Chapter 6, Problem 6.20QP

(a)

Interpretation Introduction

Interpretation:

The given bonds should be classified as ionic, polar covalent, nonpolar covalent.

Concept introduction:

• Ionic compound: Ionic compounds are those which consist of charged ions. These have opposite charges
• Covalent compounds: covalent compounds are non-metals which are bound together and consist of two electrons that are shared between two atoms.
• Electronegativity can be defined as the ability of an atom in a compound to attract the shared pair of electron towards itself.
• The larger the electronegativity difference, the greater the ionic character.
• The larger the electronegativity difference, lesser the covalent character
• The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

  $\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

Answer to Problem 6.20QP

The bond is non polar covalent

Explanation of Solution

The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

$\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

So it is needed to find the electronegativity difference for the given each bonds and it should be mentioned the types of by looking the electronegativity difference range for each bond.

$\text{C-C}$

Electronegativity of Carbon is 2.5

Here there are two carbon atoms. Electronegativity of carbon is 2.55. Since both shared pair of atoms are same, the given bond is non-polar covalent

Electronegativity difference $=\text{ 2}\text{.5}-2.5=\text{ 0}$

So, the given bond is non polar.

(b)

Interpretation Introduction

Interpretation:

The given bonds should be classified as ionic, polar covalent, nonpolar covalent.

Concept introduction:

• Ionic compound: Ionic compounds are those which consist of charged ions. These have opposite charges
• Covalent compounds: covalent compounds are non-metals which are bound together and consist of two electrons that are shared between two atoms.
• Electronegativity can be defined as the ability of an atom in a compound to attract the shared pair of electron towards itself.
• The larger the electronegativity difference, the greater the ionic character.
• The larger the electronegativity difference, lesser the covalent character
• The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

  $\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

Answer to Problem 6.20QP

The bond is ionic

Explanation of Solution

The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

$\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

So it is needed to find the electronegativity difference for the given each bonds and it should be mentioned the types of by looking the electronegativity difference range for each bond.

$\text{K-I}$

Electronegativity of Potassium is 0.82

Electronegativity of iodine is 2.66

Electronegativity difference=$2.66-0.82=1.84$

So, the given bond is ionic

(c)

Interpretation Introduction

Interpretation:

The given bonds should be classified as ionic, polar covalent, nonpolar covalent.

Concept introduction:

• Ionic compound: Ionic compounds are those which consist of charged ions. These have opposite charges
• Covalent compounds: covalent compounds are non-metals which are bound together and consist of two electrons that are shared between two atoms.
• Electronegativity can be defined as the ability of an atom in a compound to attract the shared pair of electron towards itself.
• The larger the electronegativity difference, the greater the ionic character.
• The larger the electronegativity difference, lesser the covalent character
• The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

  $\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

To classify: The given bond as ionic, polar covalent, nonpolar

Answer to Problem 6.20QP

The bond is polar covalent

Explanation of Solution

The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

$\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

So it is needed to find the electronegativity difference for the given each bonds and it should be mentioned the types of by looking the electronegativity difference range for each bond.

1. a. $\text{C-C}$

Electronegativity of Carbon is 2.5

Here there are two carbon atoms. Electronegativity of carbon is 2.55. Since both shared pair of atoms are same, the given bond is non-polar covalent

Electronegativity difference $=\text{ 2}\text{.5}-2.5=\text{ 0}$

So, the given bond is non polar.

$\text{N-B}$

Electronegativity of nitrogen is 3.04

Electronegativity of boron is 2.04

Electronegativity difference=$\text{3}\text{.04-2}\text{.01=1}$

So, the given bond is polar covalent

(d)

Interpretation Introduction

Interpretation:

The given bonds should be classified as ionic, polar covalent, nonpolar covalent.

Concept introduction:

• Ionic compound: Ionic compounds are those which consist of charged ions. These have opposite charges
• Covalent compounds: covalent compounds are non-metals which are bound together and consist of two electrons that are shared between two atoms.
• Electronegativity can be defined as the ability of an atom in a compound to attract the shared pair of electron towards itself.
• The larger the electronegativity difference, the greater the ionic character.
• The larger the electronegativity difference, lesser the covalent character
• The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

  $\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

Answer to Problem 6.20QP

The bond is polar covalant

Explanation of Solution

The electronegativity difference range is used to find whether the given bond is ionic, non-polar covalent and polar covalent. That is,

$\begin{array}{l}\text{Electronegativity}\text{differnece}\text{Types}\text{of bond}\\ \text{0 to 0}\text{.4}\text{Non- polar covalent}\\ \text{0}\text{.4 to 1}\text{.7}\text{Polar}\text{covalent}\\ \text{1}\text{.7 to }\mathrm{...}\text{Ionic}\text{bond}\end{array}$

So it is needed to find the electronegativity difference for the given each bonds and it should be mentioned the types of by looking the electronegativity difference range for each bond.

So, the given bond is polar covalent

$\text{C-F}$

Electronegativity of carbon is 2.5

Electronegativity of fluorine is 3.98

Electronegativity difference = $\text{3}\text{.98-2}\text{.5 =1}\text{.48}$

So, the given bond is polar covalent.