Chapter 4, Problem 4.123EP

(a)

Interpretation Introduction

Interpretation:

Completion of the following table of information about selected ionic compounds.

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & {\text{BaBr}}_{\text{2}}& \end{array}$

Concept Introduction:

Chemical formulas and names for ionic compounds containing polyatomic ions:

Ionic compounds containing positive and negative charges present must add to zero.

If more than one polyatomic ion is present in a compound, the chemical formula can be written as a polyatomic ion is enclose in parentheses and a subscript, placed outside of the parentheses.  Example- ${\text{Fe(OH)}}_{\text{3}}$.

Sometimes same element is present in two different locations.  For example, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$- in this chemical formula “N” is present in two different locations as ${\text{NH}}_{\text{4}}^{\text{+}}{\text{ and NO}}_{\text{3}}^{\text{-}}$ ions.

Answer to Problem 4.123EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Ba}}^{\text{2+}}& {\text{Br}}^{\text{-}}& {\text{BaBr}}_{\text{2}}& \text{Barium bromide}\end{array}$

Explanation of Solution

The given name of the compound has ${\text{BaBr}}_{\text{2}}$ chemical formula.

The dissociation of ${\text{BaBr}}_{\text{2}}$ is as follows:

    ${\text{BaBr}}_{\text{2}}\to {\text{ Ba}}^{\text{2+}}{\text{+2Br}}^{\text{-}}$

From the above reaction, ${\text{BaBr}}_{\text{2}}$ has positive ion -${\text{Ba}}^{\text{2+}}$ and negative ion -${\text{Br}}^{\text{-}}$.

Barium is a metal cation and bromide is a nonmetal anion.

Hence, the compound name is Barium bromide.

Therefore, the completed table is shown below:

$\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Ba}}^{\text{2+}}& {\text{Br}}^{\text{-}}& {\text{BaBr}}_{\text{2}}& \text{Barium bromide}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Completion of the following table of information about selected ionic compounds.

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Zn}}^{\text{2+}}& {\text{NO}}_{\text{3}}^{\text{-}}& & \end{array}$

Concept Introduction:

Chemical formulas and names for ionic compounds containing polyatomic ions:

Ionic compounds containing positive and negative charges present must add to zero.

If more than one polyatomic ion is present in a compound, the chemical formula can be written as a polyatomic ion is enclose in parentheses and a subscript, placed outside of the parentheses.  Example- ${\text{Fe(OH)}}_{\text{3}}$.

Sometimes same element is present in two different locations.  For example, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$- in this chemical formula “N” is present in two different locations as ${\text{NH}}_{\text{4}}^{\text{+}}{\text{ and NO}}_{\text{3}}^{\text{-}}$ ions.

Answer to Problem 4.123EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Zn}}^{\text{2+}}& {\text{NO}}_{\text{3}}^{\text{-}}& {\text{Zn(NO}}_{\text{3}}{\text{)}}_{\text{2}}& \text{Zinc nitrate}\end{array}$

Explanation of Solution

The ionic compound has positive zinc ion ${\text{Zn}}^{\text{2+}}$ and negative nitrate ion ${\text{NO}}_{\text{3}}^{\text{-}}$.  Both these two ions combine to form ${\text{Zn(NO}}_{\text{3}}{\text{)}}_{\text{2}}$.

The name of the compound ${\text{Zn(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is Zinc nitrate.

Therefore, the completed table is shown below:

$\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Zn}}^{\text{2+}}& {\text{NO}}_{\text{3}}^{\text{-}}& {\text{Zn(NO}}_{\text{3}}{\text{)}}_{\text{2}}& \text{Zinc nitrate}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Completion of the following table of information about selected ionic compounds.

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & & \begin{array}{l}\text{Iron(III)}\\ \text{carbonate}\end{array}\end{array}$

Concept Introduction:

Chemical formulas and names for ionic compounds containing polyatomic ions:

Ionic compounds containing positive and negative charges present must add to zero.

If more than one polyatomic ion is present in a compound, the chemical formula can be written as a polyatomic ion is enclose in parentheses and a subscript, placed outside of the parentheses.  Example- ${\text{Fe(OH)}}_{\text{3}}$.

Sometimes same element is present in two different locations.  For example, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$- in this chemical formula “N” is present in two different locations as ${\text{NH}}_{\text{4}}^{\text{+}}{\text{ and NO}}_{\text{3}}^{\text{-}}$ ions.

Answer to Problem 4.123EP

The completed table is as follows.

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Fe}}^{\text{3+}}& {\text{CO}}_{\text{3}}^{\text{2-}}& {\text{Fe}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}& \begin{array}{l}\text{Iron(III)}\\ \text{carbonate}\end{array}\end{array}$

Explanation of Solution

The given compound name has Iron (III) carbonate.

It indicates compound has positive ferric ion ${\text{Fe}}^{\text{3+}}$ and carbonate ion ${\text{CO}}_{\text{3}}^{\text{2-}}$.  Both ions combine to form a compound with chemical formula is ${\text{Fe}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}$.

Therefore, the completed table is shown below:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Fe}}^{\text{3+}}& {\text{CO}}_{\text{3}}^{\text{2-}}& {\text{Fe}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3}}& \begin{array}{l}\text{Iron(III)}\\ \text{carbonate}\end{array}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Completion of the following table of information about selected ionic compounds.

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & {\text{PbO}}_{\text{2}}& \end{array}$

Concept Introduction:

Chemical formulas and names for ionic compounds containing polyatomic ions:

Ionic compounds containing positive and negative charges present must add to zero.

If more than one polyatomic ion is present in a compound, the chemical formula can be written as a polyatomic ion is enclose in parentheses and a subscript, placed outside of the parentheses.  Example- ${\text{Fe(OH)}}_{\text{3}}$.

Sometimes same element is present in two different locations.  For example, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$- in this chemical formula “N” is present in two different locations as ${\text{NH}}_{\text{4}}^{\text{+}}{\text{ and NO}}_{\text{3}}^{\text{-}}$ ions.

Answer to Problem 4.123EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Pb}}^{\text{4+}}& {\text{O}}^{\text{2-}}& {\text{PbO}}_{\text{2}}& \begin{array}{l}\text{Lead(IV)}\\ \text{oxide}\end{array}\end{array}$

Explanation of Solution

The given name of the compound has ${\text{PbO}}_{\text{2}}$ chemical formula.

The dissociation of ${\text{PbO}}_{\text{2}}$ is as follows.

    ${\text{PbO}}_{\text{2}}\to {\text{ Pb}}^{\text{4+}}{\text{+2O}}^{\text{2-}}$

From the above reaction, ${\text{PbO}}_{\text{2}}$ has positive ion -${\text{Pb}}^{\text{4+}}$ and negative ion -${\text{O}}^{\text{2-}}$.  Lead is a metal cation with $\text{+4}$ oxidation state and oxide is a nonmetal anion.

Hence, the compound name is Lead(IV) oxide.

Therefore, the completed table is shown below:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Pb}}^{\text{4+}}& {\text{O}}^{\text{2-}}& {\text{PbO}}_{\text{2}}& \begin{array}{l}\text{Lead(IV)}\\ \text{oxide}\end{array}\end{array}$