Chapter 4, Problem 14Q

1. a. Draw the Lewis structure for ethene (ethylene), H₂CCH₂, a small hydrocarbon with a double bond.
2. b. Based on this structure, predict the H–C–H bond angle. Explain your reasoning.
3. c. Sketch the molecule showing the predicted bond angles.

Interpretation Introduction

Interpretation:

The Lewis structure for ethylene has to be drawn.

Concept Introduction:

Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.

The chemical bonding present in covalently bonded molecules and in coordination compounds are represented using Lewis structures.

It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.

The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

Sometimes the chemical bonding of a molecule cannot be represented using a single Lewis structure. In these cases, the chemical bonding are described by delocalization of electrons and is known as resonance.

All the possible resonance structures are imaginary whereas the resonance hybrid is real.

These structures will differ only in the arrangement of the electrons not in the relative position of the atomic nuclei.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed such that each atom contains eight electrons in its valence shell.

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 1& Tetrahedral& Trigonalpyramidal\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

Explanation of Solution

The Lewis electron dot structure for given molecules are determined by first drawing the skeletal structure for the given molecules, then the total number of valence electrons for all atoms present in the molecules are determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

The given moleucle is ethylene $C{H}_{2}C{H}_2$

$\begin{array}{c}C=2\times 4\\ 4=4\times 1\\ Total=12\\ 6bonds=6\times 2=12\\ =12-12\\ =0\end{array}$

Thus, the Lewis structure of given compound is,

Interpretation Introduction

Interpretation:

The bond angle value in $H-C-H$ bond in methanol has to be predicted.

Concept Introduction:

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Lewis structure for the given compound ethylene is,

Here, the carbon atom does not have lone pair of electrons and the geometry around the carbon atom is triogonal planar, thus the bond angle in between $H-C-H$ bond in this molecule is $120{}^o$

Interpretation Introduction

Interpretation:

The ethylene molecule has to be sketched and the predicted bond angle should be shown in the structure.

Concept Introduction:

Molecular geometry is the shape of a molecule predicted by considering only bond pair of electrons.

Geometry of different type of molecules with respect to the number of electron pairs are mentioned below,

$\begin{array}{ccccc}\begin{array}{l}Typeof\\ Molecule\end{array}& \begin{array}{l}No.ofatoms\\ bondedtocentralatoms\end{array}& \begin{array}{l}No.oflonepairs\\ oncentralatom\end{array}& \begin{array}{l}Arrangementof\\ electronpairs\end{array}& \begin{array}{l}Molecular\\ Geometry\end{array}\\ A{B}_2& 2& 0& Linear& Linear\\ A{B}_3& 3& 0& Trigonalplanar& Trigonalplanar\\ A{B}_4& 4& 0& Tetrahedral& Tetrahedral\\ A{B}_5& 5& 0& Trigonalbipyramidal& Trigonalbipyramidal\\ A{B}_6& 6& 0& Octahedral& Octahedral\end{array}$

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Lewis structure for the given compound ethylene is,

Here, the carbon atom does not have lone pair of electrons and the geometry around the carbon atom is triogonal planar, thus the bond angle in between $H-C-H$ bond in this molecule is $120{}^o$

According to the predicted bond angle the structure can be sketched as follows,