Problem 24
Question
The structure of the molecule allene, \(\mathrm{CH}_{2} \mathrm{CCH}_{2}\), is shown here. Propose hybridization schemes for the \(C\) atoms in this molecule.
Step-by-Step Solution
Verified Answer
The end carbon atoms in allene are sp3 hybridized and the middle carbon atom is sp hybridized.
1Step 1: Identify Electron Dense Regions Around Carbon Atoms
Start by observing the molecule's structure and identifying the regions around each carbon atom which are dense in electrons. These can be made up by single, double, or triple bonds, as well as lone electron pairs. In allene \(\mathrm{CH}_{2} \mathrm{CCH}_{2}\), the first and last carbon have 4 regions (4 single bonds), and the middle carbon has three regions (one single and one double bond).
2Step 2: Determine the Hybridization Scheme
Knowing the number of electron dense regions, you can now determine the hybridization of each carbon atom. If there are four regions the atom is sp3 hybridized, if three - sp2 hybridized, and if two - it's sp. Thus, in allene the end carbon atoms are sp3 hybridized and the middle carbon atom is sp2 hybridized.
3Step 3: Account for the Double Bond
Noting that the double bond in allene is unusually structured, it should be concluded that the two 'p' orbitals on the central carbon atom, which have not participated in the hybridization, form a 'p' bond on either side of the molecule. This structure results in 'p' orbitals being perpendicular to each other, consistent with the linear structure of allene. Thus, the middle carbon atom is better described as being sp hybridized.
Key Concepts
Allene StructureCarbon HybridizationLinear Molecular Geometry
Allene Structure
The allene molecule, represented as \( \mathrm{CH}_{2}\mathrm{CCH}_{2} \), is a fascinating example of organic molecular geometry. Allene consists of three carbon atoms with the central carbon atom forming a double bond with each of the surrounding carbon atoms. This particular geometric arrangement leads to an interesting molecular structure.The key feature of allene is its linear connectivity along the central C atoms, which holds a unique linear shape. The two end C-H portions of the molecule are aligned perpendicular to each other.The structure can be visualized as a two-dimensional plane where each of the hydrogen atoms attached to the terminal carbon atoms are in different planes generated due to the perpendicular 'p' orbitals on the middle carbon. This gives allene a distinctive structural characteristic that plays a critical role in determining its behavior and reactivity.
Carbon Hybridization
Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals, which influences molecular geometry and bonding properties. In allene, each carbon atom exhibits a different hybridization state: sp for the central carbon and sp\(^2\) for the terminal carbon atoms.
- Central Carbon: In the middle carbon, the atom is involved in two double bonds, resulting in two regions of electron density. These regions dictate an sp hybridization state, which involves mixing one s orbital with one p orbital, leaving the other two p orbitals untouched and perpendicular.
- Terminal Carbons: Each of these has two single C-H bonds and one single C-C bond, resulting in three regions of electron density. This configuration corresponds to an sp\(^2\) hybridization, where one s orbital combines with two p orbitals, leading to a trigonal planar arrangement.
Linear Molecular Geometry
Linear molecular geometry is a key characteristic feature of molecules like allene. In linear geometry, atoms are arranged in a straight line, which in the case of organic compounds, often involves a head-to-head arrangement of p orbitals along a single axis. For allene, the central carbon atom uses its two unhybridized p orbitals to form \(\pi\) bonds with each terminal carbon atom. These \(\pi\) bonds are perpendicular to one another, enabling a linear arrangement.This linear setup is critical because:
- It ensures that the bond angles around the central carbon are 180°.
- The perpendicular \(\pi\) orbitals limit rotation around the double bonds, maintaining the rigidity and planarity of the molecule.
Other exercises in this chapter
Problem 15
Write Lewis structures for the following molecules, and then label each \(\sigma\) and \(\pi\) bond. (a) \(\mathrm{HCN} ;\) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}\
View solution Problem 16
Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identi
View solution Problem 27
Explain the essential difference in how the valencebond method and molecular orbital theory describe a covalent bond.
View solution Problem 28
Describe the bond order of diatomic carbon, \(\mathrm{C}_{2},\) with Lewis theory and molecular orbital theory, and explain why the results are different.
View solution