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Chapter 9: Problem 31

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

Short Answer

Expert verified
In the ethylene molecule (\(\mathrm{C}_{2} \mathrm{H}_{4}\)), each carbon atom is sp2 hybridized, forming a trigonal planar geometry with bond angles of approximately 120 degrees. All atoms connected by sp2 hybrid orbitals lie in the same plane. The pi bond between the two carbon atoms exists above and below this plane, but does not affect the planarity of the molecule. Therefore, all six atoms in the \(\mathrm{C}_{2} \mathrm{H}_{4}\) molecule must lie in the same plane.

Step by step solution

01

Understand the structure of the C2H4 molecule

The C2H4 molecule is an organic compound known as ethylene. It consists of two carbon (C) atoms, each bonded to two hydrogen (H) atoms, and the two carbon atoms are also bonded to each other.
02

Determine hybridization of carbon atoms

Each carbon atom in C2H4 has sp2 hybridization. This is because they form one sigma bond with another carbon atom and two sigma bonds with hydrogen atoms, forming 3 sigma bonds in total. In sp2 hybridization, one s-orbital and two p-orbitals from the carbon atom mix to form three new sp2 hybrid orbitals.
03

Recognize the geometry of sp2 hybridized bonds

The geometry of sp2 hybrid orbitals is trigonal planar, with bond angles of roughly 120 degrees between each bond. In trigonal planar geometry, all atoms connected by sp2 hybrid orbitals lie in the same plane.
04

Consider the pi bond between carbon atoms

In C2H4, the two carbon atoms are not only bonded through a sigma bond but also through a pi bond. A pi bond is formed by the side-by-side overlap of two p-orbitals from each carbon atom that was not involved in the formation of sp2 hybrid orbitals. The electron cloud in a pi bond exists above and below the plane of the molecule.
05

Conclude the planar structure of C2H4

Since all four sigma bonds with hydrogen and the sigma bond between the carbon atoms involve sp2 hybrid orbitals, they are all in the same plane due to the trigonal planar geometry of sp2 hybridized bonds. As the pi bond exists above and below this plane, it does not impact the planarity of the molecule. Therefore, all six atoms in the C2H4 molecule must lie in the same plane.

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Most popular questions from this chapter

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homonuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?

Arrange the following from lowest to highest ionization energy: \(\mathrm{O}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

Complete the following resonance structures for \(\mathrm{POCl}_{3}\). a. Would you predict the same molecular structure from each resonance structure? b. What is the hybridization of \(\mathrm{P}\) in each structure? c. What orbitals can the \(\mathrm{P}\) atom use to form the \(\pi\) bond in struc- ture \(\mathrm{B}\) ? d. Which resonance structure would be favored on the basis of formal charges?

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-}\), and NO using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? a. \(\mathrm{Li}_{2}\) b. \(\mathrm{C}_{2}\) c. \(\mathrm{S}_{2}\)
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