Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN},\) (b) \(\mathrm{SO}_{3}^{2-},\) (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{PF}_{6}^{-}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of \(\mathrm{NO} ?\) (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) How do the \(\pi_{2 p}^{*}\) MOs formed from the \(2 p_{x}\) and \(2 p_{y}\) atomic orbitals differ from the \(\pi_{2 p}\) MOs in terms of energies and electron distributions?

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free