Chapter 12, Problem 53E

Interpretation Introduction

(a)

Interpretation:

The labeling of each atom in the polar covalent bond $\text{C}-\text{H}$ is to be done using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Answer to Problem 53E

The polar $\text{C}-\text{H}$ bond using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$).is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{C}-\text{H}\end{array}$

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure $12.9$ the electronegativity value of $\text{C}$ is $2.5$ and that of $\text{H}$ is $2.1$. As carbon is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on carbon and a corresponding positive charge on hydrogen. The polar $\text{C}-\text{H}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{C}-\text{H}\end{array}$

Conclusion

The polar $\text{C}-\text{H}$ bond using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$) is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{C}-\text{H}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The labeling of each atom in the polar covalent bond $\text{Se}-\text{O}$ is to be done using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Answer to Problem 53E

The polar $\text{Se}-\text{O}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{O}-\text{Se}\end{array}$

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure $12.9$ the electronegativity value of $\text{Se}$ is $2.4$ and that of $\text{O}$ is $3.5$. As oxygen is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on oxygen and a corresponding positive charge on selenium. The polar $\text{Se}-\text{O}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{O}-\text{Se}\end{array}$

Conclusion

The polar $\text{Se}-\text{O}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{O}-\text{Se}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The labeling of each atom in the polar covalent bond $\text{P}-\text{I}$ is to be done using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Answer to Problem 53E

The polar $\text{P}-\text{I}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{I}-\text{P}\end{array}$

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure $12.9$ the electronegativity value of $\text{P}$ is $2.1$ and that of $\text{I}$ is $2.5$. As iodine is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on iodine and a corresponding positive charge on phosphorous. The polar $\text{P}-\text{I}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{I}-\text{P}\end{array}$

Conclusion

The polar $\text{P}-\text{I}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{I}-\text{P}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The labeling of each atom in the polar covalent bond $\text{H}-\text{Br}$ is to be done using delta notation (${\text{δ}}^{\text{+}}$ and ${\text{δ}}^{-}$).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Answer to Problem 53E

The polar $\text{H}-\text{Br}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{Br}-\text{H}\end{array}$

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure $12.9$ the electronegativity value of $\text{Br}$ is $2.8$ and that of $\text{H}$ is $2.1$. As bromine is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on bromine and a corresponding positive charge on hydrogen. The polar $\text{H}-\text{Br}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{Br}-\text{H}\end{array}$

Conclusion

The polar $\text{H}-\text{Br}$ bond using delta notation is shown below.

$\begin{array}{l}{\text{δ}}^{-}{\text{δ}}^{\text{+}}\\ \text{Br}-\text{H}\end{array}$