$ is potassium cation $\left({\text{K}}^{+}\right)$ and the anion is permanganate anion $\left({\text{MnO}}_4^{-}\right)$.

The charge on potassium cation is $+1$ and that on permanganate anion is $-1$ giving an overall neutral compound.

Hence, the compound ${\text{KMnO}}_{\text{4}}$ has the name potassium permanganate.

(d)

Interpretation Introduction

Interpretation:The name of the compound ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is to be predicted.

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. The cation’s name is always written first followed by anion’s name. Cation’s name is same as its elemental name however in anion’s name, the suffix “ide” is substituted at the end part of its element’s name.

Answer to Problem 52E

The compound, ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ has the namepotassium dichromate.

Explanation of Solution

The cation in ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is potassium cation $\left({\text{K}}^{+}\right)$ and the anion dichromate anion $\left({\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}\right)$.

The charge on potassium cation is $+1$ and that on dichromate anion is $-2$. In order to give a neutral compound, $2{\text{K}}^{+}$ combines with $1{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ to give a neutral compound.

Hence, the compound ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ has the name potassium dichromate.

(e)

Interpretation Introduction

Interpretation:The formula of the compound chromium (III)hydroxide is to be predicted.

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. While writing formula for any ionic compound, the cation’s charge becomes the subscript for respective anion and vice versa generating an overall neutral compound.

Answer to Problem 52E

The compound,chromium (III) hydroxide has the formula $\text{Cr}{\left(\text{OH}\right)}_{\text{3}}$.

Explanation of Solution

The cation in Chromium (III) hydroxide is chromium cation $\left({\text{Cr}}^{3+}\right)$ and the anion is hydroxide anion $\left({\text{OH}}^{-}\right)$.

The charge on chromium cation is $+3$ which become subscript for hydroxide anion and that on hydroxide anion is $-1$ which becomes subscript for chromium cation.

Thus, 1 ${\text{Cr}}^{3+}$ combines with $3{\text{OH}}^{-}$ to give a neutral compound.

Hence, the compound chromium (III) hydroxide has the formula $\text{Cr}{\left(\text{OH}\right)}_{\text{3}}$.

(f)

Interpretation Introduction

Interpretation:The formula of the compound magnesium cyanide is to be predicted.

Concept introduction:Combination of cation with respective anion generates an ionic compound. While writing formulas for these ionic compounds, one has to follow certain simple rules. While writing formula for any ionic compound, the cation’s charge becomes the subscript for respective anion and vice versa generating an overall neutral compound.

Answer to Problem 52E

The compound,magnesium cyanide has the formula $\text{Mg}{\left(\text{CN}\right)}_{\text{2}}$.

Explanation of Solution

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

The charge on magnesium cation is $+2$ which become subscript for cyanide anion and that on cyanide anion is $-1$ which becomes subscript for magnesium cation.

Thus, 1 ${\text{Mg}}^{2+}$ combines with $2{\text{CN}}^{-}$ to give a neutral compound.

Hence, the compound magnesium cyanide has the formula $\text{Mg}{\left(\text{CN}\right)}_{\text{2}}$.

(g)

Interpretation Introduction

Interpretation:The formula of the compound lead (IV) carbonate is to be predicted.

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. While writing formula for any ionic compound, the cation’s charge becomes the subscript for respective anion and vice versa generating an overall neutral compound.

Answer to Problem 52E

The compound,lead (IV) carbonate has the formula $\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_{\text{2}}$.

Explanation of Solution

The cation in lead (IV) carbonate is lead cation $\left({\text{Pb}}^{4+}\right)$ and the anion is carbonate anion $\left({\text{CO}}_3^{2-}\right)$.

The charge on lead cation is $+4$ which become subscript for carbonate anion and that on carbonate anion is $-2$ which becomes subscript for lead cation.

These subscript when reduced to the lowest ratio becomes 1:2.

Thus, 1 ${\text{Pb}}^{4+}$ combines with $2{\text{CO}}_3^{2-}$ to give a neutral compound.

Hence, the compound Lead (IV) carbonate has the formula $\text{Pb}{\left({\text{CO}}_{\text{3}}\right)}_{\text{2}}$.

(h)

Interpretation Introduction

Interpretation:The formula of the compound ammonium acetate is to be predicted.

Concept introduction:Combination of cation with respective anion generates an ionic compound. While giving names these ionic compounds, one has to follow certain simple rules. While writing formula for any ionic compound, the cation’s charge becomes the subscript for respective anion and vice versa generating an overall neutral compound.

Answer to Problem 52E

The compound,ammonium acetate has the formula ${\text{NH}}_{\text{4}}{\text{CH}}_{\text{3}}\text{COO}$.

Explanation of Solution

The cation in ammonium acetate is ammonium cation $\left({\text{NH}}_4^{+}\right)$ and the anion is acetate anion $\left({\text{CH}}_3{\text{COO}}^{-}\right)$.

The charge on ammonium cation is $+1$ which become subscript for acetate anion and that on acetate anion is $-1$ which becomes subscript for ammonium cation.

Thus, 1 ${\text{NH}}_4^{+}$ combines with $1{\text{CH}}_3{\text{COO}}^{-}$ to give a neutral compound.

Hence, the compound ammonium acetate has the formula ${\text{NH}}_{\text{4}}{\text{CH}}_{\text{3}}\text{COO}$.