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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 (H2C=C=CH2) is not planar due to the trigonal planar molecular geometry of the terminal carbon atoms. It has a zero dipole moment as it has a symmetric electron distribution. The bonding in allene is localized, not delocalized, as the double bonds between the central and terminal carbon atoms involve direct sharing of electrons between the carbon atoms.

Step by step solution

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1. Drawing a Lewis structure of Allene

First, let's identify the atomic structure of allene. Allene (C3H4) has three carbon atoms and four hydrogen atoms. We need to arrange the carbon atoms in a straight line and bond two hydrogen atoms to the first carbon atom and the last carbon atom.
02

2. Sketch the Allene molecule

The Lewis structure of allene is H2C=C=CH2, where the double bonds are between the central carbon atom and the two terminal carbon atoms. The structure looks like this: ``` H H \| |/ C=C ```
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3. Determine if the molecule is planar

The allene molecule is not planar because the two terminal carbon atoms are not in the same plane as the central carbon atom. The double bond between the central and terminal carbon atoms causes the molecular geometry of the terminal carbon atoms to be trigonal planar. As a result, the hydrogen atoms that are bonded to these terminal carbon atoms are placed on different planes.
04

4. Determine if the molecule has a nonzero dipole moment

A molecule has a nonzero dipole moment if its charge distribution is asymmetric. In allene, H2C=C=CH2, the double bonds between the central and terminal carbon atoms cause the terminal carbon atoms to have an electron-rich side with two pairs of bonding electrons. Therefore, the molecule has a symmetric electron distribution, resulting in no net dipole moment. Thus, the molecule has a zero dipole moment.
05

5. Determine if the bonding in allene is delocalized

A delocalized bond is a bond in which the electrons are spread out over multiple atoms. In other words, electrons do not only belong to two specific atoms but participate in a collective sharing of electrons over several atoms. In the allene molecule, the double bonds between the central and terminal carbon atoms are localized, and the electrons are directly shared between carbon atoms. Therefore, the bonding in allene is not considered delocalized. In conclusion, the allene molecule is not planar, has a zero dipole moment, and its bonding is localized, not delocalized.

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Key Concepts

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

Molecular Geometry
Molecular geometry is a fundamental concept in chemistry that describes the arrangement of atoms in a molecule. Considering allene, a molecule with the chemical formula C3H4, we learn that it exhibits a linear structure at its core, with the central carbon atom doubly bonded to two other carbon atoms. However, because of these double bonds, the surrounding hydrogen atoms are not all in the same plane, leading to a non-planar 3D structure.

Each carbon atom involved in a double bond assumes a trigonal planar geometry. This reflects how the p-orbitals overlap to form the pi bonds in the carbon double bonds. For allene, the central carbon is linked symmetrically to two other carbons in a straight-line, but the 'wings' or the hydrogen atoms bonded to the end carbons are perpendicular to each other. This three-dimensional aspect can be confusing, but imagine a propeller with its blades at right angles; this is similar to allene's overall shape.
Dipole Moment
The dipole moment of a molecule gives us insight into its charge distribution and indicates if it's a polar molecule. A nonzero dipole moment occurs when there is an uneven distribution of electrons within the molecule. For allene, H2C=C=CH2, the symmetrical linear structure with identical substituents on both ends means that the individual bond dipoles cancel each other out.

While the concept of dipole moments might bring to mind polar molecules with clear negative and positive ends, allene challenges this notion. The linear symmetry ensures that any potential for polarity within the double bonds is counteracted by the molecule's mirror-image symmetry. Hence, allene has a zero dipole moment, categorizing it as a nonpolar molecule despite having polar bonds.
Delocalized Bonding
Delocalized bonding involves electrons that are shared among more than two atoms, often seen in structures with resonance or in aromatic compounds. Many students get tripped up by the idea of delocalized electrons because they're used to visualizing electrons as being strictly between two atoms. However, delocalization refers to the pi electrons in a molecule being spread out across multiple atoms, creating a sort of electron cloud that isn't confined.

In the context of allene, the pi bonds are between the central carbon and each terminal carbon but they are not delocalized because they are localized to the two adjacent carbon atoms without spread over the entire molecule. Thus, allene does not exhibit delocalized bonding; its double bonds are considered 'fixed' or 'localized,' contributing to the molecule's unique structure and stability.

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Most popular questions from this chapter

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are\(0.96 \mathrm{A},\) and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is 1.85 \(\mathrm{D}\) . (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole-of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this. \((\mathbf{c})\) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3\() .\) Is your answer in accord with the relative electronegativity of oxygen?

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

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\operatorname{SiH}_{4,}(\mathbf{b}) \mathrm{PF}_{3},(\mathbf{c}) \mathrm{HBr},(\mathbf{d}) \mathrm{HCN},(\mathbf{e}) \mathrm{SO}_{2}\)

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

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: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ 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 \(\mathrm{OSF}_{4}\), 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.

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