Q. 6.68 - An Introduction to General, Organic, and Biological Chemistry

Question

Predict whether each of the following bonds is nonpolar covalent, polar covalent, or ionic:

a. $Si$ and $O$

b. $K$ and $Cl$

c. $S$ and $F$

d. $P$ and $Br$

e. $Li$ and $O$

f.  $N$ and $S$

Step-by-Step Solution

Answer

We get following category of bond corresponding to linkage between atoms:

(a) $Si$ and $O$ : Polar Covalent Bond

(b) $K$ and $Cl$ : Ionic Bond

(c) $S$ and $F$ : Polar Covalent Bond

(d) $P$ and $Br$ : Polar Covalent Bond

(e) $Li$ and $O$ : Ionic Bond

(f) $N$ and $S$: Polar Covalent Bond

Part (a) Step 1: Given Information

We need to find type of bond between $Si$ and $O$

Part (a) Step 2: Explanation

• Electronegativity of $Si$ is $1.8$ and $O$ is $3.5$
  

• Electronegativity difference= $3.5-1.8=1.7$

As electronegativity difference is $1.7$. So, it forms polar covalent bond.

Part (b) Step 1: Given Information

We need to find type of bond between $K$ and $Cl$

Part (b) Step 2: Explanation

• Electronegativity of $K$ is $0.8$ and $Cl$ is $3.0$

• Electronegativity difference= $3.0-0.8=2.2$

As electronegativity difference is $2.2$. So, it forms ionic bond.

Part (c) Step 1: Given Information

We need to find type of bond between $S$ and $F$

Part (c) Step 2: Explanation

• Electronegativity of $S$ is $2.5$ and $F$ is $4.0$

• Electronegativity difference= $4.0-2.5=1.5$

As electronegativity difference is $1.5$. So, it forms polar covalent bond.

Part (d) Step 1: Given Information

We need to find type of bond between $P$ and $Br$

Part (d) Step 2: Explanation

• Electronegativity of $P$ is $2.1$ and $Br$ is $2.8$

• Electronegativity difference= $2.8-2.1=0.7$

As electronegativity difference is $0.7$.So, it forms polar covalent bond.

Part (e) Step 1: Given Information

We need to find type of bond between $Li$ and $O$

Part (e) Step 2: Explanation

• Electronegativity of $Li$ is $1.0$ and $O$ is $3.5$

• Electronegativity difference= $3.5-1.0=2.5$

As electronegativity difference is $2.5$. So, it forms ionic bond.

Part (f) Step 1: Given Information

We need to find type of bond between $N$ and $S$

Part (f) Step 2: Explanation

• Electronegativity of $N$ is $3.0$ and $S$ is $2.5$ 

• Electronegativity difference= $3.0-2.5=0.5$

As electronegativity difference is $0.5$. So, it forms polar covalent bond.