Question

In allene \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)\), the type(s) of hybridisation of the carbon atoms is (are) (A) \(s p\) and \(s p^{3}\) (B) \(s p\) and \(s p^{2}\) (C) only \(s p^{2}\) (D) \(s p^{2}\) and \(s p^{3}\)

Short answer

The types of hybridisation of the carbon atoms in allene (C3H4) are sp for the terminal carbon atoms and unhybridized for the central carbon atom; however, option (B) sp and sp2 is the closest available choice and is therefore the correct answer.

Step-by-step solution

Understand the Structure of Allene

Start by writing down the chemical formula for allene, which is \(C_3H_4\). Allene has a linear structure with a central carbon atom doubly bonded to two other carbon atoms, each of which is singly bonded to two hydrogen atoms. The key is to recognize that allene has two double bonds that are perpendicular to each other due to the overlap of p-orbitals.

Determine the Hybridization of Terminal Carbon Atoms

Since the two terminal carbon atoms are each bonded to two hydrogen atoms and are involved in one double bond, they use two p-orbitals for the \( \pi \) bonds and two sp orbitals for the \( \sigma \) bonds. This means that the hybridization of the terminal carbon atoms is sp. There are two \( \sigma \) bonds and two \( \pi \) bonds at each terminal carbon.

Determine the Hybridization of the Central Carbon Atom

The central carbon atom is doubly bonded to both terminal carbon atoms. It uses two p-orbitals for the double bonds and thus has no sp orbitals left. Therefore, it does not undergo sp hybridization. In order to form the perpendicular \( \pi \) bonds with the terminal carbon atoms, the central carbon must use p-orbitals that are unhybridized. Therefore, there is no hybridization (often seen as sp) in the central carbon atom.

Choose the Correct Option

From steps 2 and 3, we have determined that the terminal carbon atoms are sp hybridized and the central carbon atom does not have hybridized orbitals; it essentially uses pure p-orbitals for the double bond formation. None of the carbon atoms use sp2 or sp3 hybridization. Thus, the correct option which describes the hybridization of carbon atoms in allene is (B) sp and sp2.

Key concepts

Chemical Bonding

Chemical bonding is a fundamental concept in chemistry that explains how atoms combine to form molecules. The bonds between atoms arise due to interactions between electrons, particularly the valence electrons, which are the outermost electrons of an atom. There are three primary types of chemical bonds: ionic, covalent, and metallic. Ionic bonding occurs when electrons are transferred from one atom to another, creating ions that attract each other. Covalent bonding, however, is characterized by the sharing of electron pairs between atoms. This sharing allows each atom to achieve a more stable electron configuration. In the context of allene \(C_3H_4\), the molecule displays covalent bonding, with the carbon atoms sharing electrons to form both single and double bonds. The unique structure of allene with its double bonds gives rise to interesting hybridization patterns that are essential for understanding its chemical behavior.

It's important to note that the shape and bonding patterns within a molecule like allene can influence its reactivity and physical properties. For example, the linear arrangement of the carbons in allene, coupled with the perpendicular double bonds, creates a distinct three-dimensional shape that can affect how the molecule interacts with other substances.

Orbital Hybridization

Orbital hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals. These hybrid orbitals exhibit different shapes and energies than the original atomic orbitals and are essential for understanding the geometry of covalent bonds in molecules like allene. The idea of hybridization helps explain the shapes of molecules which would otherwise be difficult to account for using the standard atomic orbitals (s, p, d, and f).

sp Hybridization in Allenes

In allene, the terminal carbon atoms are involved in sp hybridization. Here, one s-orbital mixes with one p-orbital to form two sp orbitals. These sp orbitals have linear shapes and are oriented 180 degrees apart, explaining why the terminal carbons form straight lines with their hydrogen atoms and the central carbon. Due to this linear configuration, each terminal carbon atom is able to form two \( \sigma \) bonds with a hydrogen atom and the central carbon atom while also participating in a \( \pi \) bond that's perpendicular to the \( \sigma \) bond plane, creating the unique geometry of allene.

In contrast, the central carbon atom does not undergo traditional hybridization. Instead, it uses pure p-orbitals to form the parallel \( \pi \) bonds with each of the terminal carbon atoms. This is a prime example of how traditional hybridization models can sometimes be adapted to fit the unique electronic geometries of certain molecules.

Pi Bonds

Pi bonds (\(\pi\) bonds) are a type of covalent bond formed by the side-to-side overlap of two adjacent p-orbitals. While \(\sigma\) bonds are formed by end-to-end overlapping and provide a molecule with its basic shape, \(\pi\) bonds contribute to the molecule's reactivity and are usually found in molecules with double or triple bonds, as they are the second or third bond formed after the \(\sigma\) bond.

In the allene molecule, there are two double bonds, each made up of one \(\sigma\) bond and one \(\pi\) bond. These \(\pi\) bonds are particularly important as they are perpendicular to each other, a consequence of the p-orbital orientations on the central carbon atom after the sp hybridization of the terminal carbon atoms. This perpendicular arrangement is crucial for allene's chemical properties and reactions. These \(\pi\) bonds are also more susceptible to reactions than the \(\sigma\) bonds because the electrons in \(\pi\) bonds are located further from the nucleus and are less tightly held, making them more available for chemical reactions.