Question

Question: What orbitals are used to form the labeled bonds in the following molecule? Of the labeled bonds, which is the shortest?

Short answer

Answer

1. $H_{\left(1s\right)}-C_{\left(s{p}^2\right)}$
2. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^3\right)}$
3. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$(Double bond outside ring)
4. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$(Single bond)
5. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$ (Double bond inside ring)

Shortest bond: Bond labeled 3

Step-by-step solution

Hybridization orbitals

Bonds are formed due to the mixing of pure orbitals and forming hybrid orbitals. Depending upon the number of groups surrounding carbon hybridization is fixed.

When carbon is surrounded by four groups it is $s{p}^3$hybridized and when attached to three groups it is $s{p}^2$hybridized.

Effect of position on the double bond

As the bond order increases, bond length decreases. Therefore, double bonds are shorter in length as compared to single bonds.

C-C bonds inside the benzene ring undergo resonance due to which all C-C bond length lies between a double-bond and a single bond.

Orbitals involved in bonds

Given compound

1. $H_{\left(1s\right)}-C_{\left(s{p}^2\right)}$
2. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^3\right)}$
3. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$(Double bond outside ring)
4. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$(Single bond)
5. $C_{\left(s{p}^2\right)}-C_{\left(s{p}^2\right)}$(Double bond inside ring)

Bond 3 in the structure is a double bond outside the ring and is the shortest among all bonds labeled.