Chapter 5.7, Problem 9PPB

(a)

Interpretation Introduction

Interpretation: The respective formulae of the compounds to be identified.

Concept Introduction:

In the formula of the compound, the sum of the charge of cation and anion must be equal to zero. The ration of cation and anion to be known in order to arrive the formula of a compound.

Rules for Nomenclature of ionic compounds:

• The name of the cation is followed by the name of the anion.

Answer to Problem 9PPB

$\begin{array}{l}\text{ a) Name : Potassium dichromate}\text{. }\\ {\text{Formula : K}}_{\text{2}}{\text{(Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{)}\text{.}\\ \end{array}$

Explanation of Solution

Predict the Formula.

$\begin{array}{l}\text{a) Name : Potassium dichromate}\text{.}\\ \text{ Cation: Potassium}\\ \text{ Anion: dichromate}\text{.}\\ \text{Instruction follows:}\\ \text{ 1) Write}\text{cation's}2)\text{Write}\text{anion's}3)\text{Combine}\text{ions}\\ \text{symbol}\text{and}\text{symbol}\text{and}\text{in}\text{neutral}\text{ratio}\text{.}\\ \text{charge}\text{.}\text{charge}\text{.}\\ {\text{ K}}^{\text{+}}{{\text{(Cr}}_{\text{2}}}^{{\text{O}}_{\text{7}}}2{\text{K}}^{\text{+}}+{{\text{(Cr}}_{\text{2}}}^{{\text{O}}_{\text{7}}}\text{ =}{\text{K}}_{\text{2}}{\text{(Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{)}\\ \\ \boxed{\text{The formula is: }{\text{K}}_{\text{2}}{\text{(Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{)}.}\end{array}$

The cation and anion along with the charges are predicted. Finally combine the ions in neutral ratio. The sum of the charge of both cation and anion is equal to zero when the ratio of combination is $\text{2:1}$. Two potassium cation and dichromate anions are present.

(b)

Interpretation Introduction

Interpretation: The respective formulae of the compounds to be identified.

Concept Introduction:

In the formula of the compound, the sum of the charge of cation and anion must be equal to zero. The ration of cation and anion to be known in order to arrive the formula of a compound.

Rules for Nomenclature of ionic compounds:

• The name of the cation is followed by the name of the anion.

Answer to Problem 9PPB

$\begin{array}{l}\text{b) Name : Lithium oxalate}\text{.}\\ {\text{Formula : Li}}_{\text{2}}{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}\text{.}\end{array}$

Explanation of Solution

Predict the Formula.

$\begin{array}{l}\text{b) Name : Lithium oxalate}\text{.}\\ \text{ Cation: Lithium}\\ \text{ Anion: oxalate}\text{.}\\ \text{Instruction follows:}\\ \text{ 1) Write}\text{cation's}2)\text{Write}\text{anion's}3)\text{Combine}\text{ions}\\ \text{symbol}\text{and}\text{symbol}\text{and}\text{in}\text{neutral}\text{ratio}\text{.}\\ \text{charge}\text{.}\text{charge}\text{.}\\ {\text{ Li}}^{\text{+}}{{\text{(C}}_{\text{2}}}^{{\text{O}}_{\text{4}}}2{\text{Li}}^{\text{+}}+{{\text{(C}}_{\text{2}}}^{{\text{O}}_{\text{4}}}={\text{Li}}_{\text{2}}{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}\\ \\ \boxed{\text{The formula is: }{\text{Li}}_{\text{2}}{\text{(C}}_{\text{2}}{\text{O}}_{\text{4}}\text{)}.}\end{array}$

The cation and anion along with the charges are predicted. Finally combine the ions in neutral ratio. The sum of the charge of both cation and anion is equal to zero when the ratio of combination is $\text{2:1}$. Two lead cation and one oxalate anion are present.

(c)

Interpretation Introduction

Interpretation: The respective formulae of the compounds to be identified.

Concept Introduction:

In the formula of the compound, the sum of the charge of cation and anion must be equal to zero. The ration of cation and anion to be known in order to arrive the formula of a compound.

Rules for Nomenclature of ionic compounds:

• The name of the cation is followed by the name of the anion.

Answer to Problem 9PPB

$\begin{array}{l}\text{c) Name : Copper(II) nitrate}\\ {\text{ Formula : Cu(NO}}_{\text{3}}{\text{)}}_2\end{array}$

Explanation of Solution

Predict the Formula.

$\begin{array}{l}\text{c) Name : Copper(II) nitrate}\text{.}\\ \text{ Cation: Copper}\\ \text{ Anion: nitrate}\text{.}\\ \text{Instruction follows:}\\ \text{ 1) Write}\text{cation's}2)\text{Write}\text{anion's}3)\text{Combine}\text{ions}\\ \text{symbol}\text{and}\text{symbol}\text{and}\text{in}\text{neutral}\text{ratio}\text{.}\\ \text{charge}\text{.}\text{charge}\text{.}\\ {\text{ Cu}}^{\text{2+}}{{\text{(NO}}_{\text{3}}}^{\text{)}}{\text{Cu}}^{\text{2+}}+{\text{ 2(NO}}_{\text{3}}{\text{)}}^{-}\text{ =}{\text{Cu(NO}}_{\text{3}}{\text{)}}_2\\ \\ \boxed{\text{The formula is: }{\text{Cu(NO}}_{\text{3}}{\text{)}}_2\text{.}}\end{array}$

The cation and anion along with the charges are predicted. Finally combine the ions in neutral ratio. The sum of the charge of both cation and anion is equal to zero when the ratio of combination is $\text{1:2}$. One copper cation and two nitrate anions are present.