Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: In ${\text{CH}}_2=\text{C}=\text{CH}-{\text{CH}}_3$ molecule, the hybridisation of carbon 1, 2, 3 and 4 respectively are

When it comes to finding the hybridisation of an atom in an organic molecule, the most reliable and straightforward method is the Steric Number Rule. The steric number is simply the sum of the number of $\sigma$ (sigma) bonds formed by the atom and the number of lone pairs of electrons it possesses.

For carbon atoms in stable organic molecules, there are usually no lone pairs. Therefore, the hybridisation is entirely dictated by the number of $\sigma$ bonds: - 4 $\sigma$ bonds $\to$ Steric number 4 $\to$ $s{p}^3$ hybridisation - 3 $\sigma$ bonds $\to$ Steric number 3 $\to$ $s{p}^2$ hybridisation - 2 $\sigma$ bonds $\to$ Steric number 2 $\to$ $sp$ hybridisation

Remember, $\pi$ (pi) bonds are formed by the lateral overlap of unhybridised $p$-orbitals, so they do not contribute to the steric number or the hybridisation state.

Let's break down the given molecule, ${\text{CH}}_2=\text{C}=\text{CH}-{\text{CH}}_3$, from left to right.

Carbon 1 ($C_1$): The terminal ${\text{CH}}_2$ group is bonded to two hydrogen atoms via single bonds (2 $\sigma$ bonds) and to the adjacent carbon via a double bond (1 $\sigma$ bond + 1 $\pi$ bond). Total $\sigma$ bonds = $2+1=3$. Thus, $C_1$ is $s{p}^2$ hybridised.

Carbon 2 ($C_2$): This central carbon is fascinating. It forms double bonds with both of its neighboring carbon atoms ($=\text{C}=$). Each double bond contributes exactly one $\sigma$ bond. Total $\sigma$ bonds = $1+1=2$. Thus, $C_2$ is $sp$ hybridised.

Carbon 3 ($C_3$): This carbon is part of the $=\text{CH}-$ group. It has a double bond with $C_2$ (1 $\sigma$ bond), a single bond with a hydrogen atom (1 $\sigma$ bond), and a single bond with $C_4$ (1 $\sigma$ bond). Total $\sigma$ bonds = $1+1+1=3$. Thus, $C_3$ is $s{p}^2$ hybridised.

Carbon 4 ($C_4$): The terminal methyl group ($-{\text{CH}}_3$) forms a single bond with $C_3$ and three single bonds with three hydrogen atoms. Total $\sigma$ bonds = $1+3=4$. Thus, $C_4$ is $s{p}^3$ hybridised.

Putting it all together, the hybridisation sequence for carbons 1, 2, 3, and 4 is $s{p}^2,sp,s{p}^2,s{p}^3$. This perfectly matches option (c).

A Deeper Insight: Molecules containing consecutive double bonds (like the $C_1=C_2=C_3$ segment) are known as allenes. Because the central $sp$ hybridised carbon uses two mutually perpendicular $p$-orbitals to form its two $\pi$ bonds, the groups attached to the terminal $s{p}^2$ carbons are forced into perpendicular planes. Consequently, despite its linear-looking backbone, the entire molecule is non-planar. This is a favorite concept for JEE Advanced, so always keep an eye out for consecutive double bonds!