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The allene molecule has the following Lewis structure: Must all hydrogen atoms lie the same plane? If not, what is their spatial relationship? Explain.

Short Answer

Expert verified
In the allene molecule, the hydrogen atoms do not all lie in the same plane. Instead, they lie in two separate, perpendicular planes. This is due to the terminal carbon atoms being sp^2 hybridized with trigonal planar geometry, while the central carbon atom is sp hybridized with linear geometry.

Step by step solution

01

Identify the Lewis structure of the allene molecule

The allene molecule has the given Lewis structure: H \ H-C=C=C-H / H In this structure, there are two carbon-carbon double bonds, and each carbon atom is bonded to a hydrogen atom.
02

Determine the hybridization state of the carbon atoms

To determine the hybridization state of the carbon atoms, we first need to count the electron domains around each carbon atom. An electron domain is defined as a region where electrons are most likely found, which includes lone pairs and bonds (including double bonds). For the two terminal carbon atoms: - Each has three electron domains (1 single bond and 1 double bond). - Since three electron domains require three hybrid orbitals, these carbon atoms are sp^2 hybridized. For the central carbon atom: - It has two electron domains (2 double bonds). - Since two electron domains require two hybrid orbitals, the central carbon atom is sp hybridized.
03

Determine the molecular geometry around each carbon atom

The molecular geometry around each carbon atom can be determined based on the hybridization state found in Step 2. For the terminal sp^2 hybridized carbon atoms: - Molecular geometry: Trigonal planar - Bond angles: Approximately 120° For the central sp hybridized carbon atom: - Molecular geometry: Linear - Bond angles: 180°
04

Describe the spatial relationship of the hydrogen atoms

Now, we can determine the spatial relationship of the hydrogen atoms within the allene molecule. For the terminal sp^2 hybridized carbon atoms, the three single bonds (2 carbon-hydrogen and 1 carbon-carbon) are arranged in a trigonal planar shape around each carbon atom. The hydrogen atoms bonded to these terminal carbon atoms lie in the same plane as the carbon atoms themselves, which form two separate planes. In contrast, the central sp hybridized carbon atom doesn't lie in either of these planes due to its linear bonding arrangement with the terminal carbon atoms. Therefore, the hydrogen atoms bonded to the terminal carbon atoms do not lie in the same plane as each other and are spatially related by the following arrangement: - Two hydrogen atoms (one from each terminal carbon) and the central carbon lie in one plane. - The remaining two hydrogen atoms and the central carbon lie in another plane, perpendicular to the first plane. In conclusion, the hydrogen atoms in the allene molecule do not all lie in the same plane, instead, they have a spatial relationship in which they lie in two separate, perpendicular planes.

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Key Concepts

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

Lewis structure
The Lewis structure is a diagrammatic method for representing the electron distribution around atoms and the chemical bonds between them. It relies on the use of dots to represent valence electrons, and lines to symbolize the bonds between atoms. For understanding the structure of an allene molecule, important clues about its electronic configuration are found through the Lewis structure.

In the case of allene, as seen in the exercise, each terminal carbon is connected by a single bond to a hydrogen atom and a double bond to another carbon atom. The central carbon is connected by double bonds to each terminal carbon. This setup of bonds and atoms provides the foundational layout that determines the molecule's shape and reactivity.

Because each double bond acts as a separate electron domain, and a single bond is another domain, the exercise hints at the presence of three electron domains around each terminal carbon atom, leading to a specific hybridization and molecular geometry which we detail in subsequent sections.
Molecular hybridization
Molecular hybridization is the concept of mixing atomic orbitals to create new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in molecular orbitals. The allene molecule, containing carbon atoms with different electron domains, showcases two types of hybridization.

The terminal carbons have three electron domains and are therefore sp2 hybridized. Each uses one s and two p orbitals to form three equivalent sp2 hybrid orbitals, facilitating the trigonal planar shape around these carbons. On the other hand, the central carbon is sp hybridized, explained by its two electron domains; two hybrid orbitals are formed from one s and one p orbital, dictating a linear arrangement.
Molecular geometry
Molecular geometry refers to the three-dimensional shape of a molecule, which determines its physical and chemical properties. The geometry around each carbon atom in an allene molecule is dependent on its hybridization state.

For the terminal sp2 hybridized carbon atoms, a trigonal planar geometry is observed. This arrangement means that the atoms connected to each terminal carbon atom are spaced at approximately 120° relative to each other and lie in the same plane.

Conversely, the central sp hybridized carbon atom adopts a linear geometry, which is a direct consequence of its two-domain configuration. The two double bonds extend in opposite directions, creating a straight line with 180° bond angles.
Spatial relationship of hydrogen atoms
Understanding the spatial relationship of hydrogen atoms in an allene molecule involves considering the planes in which these atoms lie. In accordance with the provided exercise, the terminal hydrogen atoms bonded to the sp2 hybridized carbons lie in the same plane as their corresponding carbon atoms.

However, these planes are not the same for both sets of hydrogen-carbon pairs. Instead, they are perpendicular to each other with the central sp hybridized carbon atom lying at the intersection of these planes. Thus, two hydrogen atoms are found in one plane, while the other two lie in the orthogonal plane, and only two hydrogen atoms from each terminal carbon can lie in the same plane as the central carbon atom.

This perpendicular arrangement is essential for understanding the molecular structure and reactivity of allenes, as it has a direct impact on the molecule's interaction with light, other molecules, and its overall stability.

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Most popular questions from this chapter

Consider three molecules: A, B, and C. Molecule A has a hybridization of \(s p^{3}\). Molecule \(\mathrm{B}\) has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule \(\mathrm{C}\) consists of two \(\sigma\) bonds and two \(\pi\) bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{H}_{2}+, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\) b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d\), or \(6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

One of the first drugs to be approved for use in treatment of acquired immune deficiency syndrome (AIDS) was azidothymidine (AZT). Complete the Lewis structure for AZT. a. How many carbon atoms are \(s p^{3}\) hybridized? b. How many carbon atoms are \(s p^{2}\) hybridized? c. Which atom is \(s p\) hybridized? d. How many \(\sigma\) bonds are in the molecule? e. How many \(\pi\) bonds are in the molecule? f. What is the \(\mathrm{N}=\mathrm{N}=\mathrm{N}\) bond angle in the azide \(\left(-\mathrm{N}_{3}\right)\) group? g. What is the \(\mathrm{H}-\mathrm{O}-\mathrm{C}\) bond angle in the side group attached to the five-membered ring? h. What is the hybridization of the oxygen atom in the \(-\mathrm{CH}_{2} \mathrm{OH}\) group?

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