Chapter 7, Problem 100QRT

Ketene, C₂H₂O, is a reactant for synthesizing cellulose acetate, which is used to make films, fibers, and fashionable clothing.

1. (a) Write the Lewis structure of ketene. Ketene does not contain an —OH bond.
2. (b) Identify the electron-region geometry and the molecular geometry around each carbon atom and all the bond angles in the molecule.
3. (c) Identify the hybridization of each carbon and oxygen atom.
4. (d) Is the molecule polar or nonpolar? Use appropriate data to support your answer.

Interpretation Introduction

Interpretation:

Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}^{}\text{O}$ has to be described.

Concept Introduction:

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

Explanation of Solution

Given molecule is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}^{}\text{O}$

The Lewis electron dot structure for given molecule can be determined by first drawing the skeletal structure. 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 considering each atom contains eight electrons in its valence shell.

Total number of valence electrons is given below:

  $\left(2\times 4\right)+\left(1\times 6\right)+\left(1\times 2\right)=16$.

Total number of electrons in bonds present is given below:

  $\left(2\times 6\right)=12$.

The four electrons remaining will be distributing in such a way that each atom should have 8 electrons in its outer shell.

Therefore, the Lewis structure is given below:

Interpretation Introduction

Interpretation:

The electron-region geometry and the molecular-geometry around each atom and the bond angles in the molecule have to be described.

Concept Introduction:

Electron geometry is the shape of a molecule predicted by considering both bond electron pairs and lone pair of electrons.

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,

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

Explanation of Solution

Given molecule is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}^{}\text{O}$.

The Lewis electron dot structure is given below:

  .

Consider the first carbon atom.

The molecular geometry and the electron-pair geometry will be trigonal planar because of the presence of three bond pairs around the carbon atom. The $\text{H-C-C}\text{and}\text{H-C-H}$ bond angle will be $120°$.

Consider the second carbon atom.

The molecular geometry and the electron-pair geometry will be linear because of the presence of two bond pairs around the carbon atom. The $\text{C-C-O}$ bond angle will be $180°$.

Interpretation Introduction

Interpretation:

The hybridization of each carbon and oxygen has to be described.

Explanation of Solution

Given molecule is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}^{}\text{O}$.

The Lewis electron dot structure is given below:

  .

Consider the first carbon atom.

The molecular geometry and the electron-pair geometry will be trigonal planar because of the presence of three bond pairs around the carbon atom. Therefore, the hybridisation will be ${\text{sp}}^2$.

Consider the second carbon atom.

The molecular geometry and the electron-pair geometry will be linear because of the presence of two bond pairs around the carbon atom. Therefore, the hybridisation will be will be $\text{sp}$.

Consider the oxygen atom.

The electron-pair geometry around oxygen will be trigonal planar because of the presence of one bond pair and two lone pair around the oxygen atom. Therefore, the hybridisation will be ${\text{sp}}^2$.

Interpretation Introduction

Interpretation:

Whether the molecule is polar or non-polar has to be given.

Explanation of Solution

Given molecule is ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}^{}\text{O}$.

The Lewis electron dot structure is given below:

  .

The $\text{C-O}$ bond present in the molecule is highly polar and thus the compound is a polar one.