Question

The hybridization of carbon atoms in \(\mathrm{C}-\mathrm{C}\) single bond of \(\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) is (a) \(\mathrm{sp}^{3}-\mathrm{sp}^{3}\) (b) \(\mathrm{sp}^{2}-\mathrm{sp}^{3}\) (c) \(\mathrm{sp}-\mathrm{sp}^{2}\) (d) \(\mathrm{sp}^{3}-\mathrm{sp}\)

Short answer

The correct hybridization of the carbon atoms in the single bond is \(\text{sp}-\text{sp}^2\). Option (c).

Step-by-step solution

Identify Carbon Atoms in the Single Bond

The question asks for the hybridization of carbon atoms involved in a single bond. In the compound \( ext{HC} \equiv \text{C}-\text{CH}=\text{CH}_2\), there is a single bond between the second and third carbon atoms: \(\text{C}-\text{CH}\).

Determine Hybridization of Each Carbon

Each carbon's hybridization depends on the number of atoms it is bonded to and the bonds formed. The second carbon \(\text{C}\) is part of a triple bond (\(\text{C}\equiv\text{C}\)) and a single bond, making it \(\text{sp}\) hybridized. The third carbon \(\text{CH}\) forms a double bond (\(\text{CH}=\text{CH}_2\)) and a single bond, which corresponds to \(\text{sp}^2\) hybridization.

Match to the Correct Option

The task is to identify the type of hybridization for the \(\text{C}-\text{CH}\) single bond. For this bond, the second carbon is \(\text{sp}\) hybridized, and the third carbon is \(\text{sp}^2\) hybridized. Therefore, the correct answer is \(\text{sp}-\text{sp}^2\).

Key concepts

Carbon-Carbon Single Bonds

In organic chemistry, carbon-carbon single bonds are important for constructing the framework of molecules. These single bonds are formed when two carbon atoms each contribute one electron to form a pair. This sharing of electrons constitutes a sigma (σ) bond, which is the strongest type of covalent bond found between carbon atoms.
Single bonds allow for the free rotation of the bonded atoms, offering molecules a degree of flexibility in their structure. This is different from double bonds, where restricted rotation occurs due to the presence of pi (π) bonds.
The hybridization of carbon atoms involved in single bonds can vary, depending on the specific molecular structure. It primarily affects the geometry and reactivity of organic molecules, leading to diverse chemical behaviors and physical properties. Understanding this concept is key for predicting how these molecules interact in various chemical environments.

SP Hybridization

SP hybridization occurs when one s orbital and one p orbital from the same atom combine to form two sp orbitals. This type of hybridization is seen when carbon forms a linear geometry, such as in molecules with triple bonds or two double bonds.
In an sp hybridized carbon, two p orbitals remain unhybridized and are oriented at 90 degrees to the linear axis formed by the sp orbitals. These unhybridized p orbitals participate in forming the pi (π) bonds in triple bonds, giving rise to molecules like acetylene ( -HC≡CH).
SP hybridized carbons form molecules that are linear, meaning that all atoms connected in this structure lie along a straight line. This hybridization is crucial in understanding the shape and bonding in highly unsaturated hydrocarbons.

SP2 Hybridization

SP2 hybridization occurs when one s orbital combines with two p orbitals to form three equivalent sp2 orbitals. This hybridization results in a trigonal planar geometry around the carbon atom, often observed in molecules with double bonds.
The remaining p orbital, which is not hybridized, remains perpendicular to the plane of the sp2 orbitals. This p orbital participates in forming a pi (π) bond alongside the sigma bond formed by overlapping sp2 orbitals. This geometry allows the formation of structures like ethylene ( CH2=CH2), where each carbon is sp2 hybridized.
SP2 hybridization leads to planar structures that restrict rotation around the double bond, affecting the molecule's reactivity and interaction capabilities. Understanding sp2 hybridization is fundamental to predicting the reactivity and stability of compounds containing carbon-carbon double bonds.