Chapter 4, Problem 4.124EP

(a)

Interpretation Introduction

Interpretation:

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

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & {\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}& \end{array}$

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 4.124EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Cu}}^{\text{+}}& {\text{SO}}_{\text{4}}^{\text{2-}}& {\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}& \begin{array}{l}\text{Copper(I)}\\ \text{Sulfate}\end{array}\end{array}$

Explanation of Solution

The given name of the compound has ${\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}$ chemical formula.

The dissociation of ${\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}$ is as follows.

    ${\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{ 2Cu}}^{\text{+}}{\text{+ SO}}_{\text{4}}^{\text{2-}}$

From the above reaction, ${\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}$ has positive ion -${\text{Cu}}^{\text{+}}$ and negative ion -${\text{SO}}_{\text{4}}^{\text{2-}}$.

Copper is a metal cation with $\text{+1}$ charge and sulphate is a nonmetal anion.

Hence, the compound name is Copper (I) sulfate.

Therefore, the completed table is shown below:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Cu}}^{\text{+}}& {\text{SO}}_{\text{4}}^{\text{2-}}& {\text{Cu}}_{\text{2}}{\text{SO}}_{\text{4}}& \begin{array}{l}\text{Copper(I)}\\ \text{Sulfate}\end{array}\end{array}$

(b)

Interpretation Introduction

Interpretation:

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

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & & \begin{array}{l}\text{Lithium}\\ \text{nitride}\end{array}\end{array}$

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 4.124EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Li}}^{\text{+}}& {\text{N}}^{\text{3-}}& {\text{Li}}_{\text{3}}\text{N}& \begin{array}{l}\text{Lithium}\\ \text{nitride}\end{array}\end{array}$

Explanation of Solution

The given compound name has Lithium nitride.

It indicates that compound has positive lithium ion ${\text{Li}}^{\text{+}}$ and negative nitride ion ${\text{N}}^{\text{3-}}$.  Both ions combine to form a compound with chemical formula is ${\text{Li}}_{\text{3}}\text{N}$.

Therefore, the completed table is shown below:

$\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Li}}^{\text{+}}& {\text{N}}^{\text{3-}}& {\text{Li}}_{\text{3}}\text{N}& \begin{array}{l}\text{Lithium}\\ \text{nitride}\end{array}\end{array}$

(c)

Interpretation Introduction

Interpretation:

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

$\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Al}}^{\text{3+}}& {\text{S}}^{\text{2-}}& & \end{array}$

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 4.124EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Al}}^{\text{3+}}& {\text{S}}^{\text{2-}}& {\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}& \text{Aluminium sulphide}\end{array}$

Explanation of Solution

The ionic compound has positive aluminium ion ${\text{Al}}^{\text{3+}}$ and negative sulphide ion ${\text{S}}^{\text{2-}}$.  Both these two ions combine to form ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$.

The name of the compound ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is Aluminium sulphide.

Therefore, the completed table is shown below:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Al}}^{\text{3+}}& {\text{S}}^{\text{2-}}& {\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}& \text{Aluminium sulphide}\end{array}$

(d)

Interpretation Introduction

Interpretation:

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

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ & & {\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}& \end{array}$

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 4.124EP

The completed table is as follows:

  $\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Co}}^{\text{2+}}& {\text{PO}}_{\text{4}}^{\text{3-}}& {\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}& \text{Cobalt(II) phosphate}\end{array}$

Explanation of Solution

The given name of the compound has ${\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ as chemical formula.

The dissociation of ${\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is as follows:

    ${\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}\to {\text{ 3Co}}^{\text{2+}}{\text{ + 2PO}}_{\text{4}}^{\text{3-}}$

From the above reaction, ${\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ has positive ion -${\text{Co}}^{\text{2+}}$ and negative ion -${\text{PO}}_{\text{4}}^{\text{3-}}$.  Cobalt is a metal cation with $\text{+2}$ charge and phosphate is a nonmetal anion.

Hence, the compound name is Cobalt (II) phosphate.

Therefore, the completed table is shown below:

$\begin{array}{cccc}\begin{array}{l}\text{Formula of}\\ \text{Positive ion}\end{array}& \begin{array}{l}\text{Formula of}\\ \text{Negative ion}\end{array}& \begin{array}{l}\text{Chemical}\\ \text{Formula}\end{array}& \begin{array}{l}\text{Compound}\\ \text{Name}\end{array}\\ {\text{Co}}^{\text{2+}}& {\text{PO}}_{\text{4}}^{\text{3-}}& {\text{Co}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}& \text{Cobalt(II) phosphate}\end{array}$