Q. 6.67 - 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$Br$

b. $Li$and$F$

c. $Br$and$F$

d. $I$and$I$

e. $N$and$P$

f.  $C$and$P$

Step-by-Step Solution

Answer

We get following bonds corresponding to bond between atom:

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

(b)$Li$and $F$: Ionic Bond

(c) $Br$and$F$: Polar Covalent Bond

(d) $I$ and$I$: Non-Polar Covalent Bond

(e) $N$and$P$: Polar Covalent Bond

(f) $C$and$P$: Non-Polar Covalent Bond

Part (a) Step 1: Given Information

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

Part (a) Step 2: Explanation

• Electronegativity of $Si$ is $1.8$ and $Br$ is $2.8$

• Electronegativity difference= $2.8-1.8=1$

As electronegativity difference is $1$. So, it forms polar covalent bond.

Part (b) Step 1: Given Information

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

Part (b) Step 2: Explanation

• Electronegativity of $Li$ is $1.0$ and $F$ is $4.0$

• Electronegativity difference= $4.0 - 1.0 =3.0$

As electronegativity difference is $3.0$. So, it forms ionic bond.

Part (c) Step 1: Given Information

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

Part (c) Step 2: Explanation

• Electronegativity of $Br$ is $2.8$ and $F$ is $4.0$
  

• Electronegativity difference= $4.0-2.8=1.2$
  

As electronegativity difference is $1.2$. So, it forms polar covalent bond.

Part (d) Step 1: Given Information

We need to find type of bond between $I$ and $I$

Part (d) Step 2: Explanation

• Electronegativity of $I$ is $2.5$.

• Electronegativity difference of $I-I$ bond = $2.5-2.5=0$ 

As electronegativity difference is $0$. So, it forms non polar covalent bond.

Part (e) Step 1: Given Information

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

Part (e) Step 2: Explanation

• Electronegativity of $N$ is $3.0$ and $P$ is $2.1$

• Electronegativity difference= $3.0-2.1=0.9$

As electronegativity difference is $0.9$. So, it forms polar covalent bond.

Part (f) Step 1: Given Information

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

Part (f) Step 2: Explanation

• Electronegativity of $C$ is $2.5$ and $P$ is $2.1$

• Electronegativity difference= $2.5-2.1=0.4$

As electronegativity difference is $0.4$. So, it forms non polar covalent bond.