Question

Question: Rank the following bonds in order of increasing bond length.

Short answer

Answer

The bonds highlighted in the given structure are arranged in the increasing order of the bond length as shown below:

$\mathrm{Bond}\left[1\right]<\mathrm{Bond}\left[2\right]<\mathrm{Bond}\left[3\right]$

Step-by-step solution

Step-by-Step SolutionStep 1: Bond length

Bond length is the average length/distance starting from the center of one atom's nucleus to the center of the nucleus of the other bonded atom.

The factors affecting the length of a bond

The length of a bond depends upon the following factors:

• The number of electrons: With the increase in the number of electrons between the two nuclei, the strength of the covalent bonds also increases. Due to the increase in the strength of the bond, the bond length decreases.

• Bond order: It is defined as the number of bonds formed between atoms. With the increase in the bond order, the strength of the covalent bond also increases. As a result, the bond length decreases.

The single bonds are longer and weaker than the double bonds, which are further longer and weaker than the triple bonds. This implies that the triple bonds are the strongest.

Type of hybridization: The strength of the bond increases if the hybridized orbitals have more s character. The bond length of the carbon-hydrogen bonds increases with the decrease in s character of the hybridized orbitals

The increasing order of the bond length for the highlighted bonds

The order of the increasing bond length for the bonds in the given structure is as follows:

$\mathrm{Bond}\left[1\right]<\mathrm{Bond}\left[2\right]<\mathrm{Bond}\left[3\right]$

Bond [3] is the longest, and bond [1] is the shortest among the highlighted bonds in the given structure.

Bond 1 is formed by the $\mathrm{sp}$-hybridized orbitals of C and 1s orbital of the hydrogen atom, i.e., $\mathrm{C}\left(\mathrm{sp}\right)-\mathrm{H}\left(1\mathrm{s}\right)$. The sp orbitals have 50% s character. Therefore, it is shorter and stronger.

Due to more s character in the hybrid orbitals, the electron density becomes closer to the nucleus, increasing the bond strength. Consequently, the bond length decreases.

Bond 2 is formed between the ${\mathrm{sp}}^2$-hybridized orbitals of carbon. The ${\mathrm{sp}}^2$-hybridized orbitals have 33.33% s character, which is lower than the $\mathrm{sp}$-hybridized orbitals.

The bond [3] is the longest. Bond [3] is formed by the ${\mathrm{sp}}^3$-hybridized orbitals of the carbon atom and 1s orbital of the hydrogen atom. The${\mathrm{sp}}^3$ -hybridized orbitals of carbon have 25% s character. Therefore, bond [3] is the longest and weakest of all the bonds.