Chapter 12.1, Problem 7RQ

(a)

Interpretation Introduction

Interpretation:

The more polar bond is to be found out in the given pair of H-F and H-S.

Concept introduction:

A bond is polar, if there is difference in electronegativity values of atoms forming the bond.

More the difference, more is polarity of bond.

  $\begin{array}{l}{\chi }_{A-B}={\chi }_A-{\chi }_{B}if{\chi }_A>{\chi }_B\\ and{\chi }_B-{\chi }_{A}if{\chi }_B>{\chi }_A\end{array}$

Answer to Problem 7RQ

  $H-F$ is more polar than $H-S$

Explanation of Solution

  $\begin{array}{l}{\chi }_H=2.2\\ {\chi }_S=2.58\\ {\chi }_F=3.98\end{array}$

Thus,

$\begin{array}{c}{\chi }_{H-S}=2.58-2.2\\ =0.38\\ {\chi }_{H-F}=3.98-2.2\\ =1.78\end{array}$

Electronegativity difference of $H-F$ is more than $H-S$ ,therefore $H-F$ is more polar than $H-S$

(b)

Interpretation Introduction

Interpretation:

The more polar bond is to be found out in the given pair of O-S and O-F.

Concept introduction:

A bond is polar if there is difference in electronegativity values of atoms forming the bond.

More the difference more is polarity of bond.

  $\begin{array}{l}{\chi }_{A-B}={\chi }_A-{\chi }_{B}if{\chi }_A>{\chi }_B\\ and{\chi }_B-{\chi }_{A}if{\chi }_B>{\chi }_A\end{array}$

Answer to Problem 7RQ

  $O-S$ is more polar than $O-F$

Explanation of Solution

  $\begin{array}{l}{\chi }_O=3.44\\ {\chi }_S=2.58\\ {\chi }_F=3.98\end{array}$

Thus,

$\begin{array}{c}{\chi }_{O-S}=3.44-2.58\\ =0.86\\ {\chi }_{O-F}=3.98-3.44\\ =0.54\end{array}$

Electronegativity difference of $O-S$ is more than $O-F$ ,therefore $O-S$ is more polar than $O-F$

(c)

Interpretation Introduction

Interpretation:

The more polar bond is to be found out in the given pair of N-S and N-Cl.

Concept introduction:

A bond is polar if there is difference in electronegativity values of atoms forming the bond.

More the difference, more is polarity of bond.

  $\begin{array}{l}{\chi }_{A-B}={\chi }_A-{\chi }_{B}if{\chi }_A>{\chi }_B\\ and{\chi }_B-{\chi }_{A}if{\chi }_B>{\chi }_A\end{array}$

Answer to Problem 7RQ

  $N-S$ is more polar than $N-Cl$

Explanation of Solution

  $\begin{array}{l}{\chi }_S=2.58\\ {\chi }_N=3.04\\ {\chi }_{Cl}=3.16\end{array}$

Thus,

$\begin{array}{c}{\chi }_{N-Cl}=3.16-3.04\\ =0.12\\ {\chi }_{N-S}=3.04-2.58\\ =0.46\end{array}$

Electronegativity difference of $N-S$ is more than $N-Cl$ ,therefore $N-S$ is more polar than $N-Cl$

(d)

Interpretation Introduction

Interpretation:

The more polar bond is to be found out in the given pair of C-Cl and C-S.

Concept introduction:

A bond is polar if there is difference in electronegativity values of atoms forming the bond.

More the difference, more is polarity of bond.

  $\begin{array}{l}{\chi }_{A-B}={\chi }_A-{\chi }_{B}if{\chi }_A>{\chi }_B\\ and{\chi }_B-{\chi }_{A}if{\chi }_B>{\chi }_A\end{array}$

Answer to Problem 7RQ

  $C-Cl$ is more polar than $C-S$

Explanation of Solution

  $\begin{array}{l}{\chi }_S=2.58\\ {\chi }_C=2.55\\ {\chi }_{Cl}=3.16\end{array}$

Thus,

$\begin{array}{c}{\chi }_{C-Cl}=3.16-2.55\\ =0.61\\ {\chi }_{C-S}=2.58-2.55\\ =0.03\end{array}$

Electronegativity difference of $C-Cl$ is more than $C-S$ ,thereforeC-Clis more polar than $C-S$