Chapter 4, Problem 4.60EP

(a)

Interpretation Introduction

Interpretation:

Completion of the following table of chemical formulas for ionic compounds.  For each compound, the negative ion present is listed on the left side of the table and the positive ion present is listed at the top.

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{Br}}^{\text{-}}& & & & \end{array}$

Concept Introduction:

Ions are formed by the loss or gain of electrons.  The formation of ion requires the presence of two elements, these two elements are: one is metal atom and another one is non-metal atom.  Metal atom loses electron and non-metal atom accepts electron.

The ratio in which positive and negative ions combine is the ratio which achieves charge neutrality for the resulting compound.

There are three rules to remember while writing the chemical formulas.  They are as follows:

• First write the symbol for positive ion.
• The charges of the ions are not shown in the formula.
• The numbers in the formula give the combining ratio for the ions.

Answer to Problem 4.60EP

The chemical formulas of a compound when combines with given ions are as follows:

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{Br}}^{\text{-}}& \text{NaBr}& {\text{MgBr}}_{\text{2}}& {\text{AlBr}}_{\text{3}}& {\text{SiBr}}_{\text{4}}\end{array}$

Explanation of Solution

${\text{Br}}^{\text{-}}$ ion combines with  ${\text{Na}}^{\text{+}}$ ions combine in $\text{1 to 1}$ ratio because this combination causes the resultant charges to zero.  One ${\text{Br}}^{\text{-}}$ ion give one unit of positive charge.  One ${\text{Na}}^{\text{+}}$ ion results in one unit of negative charge.  Thus, the chemical formula for the compound is $\text{NaBr}$.

${\text{Br}}^{\text{-}}$ ion combines with  ${\text{Mg}}^{\text{2+}}$ ions combine in $\text{2 to 1}$ ratio because this combination causes the resultant charges to add to zero.  Two ${\text{Br}}^{\text{-}}$ ion give two units of positive charges.  One ${\text{Mg}}^{\text{2+}}$ ion results two units of negative charges.  Thus, the chemical formula for the compound is ${\text{MgBr}}_{\text{2}}$.

${\text{Br}}^{\text{-}}$ ion combines with  ${\text{Al}}^{\text{3+}}$ ions combine in $\text{3 to 1}$ ratio because this combination causes the resultant charges to zero.  Three ${\text{Br}}^{\text{-}}$ ion give three units of positive charges. One ${\text{Al}}^{\text{3+}}$ ion result in three units of negative charges.  Thus, the chemical formula for the compound is ${\text{AlBr}}_{\text{3}}$.

${\text{Br}}^{\text{-}}$ ion combines with ${\text{Si}}^{\text{4+}}$ ions combine in $\text{4 to 1}$ ratio because this combination causes the resultant charges to zero.  Four ${\text{Br}}^{\text{-}}$ ions give four units of positive charges.  One ${\text{Si}}^{\text{4+}}$ ion result in four units of negative charges.  Thus, the chemical formula for the compound is ${\text{SiBr}}_{\text{4}}$.

(b)

Interpretation Introduction

Interpretation:

Completion of the following table of chemical formulas for ionic compounds.  For each compound, the negative ion present is listed on the left side of the table and the positive ion present is listed at the top.

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{S}}^{\text{2}}{}^{\text{-}}& & & & \end{array}$

Concept Introduction:

Ions are formed by the loss or gain of electrons.  The formation of ion requires the presence of two elements, these two elements are: one is metal atom and another one is non-metal atom.  Metal atom loses electron and non-metal atom accepts electron.

The ratio in which positive and negative ions combine is the ratio which achieves charge neutrality for the resulting compound.

There are three rules to remember while writing the chemical formulas.  They are as follows:

• First write the symbol for positive ion.
• The charges of the ions are not shown in the formula.
• The numbers in the formula give the combining ratio for the ions.

