Question

Which of the following statements is incorrect?
A.$OC{l_2} > {H_2}O > O{F_2}$ order of bond angle
B.In benzyne, each carbon is $s{p^2}$ hybridized
C.In benzene, all the bonds are equal
D.In benzyne, at least two carbons are sp hybridized.

Answer 1

All of the given options are about different concepts. We have to eliminate the statements that are correct first in order to get the incorrect statement. For benzyne-related statements, we have to look at the hybridization of the molecule. In benzene, we have to look at the resonance phenomenon to check if it is true.

Complete answer:
We know that bond angle increases as the size of the bonding atoms increases. However, ${H_2}O$ has a higher bond angle than $O{F_2}$ ​due to the larger electronegativity difference between oxygen and the hydrogen atom. This leads to a concentration of electron density around the oxygen atom more in the case of ${H_2}O$ and hence, larger repulsion and larger bond angle for water.
In benzynes, all the carbons are $s{p^2}$ hybridized. The triple bond in benzyne is due to the lateral overlapping of the non-parallel p orbitals and not due to a change in hybridization.
The bond length of all Carbon-carbon bonds in benzene are equal due to resonance. Therefore we can say that option C is also correct.
Therefore we can say that the only incorrect statement is “In benzyne, at least two carbon are sp hybridized”
Hence, (D) is the correct option.

Note:
Resonance is a very important phenomenon in organic chemistry. There are many consequences of resonance in an organic molecule. It can explain the stability of a large number of molecules. It also explains bond length in molecules and how certain molecules are more reactive to a certain type of reaction.