Chapter 1, Problem 1.68P

Interpretation Introduction

(a)

Interpretation: The shortest $\text{C}-\text{C}$ single bond is to be labeled.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order, shorter is the bond length.

Answer to Problem 1.68P

The shortest $\text{C}-\text{C}$ single bond is $[2]$.

Explanation of Solution

In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital.

Figure 1

Both $[1]$ and $[2]$ are single $\text{C}-\text{C}$ bond as shown above. Bond $[1]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybrid orbitals and bond $[2]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}$ hybrid orbitals. Thus, bond $[2]$ has more percent s-character. Hence, the shortest $\text{C}-\text{C}$ single bond is $[2]$.

Conclusion

The shortest $\text{C}-\text{C}$ single bond is $[2]$.

Interpretation Introduction

(b)

Interpretation: The longest $\text{C}-\text{C}$ single bond is to be labeled.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order shorter is the bond length.

Answer to Problem 1.68P

The longest $\text{C}-\text{C}$ single bond is $[1]$.

Explanation of Solution

In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital. Both $[1]$ and $[2]$ are single bond as shown in Figure 1.Bondbut $[1]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybrid orbitals and bond $[2]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}$ hybrid orbitals. Thus, bond $[1]$ has less percent s-character. Hence the longest $\text{C}-\text{C}$ single bond is $[1]$.

Conclusion

The longest $\text{C}-\text{C}$ single bond is $[1]$.

Interpretation Introduction

(c)

Interpretation: The shortest $\text{C}-\text{C}$ bond is to be labeled.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order shorter is the bond length.

Answer to Problem 1.68P

The shortest $\text{C}-\text{C}$ bond is triple bond between two carbon atoms.

Explanation of Solution

In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital. Triple bond is formed by ${\text{C}}_{sp}-{\text{C}}_{sp}$ hybrid orbitals it has highest percent s-character among other $\text{C}-\text{C}$ bond. Hence, the shortest $\text{C}-\text{C}$ bond is triple bond.

Conclusion

The shortest $\text{C}-\text{C}$ bond is triple bond between two carbon atoms.

Interpretation Introduction

(d)

Interpretation: The weakest $\text{C}-\text{C}$ bond is to be labeled.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order shorter is the bond length.

Answer to Problem 1.68P

The weakest $\text{C}-\text{C}$ bond is $[1]$.

Explanation of Solution

In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital..

Both $[1]$ and $[2]$ are single bond as shown in Figure 1.Bond $[1]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybrid orbitals and bond $[2]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}$ hybrid orbitals. Bond $[1]$ has less percent s-character. Hence, the weakest $\text{C}-\text{C}$ bond is $[1]$.

Conclusion

The weakest $\text{C}-\text{C}$ bond is $[1]$.

Interpretation Introduction

(e)

Interpretation: The strongest $\text{C}-\text{H}$ bond is to be labeled.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order shorter is the bond length.

Answer to Problem 1.68P

The strongest $\text{C}-\text{H}$ bond is with triple $\text{C}-\text{C}$ bond.

Explanation of Solution

The length and strength of a $\text{C}-\text{H}$ bond depends on the hybridization of the carbon atom. In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital.

Triple bond is formed by ${\text{C}}_{sp}-{\text{C}}_{sp}$ it has highest percent s-character among other $\text{C}-\text{C}$ bond. Hence, $\text{C}-\text{H}$ with triple $\text{C}-\text{C}$ bond has highest percent s character and is the strongest bond.

Conclusion

The strongest $\text{C}-\text{H}$ bond is with triple $\text{C}-\text{C}$ bond.

Interpretation Introduction

(f)

Interpretation: Bond $[1]$ and $[2]$ are of different length, even though they are both $\text{C}-\text{C}$ single bond reason is to be explained.

Concept introduction: There is an inverse relationship between bond length and bond strength. Shorter bonds are stronger bonds or higher the bond order shorter is the bond length.

Answer to Problem 1.68P

Bond $[1]$ and $[2]$ are of different length, even though they are both $\text{C}-\text{C}$ single bond due to the difference in percent-s character.

Explanation of Solution

In hybridization, one $2s$ orbital is always used, but the number of $2p$ orbitals varies with the type of hybridization. The percent s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent s-character increases, the bond becomes shorter and stronger. This due to the fact that a $2s$ orbital keeps electron density closer to a nucleus as compared to a $2p$ orbital. Both $[1]$ and $[2]$ are single bond as shown in Figure 1. Bond $[1]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^3}$ hybrid orbitals and bond $[2]$ is formed by ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}$ hybrid orbitals. Hence, bond $[2]$ has more percent s character and is shorter in length than bond $[1]$.

Conclusion

Bond $[1]$ and $[2]$ are of different length, even though they are both $\text{C}-\text{C}$ single bond due to the difference in percent-s character.