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Chapter 9: Problem 35

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

Short Answer

Expert verified
In ethene (C2H4), carbon undergoes sp2 hybridization, leading to a trigonal planar geometry with bond angles of 120掳. All six atoms must lie in the same plane due to this molecular geometry and the combination of 蟽 and 蟺 bonding. The spatial arrangement of electron domains and bond angles ensures an even distribution of the carbon and hydrogen atoms within the plane.

Step by step solution

01

Identify the central atom and hybridization

In ethene, the central atom is carbon. We have two carbon atoms connected by a double bond, and each carbon atom is also connected to two hydrogen atoms. To determine the hybridization of carbon, we look at the number of electron domains (regions of high electron density) around each carbon atom: one double bond (counted as a single electron domain) and two single bonds. This adds up to three electron domains, which means carbon undergoes sp2 hybridization in ethene.
02

Determine the geometrical shape and bond angles based on hybridization

The geometry of sp2 hybridization is trigonal planar with bond angles of 120掳. This means that the three electron domains around each carbon will arrange themselves in a planar shape, equidistant from one another. In ethene, two of the domains are occupied by bonds to hydrogen atoms, while the third contains the double bond to the other carbon atom.
03

Understand the bonding in ethene as a combination of 蟽 and 蟺 bonds

In ethene, each carbon atom forms three 蟽 (sigma) bonds, two with hydrogen atoms and one with the other carbon atom (a single bond). The remaining carbon p-orbitals overlap laterally to form a 蟺 (pi) bond between the carbon atoms. 蟺 bonds aren't as symmetrical as 蟽 bonds, so their electron density is concentrated above and below the plane formed by the other atoms in the molecule.
04

Confirm the planar nature of ethene

Due to the trigonal planar geometry that arises from sp2 hybridization, the atoms in C2H4 must exist in one plane. The spatial arrangement of the electron domains around each carbon atom and the combined 蟽 and 蟺 bonding push the atoms into a flat arrangement. Furthermore, the bond angles (120掳) ensure that the carbon and hydrogen atoms position evenly within the plane. In conclusion, the planar nature of ethene (C2H4) arises from the sp2 hybridization of carbon and the associated molecular geometry, which requires that all six atoms lie in the same plane.

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Most popular questions from this chapter

The atoms in a single bond can rotate about the internuclear axis without breaking the bond. The atoms in a double and triple bond cannot rotate about the internuclear axis unless the bond is broken. Why?

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\) . Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$

Show how a hydrogen 1\(s\) atomic orbital and a fluorine 2\(p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. $$ \begin{array}{l}{\text { a. } \mathrm{SO}_{2}} \\ {\text { b. } \mathrm{SO}_{3}}\end{array} $$ $$ \text {c} \mathrm{s}_{2} \mathrm{O}_{3}^{2-}\left[\begin{array}{c}{\mathrm{o}} \\\ {\mathrm{s}-\mathrm{s}-\mathrm{o}} \\ {\mathrm{o}} \\\ {\mathrm{o}}\end{array}\right]^{2-} $$ e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}+\)
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