Chapter 4, Problem 4.59E

Use Table $4.4$ and classify the bonds in the following compounds as nonpolar covalent, polar covalent, or ionic:

a. ${\text{MgI}}_{\text{2}}$ (each $\text{I}$ is bonded to $\text{Mg}$)

b. ${\text{NCl}}_3$ (each $\text{Cl}$ is bonded to $\text{N}$)

c. ${\text{H}}_2\text{S}$ (each $\text{H}$ is bonded to $\text{S}$)

d. $\text{RbF}$

e. $\text{SrO}$

Interpretation Introduction

(a)

Interpretation:

Whether the bond in ${\text{MgI}}_{\text{2}}$ is nonpolar covalent, polar covalent, or ionic is to be stated.

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are the result of the electronegativity difference between bonding atoms. If the electronegativity difference is $0$, the bond is known as nonpolar covalent bond, if electronegativity difference is between $0$ to $2.1$, then the bond is known as polar covalent bond and if it is more than $2.1$, then the bond is known as ionic bond.

Answer to Problem 4.59E

The bonds in ${\text{MgI}}_{\text{2}}$ are polar covalent bond.

Explanation of Solution

The electronegativity values of $\text{Mg}$ atom and $\text{I}$ atom are $1.2$ and $2.5$ respectively as given in Table $4.4$. Thus, the electronegativity difference is calculated as given below.

$\begin{array}{rll}\Delta EN& =& EN\left(\text{I}\right)-EN\left(\text{Mg}\right)\\ & =& 2.5-1.2\\ & =& 1.3\end{array}$

The electronegativity difference is between $0$ to $2.1$. Therefore, ${\text{MgI}}_{\text{2}}$ has a polar covalent bond.

Conclusion

The bonds in ${\text{MgI}}_{\text{2}}$ are polar covalent bond.

Interpretation Introduction

(b)

Interpretation:

Whether the bond in ${\text{NCl}}_3$ is nonpolar covalent, polar covalent, or ionic is to be stated.

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are the result of the electronegativity difference between bonding atoms. If the electronegativity difference is $0$, the bond is known as nonpolar covalent bond, if electronegativity difference is between $0$ to $2.1$, then the bond is known as polar covalent bond and if it is more than $2.1$, then the bond is known as ionic bond.

Answer to Problem 4.59E

The bonds in ${\text{NCl}}_3$ are nonpolar covalent bond.

Explanation of Solution

The electronegativity values of $\text{N}$ atom and $\text{Cl}$ atom is $3.0$ as given in Table $4.4$. Thus, the electronegativity difference is calculated as given below.

$\begin{array}{rll}\Delta EN& =& EN\left(\text{Cl}\right)-EN\left(\text{N}\right)\\ & =& 3.0-3.0\\ & =& 0\end{array}$

The electronegativity difference is $0$. Therefore, ${\text{NCl}}_3$ has nonpolar covalent bond.

Conclusion

The bonds in ${\text{NCl}}_3$ are nonpolar covalent bond.

Interpretation Introduction

(c)

Interpretation:

Whether the bond in ${\text{H}}_2\text{S}$ is nonpolar covalent, polar covalent, or ionic is to be stated.

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are the result of the electronegativity difference between bonding atoms. If the electronegativity difference is $0$, the bond is known as nonpolar covalent bond, if electronegativity difference is between $0$ to $2.1$, then the bond is known as polar covalent bond and if it is more than $2.1$, then the bond is known as ionic bond.

Answer to Problem 4.59E

The bonds in ${\text{H}}_2\text{S}$ are polar covalent bond.

Explanation of Solution

The electronegativity values of $\text{H}$ atom and $\text{S}$ atom are $2.1$ and $2.5$ respectively as given in Table $4.4$. Thus, the electronegativity difference is calculated as given below.

$\begin{array}{rll}\Delta EN& =& EN\left(\text{S}\right)-EN\left(\text{H}\right)\\ & =& 2.5-2.1\\ & =& 0.4\end{array}$

The electronegativity difference is between $0$ to $2.1$. Therefore, ${\text{H}}_2\text{S}$ has polar covalent bond.

Conclusion

The bonds in ${\text{H}}_2\text{S}$ are polar covalent bond.

Interpretation Introduction

(d)

Interpretation:

Whether the bond in $\text{RbF}$ is nonpolar covalent, polar covalent, or ionic is to be stated.

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are the result of the electronegativity difference between bonding atoms. If the electronegativity difference is $0$, the bond is known as nonpolar covalent bond, if electronegativity difference is between $0$ to $2.1$, then the bond is known as polar covalent bond and if it is more than $2.1$, then the bond is known as ionic bond.

Answer to Problem 4.59E

The bond in $\text{RbF}$ is ionic bond.

Explanation of Solution

The electronegativity values of $\text{Rb}$ atom and $\text{F}$ atom are $0.8$ and $4.0$ respectively as given in Table $4.4$. Thus, the electronegativity difference is calculated as given below.

$\begin{array}{rll}\Delta EN& =& EN\left(\text{F}\right)-EN\left(\text{Rb}\right)\\ & =& 4.0-0.8\\ & =& 3.2\end{array}$

The electronegativity difference is more than $2.1$. Therefore, $\text{RbF}$ has ionic bond.

Conclusion

The bond in $\text{RbF}$ is found to be ionic bond.

Interpretation Introduction

(e)

Interpretation:

Whether the bond in $\text{SrO}$ is nonpolar covalent, polar covalent, or ionic is to be stated.

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are the result of the electronegativity difference between bonding atoms. If the electronegativity difference is $0$, the bond is known as nonpolar covalent bond, if electronegativity difference is between $0$ to $2.1$, then the bond is known as polar covalent bond and if it is more than $2.1$, then the bond is known as ionic bond.

Answer to Problem 4.59E

The bond in $\text{SrO}$ is ionic bond.

Explanation of Solution

The electronegativity values of $\text{Sr}$ atom and $\text{O}$ atom are $1.0$ and $3.5$ respectively as given in Table $4.4$. Thus, the electronegativity difference is calculated as given below.

$\begin{array}{rll}\Delta EN& =& EN\left(\text{O}\right)-EN\left(\text{Sr}\right)\\ & =& 3.5-1.0\\ & =& 2.5\end{array}$

The electronegativity difference is more than $2.1$. Therefore, $\text{SrO}$ has ionic bond.

Conclusion

The bond in $\text{SrO}$ is ionic bond.