Q. 6.136 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the electronegativity difference and classify each of the following bonds as nonpolar covalent, polar covalent, or ionic: 

a. $C and N$                b. $Cl and Cl$                 c. $K and Br$

d. $H and H$                e. $N and F$.

Step-by-Step Solution

Answer

a. The bond is a polar covalent bond.

b. The bond is a nonpolar covalent bond.

c. The bond is an ionic bond.

d. The bond is a nonpolar covalent bond.

e. The bond is a polar covalent bond.

Part (a) Step 1: Given Information

We need to calculate the electronegativity of $C-N$ bond and classify whether the bond is nonpolar covalent, polar covalent, or ionic: nonpolar covalent, polar covalent, or ionic.

Part (a) Step 2: Explanation

If electronegativity difference is from $0 to 0.4$ then the bond is a nonpolar covalent bond,

if the difference is from $0.4 to 1.8$ then the bond is a polar covalent bond,

and if the difference is greater than $1.8$ then the bond is an ionic bond.

Now, consider the electronegativity of $C$ is $2.5$ and of $N$ is $3$.

The electronegativity difference in $C-N$ bond is $3-2.5=0.5$

So, the $C-N$ bond is a polar covalent bond.

Part (b) Step 1: Given Information

We need to calculate the electronegativity of $Cl-Cl$ bond and classify whether the bond is nonpolar covalent, polar covalent, or ionic: nonpolar covalent, polar covalent, or ionic.

Part (b) Step 2: Explanation

Considering the electronegativity of $Cl$ is $3$.

The electronegativity difference in $Cl-Cl$ bond is zero.

So, the $Cl-Cl$ bond is a nonpolar covalent bond.

Part (c) Step 1: Given Information

We need to calculate the electronegativity of $K-Br$bond and classify whether the bond is nonpolar covalent, polar covalent, or ionic: nonpolar covalent, polar covalent, or ionic.

Part (c) Step 2: Explanation

Considering the electronegativity of $K$ is $0.8$ and of $Br$ is $2.8$

The electronegativity difference in $K-Br$ bond is $2.8-0.8=2$.

So, the $K-Br$ bond is a ionic bond.

Part (d) Step 1: Given Information

We need to calculate the electronegativity of $H-H$ bond and classify whether the bond is nonpolar covalent, polar covalent, or ionic: nonpolar covalent, polar covalent, or ionic.

Part (d) Step 2: Explanation

Considering the electronegativity of $H$ is $2.1$

The electronegativity difference of $H-H$ bond is zero.

So, the $H-H$ bond is a nonpolar covalent bond.

Part (e) Step 1: Given Information

We need to calculate the electronegativity of $N-F$ bond and classify whether the bond is nonpolar covalent, polar covalent, or ionic: nonpolar covalent, polar covalent, or ionic.

Part (e) Step 2: Explanation

Considering the electronegativity of $N$ is $3$ and of $F$ is $4$.

The electronegativity difference of $F-N$ bond is $4-3=1$.

So, the $F-N$ bond is a polar covalent bond.