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Chapter 11: Problem 24

The structure of the molecule allene, \(\mathrm{CH}_{2} \mathrm{CCH}_{2}\), is shown here. Propose hybridization schemes for the \(C\) atoms in this molecule.

Short Answer

Expert verified
The end carbon atoms in allene are sp3 hybridized and the middle carbon atom is sp hybridized.

Step by step solution

01

Identify Electron Dense Regions Around Carbon Atoms

Start by observing the molecule's structure and identifying the regions around each carbon atom which are dense in electrons. These can be made up by single, double, or triple bonds, as well as lone electron pairs. In allene \(\mathrm{CH}_{2} \mathrm{CCH}_{2}\), the first and last carbon have 4 regions (4 single bonds), and the middle carbon has three regions (one single and one double bond).
02

Determine the Hybridization Scheme

Knowing the number of electron dense regions, you can now determine the hybridization of each carbon atom. If there are four regions the atom is sp3 hybridized, if three - sp2 hybridized, and if two - it's sp. Thus, in allene the end carbon atoms are sp3 hybridized and the middle carbon atom is sp2 hybridized.
03

Account for the Double Bond

Noting that the double bond in allene is unusually structured, it should be concluded that the two 'p' orbitals on the central carbon atom, which have not participated in the hybridization, form a 'p' bond on either side of the molecule. This structure results in 'p' orbitals being perpendicular to each other, consistent with the linear structure of allene. Thus, the middle carbon atom is better described as being sp hybridized.

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Most popular questions from this chapter

Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH} ;\) (c) acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (d) carbamic acid,

Indicate several ways in which the valence-bond method is superior to Lewis structures in describing covalent bonds.

Use the valence molecular orbital configuration to determine which of the following species is expected to have the lowest ionization energy: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{C}_{2}\) (c) \(\mathrm{C}_{2}^{-}\)

Pentadiene, \(\mathrm{C}_{5} \mathrm{H}_{8},\) has three isomers, depending on the position of the two double bonds. Determine the shape of these isomers by using VSEPR theory. Describe the bonding in these molecules by using the valence-bond method. Do the shapes agree in the two theories? Use molecular orbital theory to decide which of these molecules has a delocalized \(\pi\) system. Sketch the molecular orbital and an energy-level diagram.

Construct the molecular orbital diagram for CaF. Would you expect the bond length of \(\mathrm{CaF}^{+}\) to be longer or shorter than that of CaF?
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