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Chapter 8: Q26 E (page 448)

For the molecule allene, \({{\rm{H}}_{\rm{2}}}{\rm{C = C = C}}{{\rm{H}}_{\rm{2}}}\) , give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?

Short Answer

Expert verified

Two \({\rm{\pi }}\) bonds bind the core carbon atom. \({\rm{\pi }}\) bonding requires perpendicular \({\rm{p}}\) orbitals. As a result, the \({\rm{C}}{{\rm{H}}_{\rm{2}}}\) planes must be perpendicular to one another, and the hydrogen atoms must be in those planes.

Step by step solution

01

Define molecule

A molecule is the smallest unit of a substance that holds the compound's chemical characteristics. Molecules are made up of atoms arranged in groups.

02

Explanation

Two \({\rm{\pi }}\) bonds connect the core carbon atom. In \({\rm{\pi }}\) bonding, the \({\rm{p}}\) orbitals must be perpendicular to each other. As a result, the \({\rm{C}}{{\rm{H}}_{\rm{2}}}\) planes must be perpendicular to one another, and the hydrogen atoms will be in those planes.

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Most popular questions from this chapter

Draw the orbital diagram for carbon in showing how many carbon atom electrons are in each orbital.

Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:

(a) \({\rm{H}}\) and \({{\rm{H}}_{\rm{2}}}\)

(b) \({\rm{N}}\)and \({{\rm{N}}_{\rm{2}}}\)

(c) \({\rm{O}}\)and \({{\rm{O}}_{\rm{2}}}\)

(d) \({\rm{C}}\)and \({{\rm{C}}_{\rm{2}}}\)

(e) \({\rm{B}}\)and \({{\rm{B}}_{\rm{2}}}\)

Use valence bond theory to explain the bonding in\({{\rm{O}}_{\rm{2}}}\). Sketch the overlap of the atomic orbitals involved in the bonds in \({{\rm{O}}_{\rm{2}}}\).

The bonding molecular orbital in \({{\rm{H}}_2}\) is lower in energy than an electron in \({\rm{1s}}\) orbital because the bonding molecular is more stable.

Explain how \({\rm{\sigma }}\) and \({\rm{\pi }}\) bonds are similar and how they are different.
See all solutions

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