Chapter 12, Problem 12A

(a)

Interpretation Introduction

Interpretation:

The bond with more ionic character is to be found out in the pair $\mathrm{N}\mathrm{a}-\mathrm{O}$ and $\mathrm{N}\mathrm{a}-\mathrm{N}$.

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 and more is the ionic character.

Answer to Problem 12A

  $\mathrm{N}\mathrm{a}-\mathrm{O}$ is more ionic than $\mathrm{N}\mathrm{a}-\mathrm{N}$.

Explanation of Solution

Electronegativity of sodium is $0.93$

Electronegativity of oxygen is $3.44$

Electronegativity of nitrogen is $3.04$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{O}$ $=3.44-0.93=2.51$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{N}$ $=3.04-0.93=2.11$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{O}$ is more than $\mathrm{N}\mathrm{a}-\mathrm{N}$ ,therefore $\mathrm{N}\mathrm{a}-\mathrm{O}$ is more ionic than $\mathrm{N}\mathrm{a}-\mathrm{N}$.

(b)

Interpretation Introduction

Interpretation:

The bond with more ionic character is to be found out in the pair $\mathrm{K}-\mathrm{S}$ and $\mathrm{K}-\mathrm{P}$.

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.

Answer to Problem 12A

  $\mathrm{K}-\mathrm{S}$ is more ionic than $\mathrm{K}-\mathrm{P}$.

Explanation of Solution

Electronegativity of potassium is $0.82$

Electronegativity of sulphur is $2.58$

Electronegativity of phosphorus is $2.19$

Electronegativity difference of $\mathrm{K}-\mathrm{S}$ $=2.58-0.82=1.76$

Electronegativity difference of $\mathrm{K}-\mathrm{P}$ $=2.19-0.82=1.37$

Electronegativity difference of $\mathrm{K}-\mathrm{S}$ is more than $\mathrm{K}-\mathrm{P}$ ,therefore $\mathrm{K}-\mathrm{S}$ is more ionic than $\mathrm{K}-\mathrm{P}$.

(c)

Interpretation Introduction

Interpretation:

The bond with more ionic character is to be found out in the pair $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ and $\mathrm{K}-\mathrm{C}\mathrm{l}$.

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.

Answer to Problem 12A

  $\mathrm{K}-\mathrm{C}\mathrm{l}$ is more ionic than $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$.

Explanation of Solution

Electronegativity of potassium is $0.82$

Electronegativity of Chlorine is $3.16$

Electronegativity of sodium is $0.93$

Electronegativity difference of $\mathrm{K}-\mathrm{C}\mathrm{l}$ $=3.16-0.82=2.34$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ $=3.16-0.93=2.23$

Electronegativity difference of $\mathrm{K}-\mathrm{C}\mathrm{l}$ is more than $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ ,therefore $\mathrm{K}-\mathrm{C}\mathrm{l}$ is more ionic than $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$.

(d)

Interpretation Introduction

Interpretation:

The bond with more ionic character is to be found out in the pair $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ and $\mathrm{M}\mathrm{g}-\mathrm{C}\mathrm{l}$.

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.

Answer to Problem 12A

  $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ is more ionic than $\mathrm{M}\mathrm{g}-\mathrm{C}\mathrm{l}$.

Explanation of Solution

Electronegativity of magnesium is $1.31$

Electronegativity of Chlorine is $3.16$

Electronegativity of sodium is $0.93$

Electronegativity difference of $\mathrm{M}\mathrm{g}-\mathrm{C}\mathrm{l}$ $=3.16-1.31=1.85$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ $=3.16-0.93=2.23$

Electronegativity difference of $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ is more than $\mathrm{M}\mathrm{g}-\mathrm{C}\mathrm{l}$ ,therefore $\mathrm{N}\mathrm{a}-\mathrm{C}\mathrm{l}$ is more ionic than $\mathrm{M}\mathrm{g}-\mathrm{C}\mathrm{l}$.