Q. 6.135 - 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. $Si and Cl$                  b. $C and C$                c. $Na and Cl$

d. $C and H$                  e. $F and F$.

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 nonpolar covalent bond. 

Part (a) Step 1: Given Information

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

Part (a) Step 2: Explanation

General Information

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.

We know that the electronegativity of $Si$ is $1.8$ and of $Cl$ is $3$

$Si-Cl$ electronegativity difference $3-1.8=1.2$

So the$Si-Cl$ bond is a Polar covalent bond.

Part (b) Step 1: Given Information

We need to calculate the electronegativity of $C and C$ 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 $C$ is $2.5$

In $C-C$ bond the electronegativity difference is zero.

So, the bond is a nonpolar covalent bond.

Part (c) Step 1: Given Information

We need to calculate the electronegativity of $Na and Cl$ 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 $Na$ is $0.9$ and of $Cl$ is $3$.

The electronegativity difference of $Na-Cl$ is $3-0.9=2.1$

So, the $Na-Cl$ bond is an ionic bond.

Part (d) Step 1: Given Information

We need to calculate the electronegativity of $C and H$ 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 $C$ is $2.5$ and of $H$ is $2.1$.

The electronegativity difference of $C-H$ bond is $2.5-2.1=0.4$

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

Part (e) Step 1: Given Information

We need to calculate the electronegativity of $F and F$ 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 $F$ is $4$.

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

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