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Chapter 9: Problem 99

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 (C3H4) is nonlinear and not planar due to its hybridization and geometries of the carbon atoms. It has a net dipole moment of zero as the symmetric non-planar structure results in the cancellation of individual dipole moments. The bonding in allene is not delocalized since the electrons are not shared by more than two atoms in the molecule, and all bonds are localized between neighboring carbon atoms.

Step by step solution

01

Draw the Lewis structure of allene

To draw the Lewis structure of allene, we start with the central carbon atom (C), and then we continue connecting its neighbors according to the chemical structure C = C = C. Next, we add the hydrogen (H) atoms, two bonded to each of the outer carbon atoms. \[ \text{H}\text{C}=\text{C}=\text{C}\text{H}\] Now that we have the Lewis structure for allene, we can answer the given questions.
02

Is the molecule planar?

The molecule allene is not planar. This is because the central carbon atom is sp-hybridized, with linear geometry, and forms two sigma bonds with the two outer carbon atoms. The outer carbon atoms are sp2-hybridized, which makes a trigonal planar geometry, but they are in perpendicular planes to each other. Therefore, the allene molecule adopts a non-planar arrangement in a bent shape.
03

Determine if the molecule has a nonzero dipole moment

To determine if the molecule has a nonzero dipole moment, we need to analyze the arrangement of electron densities in the molecule. Given the Lewis structure of allene, the positive end of the molecule is at one end with hydrogen atoms, and the negative end is at the other end with the carbon atoms. However, due to the symmetric non-planar structure of the molecule, the dipole moments from each side cancel out each other, resulting in a net dipole moment of zero. So, allene does not have a nonzero dipole moment.
04

Would the bonding in allene be described as delocalized?

The bonding in allene would not be described as delocalized. Delocalized bonding refers to electrons that are shared by more than two atoms in a molecule, like in benzene. In allene, the bonds between carbon atoms are double bonds, where one sigma bond and one pi bond are formed. The sigma bond is localized between the two carbon atoms, and the pi bond is between the two neighboring carbon atoms only. No electrons are shared by all three carbon atoms or delocalized across the entire molecule. Therefore, the bonding in allene is not delocalized.

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Most popular questions from this chapter

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \AA\). (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5,\) is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-}\) ? (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-}\). (e) It is often observed that \(\sigma\) bonds that involve an \(s p\) hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

Azo dyes are organic dyes that are used for many applications, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, \(\mathrm{C}_{12} \mathrm{H}_{10} \mathrm{~N}_{2}\). A closely related substance is hydrazobenzene, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{~N}_{2} .\) The Lewis structures of these two substances are (Recall the shorthand notation used for benzene.) (a) What is the hybridization at the \(\mathrm{N}\) atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the \(\mathrm{N}\) and the \(C\) atoms in each of the substances? (c) Predict the \(\mathrm{N}-\mathrm{N}-\mathrm{C}\) angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its \(\pi\) electrons than hydrazobenzene. Discuss this statement in light of your answers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.)

What property of the electron causes electron domains to have an effect on molecular shapes?

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{ }^{+}\) be expected to exist? Explain.

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)
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