Answer to Problem 4.60EP

The chemical formulas of a compound when combines given ions are as follows:

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{S}}^{\text{2}}{}^{\text{-}}& {\text{Na}}_{\text{2}}\text{S}& \text{MgS}& {\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}& {\text{SiS}}_{\text{2}}\end{array}$

Explanation of Solution

${\text{S}}^{\text{2}}{}^{\text{-}}$ ion combines with ${\text{Na}}^{\text{+}}$ ions combine in $\text{1 to 2}$ ratio because this combination causes the resultant charges to zero.  One ${\text{S}}^{\text{2}}{}^{\text{-}}$ ion give two units of positive charges.  Two ${\text{Na}}^{\text{+}}$ ions result in two units of negative charges.  Thus, the chemical formula for the compound is ${\text{Na}}_{\text{2}}\text{S}$.

${\text{S}}^{\text{2}}{}^{\text{-}}$ ion combines with ${\text{Mg}}^{\text{2+}}$ ions combine in $\text{1 to 1}$ ratio because this combination causes the charges to add to zero.  One ${\text{S}}^{\text{2}}{}^{\text{-}}$ ion gives two units of positive charges.  One ${\text{Mg}}^{\text{2+}}$ ion results in two units of negative charges.  Thus, the chemical formula for the compound is $\text{MgS}$.

${\text{S}}^{\text{2}}{}^{\text{-}}$ ion combines with  ${\text{Al}}^{\text{3+}}$ ions combine in $\text{3 to 2}$ ratio because this combination causes the resultant charges to zero.  Three ${\text{S}}^{\text{2}}{}^{\text{-}}$ ion give six units of positive charges.  Two ${\text{Al}}^{\text{3+}}$ ions result in six units of negative charges.  Thus, the chemical formula for the compound is ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$.

${\text{S}}^{\text{2}}{}^{\text{-}}$ ion combines with  ${\text{Si}}^{\text{4+}}$ ions combine ${\text{SiS}}_{\text{2}}$ ratio because this combination causes the resultant charges to zero.  Two ${\text{S}}^{\text{2}}{}^{\text{-}}$ ions give four units of positive charges.  One ${\text{Si}}^{\text{4+}}$ ion result in four units of negative charges.  Thus, the chemical formula for the compound is ${\text{SiS}}_{\text{2}}$.

(c)

Interpretation Introduction

Interpretation:

Completion of the following table of chemical formulas for ionic compounds.  For each compound, the negative ion present is listed on the left side of the table and the positive ion present is listed at the top.

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{P}}^{\text{3}}{}^{\text{-}}& & & & \end{array}$

Concept Introduction:

Ions are formed by the loss or gain of electrons.  The formation of ion requires the presence of two elements, these two elements are: one is metal atom and another one is non-metal atom.  Metal atom loses electron and non-metal atom accepts electron.

The ratio in which positive and negative ions combine is the ratio which achieves charge neutrality for the resulting compound.

There are three rules to remember while writing the chemical formulas.  They are as follows:

• First write the symbol for positive ion.
• The charges of the ions are not shown in the formula.
• The numbers in the formula give the combining ratio for the ions.

Answer to Problem 4.60EP

The chemical formulas of a compound when combines given ions are as follows:

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{P}}^{\text{3}}{}^{\text{-}}& {\text{Na}}_{\text{3}}\text{P}& {\text{Mg}}_{\text{3}}{\text{P}}_{\text{2}}& \text{AlP}& {\text{Si}}_{\text{3}}{\text{P}}_{\text{4}}\end{array}$

Explanation of Solution

${\text{P}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Na}}^{\text{+}}$ ions combine in $\text{1 to 3}$ ratio because this combination causes the resultant charges to add to zero.  One ${\text{P}}^{\text{3}}{}^{\text{-}}$ ion give three units of positive charges.  Three ${\text{Na}}^{\text{+}}$ ions result in three units of negative charges.  Thus, the chemical formula for the compound is ${\text{Na}}_{\text{3}}\text{P}$.

${\text{P}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Mg}}^{\text{2+}}$ ions combine in $\text{2 to 3}$ ratio because this combination causes the resultant charges to zero.  Two ${\text{P}}^{\text{3}}{}^{\text{-}}$ ions give six units of positive charges.  Three ${\text{Mg}}^{\text{2+}}$ ions result in six units of negative charges.  Thus, the chemical formula for the compound is ${\text{Mg}}_{\text{3}}{\text{P}}_{\text{2}}$.

