Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

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

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is $\text{d}<\text{a}<\text{c}<\text{e}<\text{b}<\text{f}$.

Explanation of Solution

The given compound is,

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage $s-$ character in orbitals is $s{p}^3<s{p}^2<sp$.

• Bond ‘a’ is present between the $s{p}^3$ hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the $s{p}^2$ hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of $s-$ character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two $s{p}^2$ hybridized carbon atoms, while bond ‘f’ is present between the two $s{p}^3$ hybridized carbon atoms. Bond ‘c’ has higher percentage of $s-$ character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one $s{p}^2$ hybridized carbon atom and one $s{p}^3$ hybridized carbon atom, while bond ‘f’ is present between the two $s{p}^3$ hybridized carbon atoms. Bond ‘b’ has higher percentage of $s-$ character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The $\text{C}\left(s{p}^2\right)-\text{H}\left(s\right)$ is shortest bond as this bond has the highest percentage $s-$ character.

From the above points, it is concluded that the order of bond-lengths in the given compound is $\text{d}<\text{a}<\text{c}<\text{e}<\text{b}<\text{f}$.

Conclusion

The increasing order of bond-lengths in the given compound is $\text{d}<\text{a}<\text{c}<\text{e}<\text{b}<\text{f}$.