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Chapter 9: Problem 32

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, all hydrogen atoms do not lie in the same plane. Instead, they are arranged in two distinct planes perpendicular to each other due to the non-planar arrangement of the molecule, which results from the π bonding between the central carbon atom and the terminal carbon atoms involving perpendicular p-orbitals.

Step by step solution

01

Identifying the Lewis structure of allene molecule

We are given the Lewis structure of the allene molecule. The Lewis structure for allene is: H2C=C=CH2
02

Determine the geometry around each carbon atom

Let's analyze the geometry of each carbon atom in the allene molecule. There are two types of carbon atoms - the central one and the two terminal ones. For terminal carbon atoms (HC=C): These carbon atoms are sp-hybridized since they form a double bond with the central carbon atom and a single bond with hydrogen. Each sp-hybridized carbon atom will have a linear geometry due to the linear arrangement of the hybrid orbitals. For the central carbon atom (C=C=C): The central carbon atom is also sp-hybridized since it forms a double bond with each of the terminal carbon atoms. The geometry around this atom will also be linear.
03

Determine the spatial arrangement of allene molecule

Now that we know the geometries around each carbon atom, we can understand the spatial arrangement of hydrogen atoms in the allene molecule. The two terminal carbon atoms (HC=C) have their hybrid orbitals (one sp-hybrid orbital and one unhybridized p-orbital) arranged linearly. The double bond between the terminal carbon atoms and the central carbon atom is formed by the overlap of an sp-hybrid orbital from each terminal carbon atom with an sp-hybrid orbital from the central carbon atom, along with a π bond involving their unhybridized p-orbitals. Since the π bond between the central carbon atom and each of the terminal carbon atoms involves p-orbitals that are perpendicular to each other, the allene molecule has a non-planar arrangement. In other words, π bonding in the allene molecule forces the groups attached to the terminal carbons to be perpendicular.
04

Determine the relationship of hydrogen atoms in allene

As a result of the non-planar arrangement of the allene molecule, the hydrogen atoms are also not in the same plane. There are two planes for hydrogen atoms: 1. The first plane contains two hydrogen atoms attached to a terminal carbon atom (on one side of the linear C=C=C unit). 2. The second plane contains the other two hydrogen atoms attached to the other terminal carbon atom (on the opposite side of the linear C=C=C unit). These two planes are perpendicular to each other. In conclusion, all hydrogen atoms in the allene molecule do not lie in the same plane. Instead, they are arranged in two distinct planes perpendicular to each other due to the non-planar arrangement of the allene molecule.

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Most popular questions from this chapter

An unusual category of acids known as superacids, which are defined as any acid stronger than \(100 \%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with \(\mathrm{SbF}_{5}\) produces the superacid \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) : $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared when \(2.93 \mathrm{~mL}\) anhydrous HF (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi\).

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \end{aligned} $$ Cyanamid Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}{ }^{2+}\)

The transport of \(\mathrm{O}_{2}\) in the blood is carried out by hemoglobin. Carbon monoxide can interfere with oxygen transport because hemoglobin has a stronger affinity for \(\mathrm{CO}\) than for \(\mathrm{O}_{2}\). If \(\mathrm{CO}\) is present, normal uptake of \(\mathrm{O}_{2}\) is prevented, depriving the body of needed oxygen. Using the molecular orbital model, write the electron configurations for \(\mathrm{CO}\) and \(\mathrm{for} \mathrm{O}_{2}\). From your configurations, give two property differences between \(\mathrm{CO}\) and \(\mathrm{O}_{2}\).
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