Chapter 2, Problem 43PS

Give the formula and the number of each ion that makes up each of the following compounds:

(a) K₂S

(b) CoSO₄

(c) KMnO₄

(d) (NH₄)₃PO₄

(e) Ca(CIO)₂

(f)NaCH₃CO₂

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

$K_{2}S$ is composed of one $K^{+}$ ion and two $S^{2-}$ ions.

Explanation of Solution

In the given ionic compound $K_{2}S$, there are $K$ ion and $S$ ion present. Since the potassium is group $\text{IA}$ element, then the charge will be $\text{1+}$ and thereby the charge of $S$ must be $\text{2-}$ to balance the $\text{2+}$ charge of two  $K$ ions.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $K^{+}$ ion and two $S^{2-}$ ions are present in $K_{2}S$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

$CoS{O}_4$ is composed of one $C{o}^{2+}$ ions and one $S{O}_4{}^{2-}$ ion.

Explanation of Solution

In the given ionic compound $CoS{O}_4$, there are $Co$ ion and $S{O}_4$ ion present. Since the charge of $S{O}_4$ is $\text{2-}$, then the charge of $Co$ must be $\text{2+}$ to balance the charge of the $S{O}_4$ ions .

The relative number of each ion is indicated in suffix of each ion.

That is,

One $C{o}^{2+}$ ion and one $S{O}_4{}^{2-}$ ion are present in $CoS{O}_4$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

$KMn{O}_4$ is composed of one $K^{+}$ ion and one $Mn{O}_4{}^{-}$ ion.

Explanation of Solution

In the given ionic compound $KMn{O}_4$, there are $K$ ion and $Mn{O}_4$ ion present. Since the potassium is group $\text{IA}$ element, then the charge will be $\text{1+}$ and also the charge of $Mn{O}_4$ is $\text{1-}$.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $K^{+}$ ion and one $Mn{O}_4{}^{-}$ ion are present in $KMn{O}_4$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

${(N{H}_4)}_{3}P{O}_4$ is composed of three $N{H}_4{}^{+}$ ion and one $P{O}_4$ ion.

Explanation of Solution

In the given ionic compound ${(N{H}_4)}_{3}P{O}_4$, there are $N{H}_4$ ion and $P{O}_4$ ion present. The charge of $N{H}_4$ is $\text{1+}$ and also the charge of $P{O}_4$ is $\text{3-}$.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $N{H}_4{}^{+}$ ion and one $P{O}_4{}^{3-}$ ion are present in ${(N{H}_4)}_{3}P{O}_4$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

$Ca{(ClO)}_2$ is composed of one $C{a}^{2+}$ ion and two $Cl{O}^{-}$ ions.

Explanation of Solution

In the given ionic compound $Ca{(ClO)}_2$, there are $Ca$ ion and $ClO$ ion present. Since the $Ca$ is group $\text{IIA}$ element, then the charge will be $\text{2+}$ and the charge of $ClO$ is $\text{1-}$. So there will be two $Cl{O}^{-}$ ions for each of the $C{a}^{2+}$ ion.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $C{a}^{2+}$ ion and one $Cl{O}^{-}$ ion are present in $Ca{(ClO)}_2$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 43PS

$NaC{H}_{3}C{O}_2$ is composed of one $N{a}^{+}$ ion and one $C{H}_{3}C{O}_2{}^{-}$ ion .

Explanation of Solution

In the given ionic compound $NaC{H}_{3}C{O}_2$, there are $Na$ ion and $C{H}_{3}C{O}_2$ ion present. Since the Na is group $\text{IA}$ element, then the charge will be $\text{1+}$ and thereby the charge $C{H}_{3}C{O}_2$ must be $\text{1-}$ to balance the two $\text{1+}$ charge of sodium ion.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $N{a}^{+}$ ion and one $C{H}_{3}C{O}_2{}^{-}$ ion are present in $NaC{H}_{3}C{O}_2$ compound.