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Chapter 9: Problem 98

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

Short Answer

Expert verified
The allene molecule is non-planar, with two perpendicular planes intersecting at the central carbon atom. It has no nonzero dipole moment, as the symmetric structure causes the dipole moment vectors to cancel out, resulting in a net dipole moment of zero. The bonding in allene is not described as delocalized because electrons are localized in the carbon-carbon double bonds and carbon-hydrogen single bonds, and not shared across multiple bonded atoms like in benzene.

Step by step solution

01

Count Valence Electrons

In order to draw the Lewis structure for allene, we first need to count the total number of valence electrons available for bonding. Allene has three carbon atoms and four hydrogen atoms. The valence electrons for each carbon atom are 4, and the valence electrons for each hydrogen atom are 1. So, the total number of valence electrons in allene is: (3 Carbon atoms x 4 valence electrons) + (4 Hydrogen atoms x 1 valence electron) = 12 + 4 = 16 valence electrons
02

Create the Skeletal Structure

Next, we will create the basic skeletal structure of allene by connecting the atoms with single bonds (2 electrons per bond). The central carbon atom is bonded with the two other carbon atoms, and each terminal carbon atom is bonded to two hydrogen atoms. - C = C = C - Each hydrogen atom is connected to one terminal carbon atom.
03

Complete the Octets

Now, we need to complete the octets of all the atoms in the structure by adding the remaining valence electrons as lone pairs. Place the remaining 8 electrons (16 - 8 = 8) around the carbon atoms to give them a complete octet. C=C=C Each carbon atom has a complete octet, and the hydrogen atoms satisfied with the two electrons they share with the carbon atoms. The Lewis structure for allene is now complete. Now that we have the Lewis structure, we can sketch the allene molecule and answer the three questions.
04

Sketch the Allene Molecule

Based on the Lewis structure and according to the Figure 9.25 style, we can draw the allene molecule like this: H \ C = C = C \ H
05

Answering Questions

(a) Is the molecule planar? No, the molecule is not planar. The two carbon atoms at the terminal ends and the hydrogen atoms attached to them form two perpendicular planes intersecting at the central carbon, forming a non-planar allene molecule. (b) Does it have a nonzero dipole moment? No, allene does not have a nonzero dipole moment. Allene is symmetric, meaning that the dipole moment vectors caused by the carbon-hydrogen and carbon-carbon bonds cancel each other out, resulting in a net dipole moment of zero. (c) Would the bonding in allene be described as delocalized? Explain. No, the bonding in allene cannot be described as delocalized. In allene, the electrons are localized in the carbon-carbon double bonds and carbon-hydrogen single bonds. In a delocalized bonding system, electrons are shared and spread out across multiple bonded atoms. This occurs in molecules like benzene where electrons are distributed over more than two atoms. Allene does not have such a bonding system, indicating that its bonding is not delocalized.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Allene Molecule
The allene molecule is an interesting compound featuring a unique arrangement of atoms. It is a small organic molecule comprising three carbon atoms connected in a linear arrangement with two double bonds. Each of the terminal carbon atoms ( C ) forms a bond with two hydrogen ( H ) atoms. The structure can be depicted as:
  • C=C=C
The central carbon is connected to the other two carbons by double bonds, forming a straight line, which is crucial for defining its geometric shape and electronic properties. With its distinctly linear form, the allene molecule presents various unique properties that are worth exploring.
Despite its simple appearance, it exhibits fascinating molecular behavior, such as interactions between the orbitals of its carbon atoms, contributing to its structural characteristics. Understanding the molecular structure and behavior of allene helps in grasping basic concepts in organic chemistry, such as hybridization and bonding.
Valence Electrons
Valence electrons are vital in determining how atoms bond to form molecules, and they are key components in understanding the chemical behavior of molecules such as allene. In the case of allene, the total number of valence electrons is determined by considering the electrons available from both carbon and hydrogen atoms.
  • Carbon: Each carbon atom has 4 valence electrons.
  • Hydrogen: Each hydrogen atom has 1 valence electron.
Consequently, when calculating for allene's valence electrons, we find: \[ \text{Valence electrons = (3 \times 4) + (4 \times 1) = 12 + 4 = 16} \]
This number of electrons helps in forming the bonds seen in allene, namely the two carbon-carbon double bonds and the carbon-hydrogen single bonds.
Understanding valence electrons allows chemists to predict the arrangement of atoms and the types of bonds formed in a molecule, establishing the foundation for Lewis structure creation and molecular geometry analysis.
Molecular Geometry
Molecular geometry is crucial for understanding the physical properties and reactivity of a molecule. When examining allene, it is important to consider that its geometry is not exactly straightforward despite its linear electron arrangement.
Allene is formed by a central carbon atom double-bonded to two terminal carbons. Due to sp hybridization of the central carbon atom, allene's molecular geometry is linear around the central carbon. However, the two outer carbons exhibit sp虏 hybridization, leading to an interesting three-dimensional structure. These characteristics give rise to two planes in which the terminal carbon atoms' substituents, the hydrogen atoms, are perpendicular to each other. The resulting non-planar structure affects the overall geometry:
  • The central carbon and the two terminal carbons form a straight line.
  • The planes formed by hydrogen atoms connected to each terminal carbon are perpendicular.
Understanding molecular geometry aids in predicting molecule behaviors, such as polarity and potential interactions with other substances.
Dipole Moment
The concept of a dipole moment is essential for understanding molecular polarity, which affects how a molecule interacts with electric fields and other molecules. A dipole moment arises when there is an uneven distribution of electron density, leading to partial charges within the molecule.
For allene, however, this is not the case. Despite its complex geometry, allene remains a nonpolar molecule. Due to its symmetrical structure, specifically the arrangement and equality in the pull of electrons around the linear central axis, any dipole moments arising from the carbon-hydrogen or carbon-carbon bonds cancel each other out.
The key points of dipole moment in allene:
  • Symmetry: The linear symmetry of allene ensures that polar bonds do not result in a polar molecule.
  • No net dipole: Because of the cancellation of molecular dipole moments, allene has a zero net dipole moment.
This absence of a net dipole moment reflects why allene doesn't exhibit polar behavior, influencing its solubility and interactions with other compounds in a range of environments.

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Most popular questions from this chapter

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

(a) Does \(C S_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{SO}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_4) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of OSF \(_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

In which of the following AF \(_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{PF}_{6}^{-}, \mathrm{SbF}_{\mathrm{s}}, \mathrm{SF}_{4} ?\)
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