# Comparison of bond length

Question: (more than one options are correct)

My reasoning:

Option A: is correct as bond length of a pi bond is less than that of a sigma bond, hence a > c as there will be a double bond character in bond a

Option B: there will be an extra resonating structure of b. Hence there will be less double bond character in b than c. So, bond length of b>c. Thus, option B should be incorrect but it is given correct.

Option A and B both are given correct. Moreover, I can't understand how to compare bond lengths in options C an D.

## 1 Answer

Option A:

$sp^2$ hybridized carbons are more electronegative than $sp^3$ hybridized carbons; hence, the$sp^2$ hybridized carbons try to pull electrons more closely compared to the $sp^3$ hybridized atoms.

In the first molecule, $a$ represents the length of the single bond. Without any doubt, single bonds are longer than double bonds. However, the single bond exists in the first molecule exists between two $sp^2$ hybridized atoms, therefore, the bond length $a$ is shorter than the bond length $b$.

The last molecule is a double bond. This will be shortest as the other two bonds are single bonds.

$$b > a > c$$

Option B:

In the first molecule, there is no resonance stabilization. The electrons of the nitrogen atom of the amine group are not delocalized. The nitrogen-carbon bond in this molecule exists as a single bond.

In the second and third molecule, there is resonance stabilization; therefore, the nitrogen-carbon bond has a partial double bond character.

From the previous two conclusions, we can infer that the $a$ will be the largest among the three.

Between the second and third molecule, the third one has a deactivating group $\ce{NO_2}$. A positive charge in the benzene ring is very destabilizing. The nitrogen atom of the amine group tries to help by delocalizing its electrons for a longer time. Therefore, the partial double bond character of nitrogen-carbon bond in the second molecule is more than that of the third molecule.

$$a > c > b$$

Option C:

While all the three bonds are certainly double bonds, they are connecting different types of atoms.

The first bond ($a$) is a bond between a $sp^2$ hybridized carbon and a $sp$ hybridized carbon.

The second bond ($b$) is a bond between two $sp$ hybridized atoms.

The third bond ($c$) is a bond between a $sp$ hybridized carbon and a $sp^2$ hybridized carbon.

The $sp$ hybridized atoms are more electronegative compared to the $sp^2$ hybridized atoms; therefore, the $sp$ hybridized atoms pull the electrons more strongly towards themselves.

From the above conclusion, we can infer that the length $b$ is going to be the smallest among the three bond lengths.

This molecule is symmetric along the vertical line. Therefore, the bond lengths $a$ and $c$ shall be equal.

$$a = c > b$$

Option D:

Unlike benzene, the carbon–carbon bonds in naphthalene are not of the same length. The bonds C1−C2, C3−C4, C5−C6 and C7−C8 are about 1.37 Å (137 pm) in length, whereas the other carbon–carbon bonds are about 1.42 Å (142 pm) long. This difference, established by X-ray diffraction,[17] is consistent with the valence bond model in naphthalene and in particular, with the theorem of cross-conjugation. This theorem would describe naphthalene as an aromatic benzene unit bonded to a diene but not extensively conjugated to it (at least in the ground state). As such, naphthalene possesses several resonance structures.

Source: https://en.wikipedia.org/wiki/Naphthalene

$$a < b$$

• Can u explain how the carbon between two benzene rings is sp in the D part . Also if u read the first paragraph of it u will want to make an edit.:) – user237650 Mar 16 '17 at 15:09
• Oh, it isn't $sp$. Let me fix it. The correct reason is cross-conjugation. – Yashas Mar 16 '17 at 15:41