${\text{P}}^{\text{3}}{}^{\text{-}}$ ions combine with  ${\text{Al}}^{\text{3+}}$ ions combine in $\text{1 to 1}$ ratio because this combination causes the resultant charges to zero.  One ${\text{P}}^{\text{3}}{}^{\text{-}}$ ion gives three units of positive charges.  One ${\text{Al}}^{\text{3+}}$ ion results in three units of negative charges.  Thus, the chemical formula for the compound is $\text{AlP}$.

${\text{P}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Si}}^{\text{4+}}$ ions combine in $\text{4 to 3}$ ratio because this combination causes the resultant charges to zero.  Four ${\text{P}}^{\text{3}}{}^{\text{-}}$ ions give twelve units of positive charges.  Three ${\text{Si}}^{\text{4+}}$ ions result in twelve units of negative charges. Thus, the chemical formula for the compound is ${\text{Si}}_{\text{3}}{\text{P}}_{\text{4}}$.

(d)

Interpretation Introduction

Interpretation:

Completion of the following table of chemical formulas for ionic compounds.  For each compound, the negative ion present is listed on the left side of the table and the positive ion present is listed at the top.

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{N}}^{\text{3}}{}^{\text{-}}& & & & \end{array}$

Concept Introduction:

Ions are formed by the loss or gain of electrons.  The formation of ion requires the presence of two elements, these two elements are: one is metal atom and another one is non-metal atom.  Metal atom loses electron and non-metal atom accepts electron.

The ratio in which positive and negative ions combine is the ratio which achieves charge neutrality for the resulting compound.

There are three rules to remember while writing the chemical formulas.  They are as follows:

• First write the symbol for positive ion.
• The charges of the ions are not shown in the formula.
• The numbers in the formula give the combining ratio for the ions.

Answer to Problem 4.60EP

The chemical formulas of a compound when combines given ions are as follows:

  $\begin{array}{ccccc}& {\text{Na}}^{\text{+}}& {\text{Mg}}^{\text{2+}}& {\text{Al}}^{\text{3+}}& {\text{Si}}^{\text{4+}}\\ {\text{N}}^{\text{3}}{}^{\text{-}}& {\text{Na}}_{\text{3}}\text{N}& {\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}& \text{AlN}& {\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}\end{array}$

Explanation of Solution

${\text{N}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Na}}^{\text{+}}$ ions combine in $\text{1 to 3}$ ratio because this combination causes the resultant charges to zero.  One ${\text{N}}^{\text{3}}{}^{\text{-}}$ ion gives three units of positive charges.  Three ${\text{Na}}^{\text{+}}$ ions result in three units of negative charges.  Thus, the chemical formula for the compound is ${\text{Na}}_{\text{3}}\text{N}$.

${\text{N}}^{\text{3}}{}^{\text{-}}$ ions combine with  ${\text{Mg}}^{\text{2+}}$ ions combine in $\text{2 to 3}$ ratio because this combination causes the resultant charges to add to zero.  Two ${\text{N}}^{\text{3}}{}^{\text{-}}$ ion give six units of positive charges.  Three ${\text{Mg}}^{\text{2+}}$ ions result in six units of negative charges.  Thus, the chemical formula for the compound is ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$.

${\text{N}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Al}}^{\text{3+}}$ ions combine in $\text{1 to 1}$ ratio because this combination causes the resultant charges to zero.  One ${\text{N}}^{\text{3}}{}^{\text{-}}$ ion give three units of positive charges.  One ${\text{Al}}^{\text{3+}}$ ion results in three units of negative charges.  Thus, the chemical formula for the compound is $\text{AlN}$.

${\text{N}}^{\text{3}}{}^{\text{-}}$ ion combines with  ${\text{Si}}^{\text{4+}}$ ions combine in $\text{4 to 3}$ ratio because this combination causes the resultant charges to zero.  Four ${\text{N}}^{\text{3}}{}^{\text{-}}$ ions give twelve units of positive charges.  Three ${\text{Si}}^{\text{4+}}$ ions result in twelve units of negative charges.  Thus, the chemical formula for the compound is ${\text{Si}}_{\text{3}}{\text{N}}_{\text{4}}$.