Chapter 12, Problem 11A

(a)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Electronegativity is denoted by $\chi$

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

More the difference, more is polarity of bond.

Answer to Problem 11A

  $H-O$ is more polar than $H-N$

Explanation of Solution

  $\begin{array}{l}{\chi }_H=2.2\\ {\chi }_O=3.44\\ {\chi }_N=3.04\end{array}$

Thus,

$\begin{array}{c}{\chi }_{H-O}=3.44-2.2\\ =1.24\\ {\chi }_{H-N}=3.04-2.2\\ =0.84\end{array}$

Electronegativity difference of $H-O$ is more than $H-N$ ,therefore $H-O$ is more polar than $H-N$

(b)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Electronegativity is denoted by $\chi$

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

More the difference more is polarity of bond.

Answer to Problem 11A

  $H-F$ is more polar than $H-N$

Explanation of Solution

  $\begin{array}{l}{\chi }_H=2.2\\ {\chi }_N=3.04\\ {\chi }_F=3.98\end{array}$

Thus,

$\begin{array}{c}{\chi }_{H-N}=3.04-2.2\\ =0.84\\ {\chi }_{H-F}=3.98-2.2\\ =1.78\end{array}$

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

(c)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Electronegativity is denoted by $\chi$

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

More the difference, more is polarity of bond.

Answer to Problem 11A

  $H-F$ is more polar than $H-O$

Explanation of Solution

  $\begin{array}{l}{\chi }_H=2.2\\ {\chi }_O=3.44\\ {\chi }_F=3.98\end{array}$

Thus,

$\begin{array}{c}{\chi }_{H-O}=3.44-2.2\\ =1.24\\ {\chi }_{H-F}=3.98-2.2\\ =1.78\end{array}$

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

(d)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Electronegativity is denoted by $\chi$

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

More the difference, more is polarity of bond.

Answer to Problem 11A

  $H-O$ is more polar than $H-Cl$

Explanation of Solution

  $\begin{array}{l}{\chi }_H=2.2\\ {\chi }_O=3.44\\ {\chi }_{Cl}=3.16\end{array}$

Thus,

$\begin{array}{c}{\chi }_{H-O}=3.44-2.2\\ =1.24\\ {\chi }_{H-Cl}=3.16-2.2\\ =0.96\end{array}$

Electronegativity difference of $H-O$ is more than $H-Cl$ ,therefore $H-O$ is more polar than $H-Cl$