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Chapter 9: Problem 36

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
The hydrogen atoms in the allene molecule do not lie in the same plane. They form two different planes perpendicular to each other, due to the orthogonal arrangement of the π orbitals in the central carbon-carbon double bond. This allows for proper overlapping of p orbitals required for the π bonds between the central carbon atoms and results in a spatial relationship where the hydrogen atoms form two sets of planes, each plane containing a terminal carbon atom and its two bonded hydrogen atoms.

Step by step solution

01

Draw the Lewis structure of the allene molecule

To determine the spatial relationship of hydrogen atoms in the allene molecule, we first need to draw its Lewis structure. For this molecule, it is given by: H₂C=C=CH₂
02

Determine the molecular geometry and hybridization

As we can see, the central carbon atoms in the allene molecule are both double-bonded to each other, forming a cumulative double bond between them. Each of them is also single-bonded to one terminal carbon atom. These carbons have two regions of electron density around them (two single bonds). Therefore, they are sp-hybridized, and the molecular geometry around these carbons is linear. The terminal carbon atoms are bonded to a central carbon and two hydrogens. These carbons have three regions of electron density around them (two single bonds and one double bond). Therefore, they are sp²-hybridized, and the molecular geometry around these carbons is trigonal planar.
03

Determine the spatial relationship of hydrogen atoms

Since the terminal carbon atoms have a trigonal planar geometry and are sp²-hybridized, that means the hydrogens bonded to them form a plane with their respective carbon atom. Also, as the central carbon-carbon double bond is linear, the planes formed by hydrogens on each end of the molecule are perpendicular to each other. So, all the hydrogen atoms do not lie in the same plane, and their spatial relationship is that they form two different planes perpendicular to each other.
04

Explain the spatial relationship of hydrogen atoms

The π orbitals of the double bonds are orthogonal to each other, resulting in the formation of two separate planes containing the hydrogen atoms. The central carbon-carbon double bond requires p orbitals from the carbons to overlap, forming "side-on" π bonds. In order to achieve this, the two sets of π orbitals must be orthogonal to each other. This arrangement results in two planes containing the hydrogen atoms, which are perpendicular to each other and ensures proper overlapping of p orbitals required for the π bonds between the central carbon atoms.

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Most popular questions from this chapter

In the hybrid orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Why must all six atoms in \(\mathrm{C}_{2} \mathrm{H}_{4}\) lie in the same plane?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: $$ \mathrm{M}-\mathrm{C} \equiv \mathrm{O} $$ a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. 9.43 and 9.44.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B. a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part \(\mathrm{b}\) ) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{kJ} / \mathrm{mol} $$ How much energy is needed to ionize 1.5 \(\mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?
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