Problem 30 - A Complete Resource Book in Chemistry for JEE Main

Question

Hybridization of 1 and 2 carbon atom in \({ }^{1} \mathrm{CH}_{2}={ }^{2} \mathrm{C}=\mathrm{CH}_{2}\) are (a) \(\mathrm{sp}, \mathrm{sp}\) (b) \(s p^{2}, s p^{2}\) (c) \(\mathrm{sp}^{2}\), sp (d) \(\mathrm{sp}^{3}, \mathrm{sp}^{2}\)

Step-by-Step Solution

Answer

The hybridization states are \(\mathrm{sp}^{2}\) for C-1 and \(\mathrm{sp}\) for C-2. Option (c) is correct.

Step 1: Understand the Structure and Bonds

The given molecule is \(^{1} \mathrm{CH}_{2}=^{2} \mathrm{C}=\mathrm{CH}_{2}\). The structure shows the presence of two double bonds spanning the three carbon atoms.

Step 2: Analyze the Carbon-1 Hybridization

Carbon-1 is attached to two hydrogen atoms and forms part of a double bond with Carbon-2. Hence, Carbon-1 engages in three sigma bonds and one pi bond, indicating \(\mathrm{sp}^{2}\) hybridization.

Step 3: Analyze the Carbon-2 Hybridization

Carbon-2 forms two double bonds, one with Carbon-1 and the other with the terminal CH group. This requires it to engage in two sigma bonds and two pi bonds, implying \(\mathrm{sp}\) hybridization.

Step 4: Choose the Correct Option

Based on the analysis, Carbon-1 has \(\mathrm{sp}^{2}\) hybridization and Carbon-2 has \(\mathrm{sp}\) hybridization. Therefore, the correct answer is option (c): \(\mathrm{sp}^{2}, \mathrm{sp}\).

Key Concepts

Carbon Atom

At the core of organic chemistry lies the carbon atom, an incredibly versatile element that forms the backbone of many molecules. Carbon is unique because it can form four covalent bonds, allowing it to create complex and diverse structures. One remarkable feature of the carbon atom is its ability to form various types of hybridizations, such as sp, sp2, and sp3. These hybridizations influence the geometry and bonding properties of carbon-containing compounds.

In molecules like the one given in the exercise, the carbon atoms play a pivotal role in determining the structure and reactivity of the molecule. By analyzing the hybridization of each carbon atom, we can infer the molecule's overall geometry and bonding interactions. This understanding is crucial when predicting the behavior and properties of organic molecules.

Double Bond

Double bonds are a key feature in organic chemistry, representing a type of covalent bond where two pairs of electrons are shared between two atoms. This results in a bond that is shorter and stronger than a single bond. Double bonds significantly impact the molecular geometry and can restrict the rotation around the bond axis due to the nature of the pi bond involved.

In the given molecule, the presence of double bonds between carbon atoms indicates that these atoms exhibit specific hybridizations and bond angles. For instance, when carbon is doubly bonded to another atom, it typically undergoes ext{sp}^2 hybridization, resulting in a planar structure with approximately 120° bond angles. This feature allows molecules with double bonds to influence chemical reactivity and physical properties like solubility and melting point.

Sigma Bond

Sigma bonds (σ-bonds) are the product of head-on overlapping of atomic orbitals, and they are among the strongest covalent bonds due to this effective overlap. In any given molecule, sigma bonds provide the fundamental framework as they are involved in all types of hybridization and are present in single, double, and triple bonds.

In a double bond, like those seen in the molecule in the exercise, there is one sigma bond formed through the overlap of sp hybrid orbitals or other hybrid orbitals, providing a stable interaction and a strong connection between the atoms. Sigma bonds are characterized by allowing free rotation around the bond axis, but this rotation is constrained in the presence of pi bonds, as seen in double bonds.

Pi Bond

Pi bonds (π-bonds) arise from the side-to-side overlap of p orbitals, creating a less strong bond than a sigma bond. They play a crucial role in defining the properties of molecules with double or triple bonds. Pi bonds prevent the rotation around the bond axis, thereby locking the atoms in a fixed position.

In the specific molecule from our exercise, the double bonds consist of one sigma bond, which provides structural stability, and one pi bond, which contributes to the rigidity of the structure. This characteristic of pi bonds affects molecular shapes and is essential in determining the reactivity patterns of alkenes and other unsaturated hydrocarbons in chemical reactions. Being able to form pi bonds is a critical factor in the conformation and functionality of many organic compounds.