Chapter 3, Problem 3.90P

Interpretation Introduction

(a)

Interpretation:

Comment on the nature of bonds present in the ${\text{Br}}_{\text{2}}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$the bond is ionic.

Answer to Problem 3.90P

Non polar covalent.

Explanation of Solution

Electronegativity of bromine is 2.96.

There are two bromine atoms in the molecule so the electro-negativity difference will be.

$\text{ΔE}\text{.N = E}\text{.N(Br) - E}\text{.N(Br) = 2}\text{.96 - 2}\text{.96 = 0}$

Hence, the molecule is non polar covalent.

Interpretation Introduction

(b)

Interpretation:

Comment on the nature of bonds present in $\text{BrCl}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is non polar covalent.

Explanation of Solution

Electronegativity of bromine is 2.96.

Electronegativity of Chlorine is 3.16.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(Cl) - E}\text{.N(Br) = 3}\text{.16 - 2}\text{.96 = 0}\text{.2}$

This is less than 0.4, hence the molecule is non polar covalent.

Interpretation Introduction

(c)

Interpretation:

Comment on the nature of bonds present in $\text{HCl}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is polar covalent.

Explanation of Solution

Electronegativity of hydrogen is 2.20.

Electronegativity of Chlorine is 3.16.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(Cl) - E}\text{.N(H) = 3}\text{.16 - 2}\text{.20 = 0}\text{.96}$

This is more than 0.4 but less than 1.7 hence the molecule is polar covalent.

Interpretation Introduction

(d)

Interpretation:

Comment on the nature of bonds present in ${\text{SrF}}_{\text{2}}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is ionic.

Explanation of Solution

Electro-negativity of Sr is 0.95.

Electronegativity of Chlorine is 3.98.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(F) - E}\text{.N(Sr) = 3}\text{.98 - 0}\text{.95 = 3}\text{.03}$

This is more than 1.7 hence the molecule is ionic.

Interpretation Introduction

(e)

Interpretation:

Comment on the nature of bonds present in ${\text{SiH}}_4$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is non polar covalent.

Explanation of Solution

Electronegativity of Si is 1.90.

Electronegativity of Chlorine is 2.20.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(H) - E}\text{.N(Si) = 2}\text{.20 - 1}\text{.90 = 0}\text{.3}$

This is less than 0.4, hence the molecule is non polar covalent.

Interpretation Introduction

(f)

Interpretation:

Comment on the nature of bonds present in $\text{CO}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is polar covalent.

Explanation of Solution

Electronegativity of C is 2.55.

Electronegativity of oxygen is 3.44.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(O) - E}\text{.N(C) = 3}\text{.44 - 2}\text{.55 = 0}\text{.89}$

This is more than 0.4 but less than 1.7 hence the molecule is polar covalent.

Interpretation Introduction

(g)

Interpretation:

Comment on the nature of bonds present in ${\text{N}}_2$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is non polar covalent.

Explanation of Solution

Electronegativity of n is 3.04.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(N) - E}\text{.N(N) = 3}\text{.04 - 3}\text{.04 = 0}$

This is less than 0.4 non polar covalent.

Interpretation Introduction

(h)

Interpretation:

Comment on the nature of bonds present in $\text{CsCl}$ as ionic, polar covalent or non polar covalent.

Concept Introduction:

A molecule has polar bond if there is electro-negativity difference between to bonded atoms.

The electro-negativity is the tendency to pull share pair of electrons towards itself.

The following criteria is used to classify the compound as ionic, polar covalent or non polar covalent.

1. If the electro-negativity difference is $0-0.4$ the bond is non polar covalent
2. If the electro-negativity difference is $0.5-1.7$ the bond is polar covalent
3. If the electro-negativity difference is more than $1.7$ the bond is ionic.

Answer to Problem 3.90P

The molecule is ionic.

Explanation of Solution

Electro-negativity of Cs is 0.79.

Electro-negativity of Cl is 3.16.

The electro-negativity difference between the two will be.

$\text{ΔE}\text{.N = E}\text{.N(Cl) - E}\text{.N(Cs) = 3}\text{.16 - 0}\text{.79 = 2}\text{.37}$

This is more than 1.7 hence ionic.