Q. 6.131 - An Introduction to General, Organic, and Biological Chemistry

Question

Select the more polar bond in each of the following pairs: ($6.7$) 

a. data-custom-editor="chemistry" $\mathrm{C}-\mathrm{N}$ or data-custom-editor="chemistry" $\mathrm{C}-\mathrm{O}$ 

b. data-custom-editor="chemistry" $\mathrm{N}-\mathrm{F}$ or data-custom-editor="chemistry" $\mathrm{N}-\mathrm{Br}$

c. data-custom-editor="chemistry" $\mathrm{Br}-\mathrm{Cl}$ or data-custom-editor="chemistry" $\mathrm{S}-\mathrm{Cl}$

d. data-custom-editor="chemistry" $\mathrm{Br}-\mathrm{Cl}$ or data-custom-editor="chemistry" $\mathrm{Br}-\mathrm{I}$

e. data-custom-editor="chemistry" $\mathrm{N}-\mathrm{F}$ or data-custom-editor="chemistry" $\mathrm{N}-\mathrm{O}$

Step-by-Step Solution

Answer

(a) The more polar bond is $\mathrm{C}-\mathrm{O}$

(b) The more polar bond is $\mathrm{N}-\mathrm{F}$

(c) The more polar bond is $\mathrm{S}-\mathrm{Cl}$

(d) The more polar bond is $\mathrm{Br}-\mathrm{I}$

(e) The more polar bond is $\mathrm{N}-\mathrm{F}$

Part (a) Step 1: Given information

We need to find the more polar bond.

Part (a) Step 2: Explanation

We know that

The greater the electronegativity difference in the elements forming the bond, the greater is the polarity of the bond 

As the electronegativity of carbon, oxygen and nitrogen respectively is $2.55, 3.5, 3.04$

So, from above electronegativities we can calculate electronegativity difference. 

Therefore, The more polar bond is,

$\mathrm{C}-\mathrm{O}$

Part (b) Step 1: Given information

We need to find the more polar bond. 

Part (b) Step 2: Explanation

We know that

The greater the electronegativity difference in the elements forming the bond, the greater is the polarity of the bond 

As the electronegativity of fluorine, bromine, and nitrogen respectively is $3.98, 2.96, 3.04$

So, from the above electronegativities, we can calculate the electronegativity difference.

Therefore, The more polar bond is,

$\mathrm{N}-\mathrm{F}$

Part (c) Step 1: Given information

We need to find the more polar bond. 

Part (c) Step 2: Explanation

We know that

The greater the electronegativity difference in the elements forming the bond, the greater is the polarity of the bond 

As the electronegativity of bromine, chlorine, sulfur respectively is $2.96, 3.16, 2.58$

So, from above electronegativities we can calculate electronegativity difference.

Therefore, The more polar bond is,

$\mathrm{S}-\mathrm{Cl}$

Part (d) Step 1: Given information

We need to find the more polar bond. 

Part (d) Step 2: Explanation

We know that

The greater the electronegativity difference in the elements forming the bond, the greater is the polarity of the bond 

As the electronegativity of bromine, chlorine, iodine respectively is $2.96, 3.16, 2.5$

So, from above electronegativities we can calculate electronegativity difference.

Therefore, The more polar bond is,

$\mathrm{Br}-\mathrm{I}$

Part (e) Step 1: Given information

We need to find the more polar bond. 

Part (e) Step 2: Explanation

We know that

The greater the electronegativity difference in the elements forming the bond, the greater is the polarity of the bond 

As the electronegativity of nitrogen, oxygen, fluorine respectively is $3.04, 3.44, 3.98$

So, from above electronegativities we can calculate electronegativity difference.

Therefore, The more polar bond is,

$\mathrm{N}-\mathrm{F}$