Chapter 10, Problem 10.120QE

Interpretation Introduction

Interpretation:

Lewis structure for ethylene oxide, ethylene glycol, and acrylonitrile has to be drawn and also the hybrid orbitals of the central atom have to be given along with the number of pi bonds present.

Concept Introduction:

Lewis structure is used for predicting the shape of molecules.  From the steric number obtained in a Lewis structure, the molecular geometry can be predicted.  VSEPR model can predict the shape of molecules considering their Lewis structure.  Certain rules has to be followed in for the VSEPR model.

• The molecule will have a shape where there is minimal electrostatic repulsion between the valence‑shell electron pairs.
• The forces of repulsion between two lone pairs of electrons will be higher than the repulsion between lone pair and bond pair of electrons.  This in turn will be higher than the bond pair‑bond pair of electrons.

The hybridized orbitals and the steric number can be related as shown below;

Steric number	Hybridized orbital
2	$sp$
3	$s{p}^2$
4	$s{p}^3$
5	$s{p}^{3}d$
6	$s{p}^3{d}^2$

Explanation of Solution

Ethylene oxide:

Formula for ethylene oxide is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{O}$.

The total number of valence electrons is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{No}\text{of}\text{valence}\text{electrons}=(2\times 4)+(4\times 1)+(1\times 6)\\ =8+4+6\\ =18\end{array}$

A total of $14$ electrons are involved in the skeletal structure.  Two lone pair of electrons are placed over the oxygen atom.  Hence, the Lewis structure of ethylene oxide is given as shown below;

Hybrid orbitals of central carbon atoms:

The carbon atom has does not have a lone pair of electrons and it is bonded to four atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+4\\ =4\end{array}$

As the steric number is four, the hybridization of carbon atom is $s{p}^3$.

Hybrid orbitals of central oxygen atom:

The oxygen atom has two lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=2+2\\ =4\end{array}$

As the steric number is four, the hybridization of oxygen atom is $s{p}^3$.

There is no unhybridized orbital left out for the formation of pi bonds.  Hence, there is no pi bonds in ethylene oxide.

Ethylene glycol:

Formula for ethylene glycol is ${\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}$.

The total number of valence electrons is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{No}\text{of}\text{valence}\text{electrons}=(2\times 4)+(6\times 1)+(2\times 6)\\ =8+6+12\\ =26\end{array}$

A total of $18$ electrons are involved in the skeletal structure.  Two lone pair of electrons are placed over each oxygen atom.  Hence, the Lewis structure of ethylene glycol is given as shown below;

Hybrid orbitals of central carbon atoms:

The carbon atom has does not have a lone pair of electrons and it is bonded to four atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+4\\ =4\end{array}$

As the steric number is four, the hybridization of carbon atom is $s{p}^3$.

Hybrid orbitals of central oxygen atoms:

The oxygen atom has two lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=2+2\\ =4\end{array}$

As the steric number is four, the hybridization of oxygen atom is $s{p}^3$.

There is no unhybridized orbital left out for the formation of pi bonds.  Hence, there is no pi bonds in ethylene glycol.

Acrylonitrile:

Formula for acrylonitrile is ${\text{CH}}_{\text{2}}\text{CHCN}$.

The total number of valence electrons is calculated as shown below;

    $\begin{array}{c}\text{Total}\text{No}\text{of}\text{valence}\text{electrons}=(3\times 4)+(3\times 1)+(1\times 5)\\ =12+3+5\\ =20\end{array}$

A total of $12$ electrons are involved in the skeletal structure.  The Lewis structure of acrylonitrile is given as shown below;

Hybrid orbitals of carbon atom C-1:

The carbon atom has does not have a lone pair of electrons and it is bonded to three atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+3\\ =3\end{array}$

As the steric number is three, the hybridization of carbon atom is $s{p}^2$.

Hybrid orbitals of carbon atom C-2:

The carbon atom has does not have a lone pair of electrons and it is bonded to three atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+3\\ =3\end{array}$

As the steric number is three, the hybridization of carbon atom is $s{p}^2$.

Hybrid orbitals of carbon atom C-3:

The carbon atom has does not have a lone pair of electrons and it is bonded to two atoms.  Therefore, the steric number is calculated as shown below;

    $\begin{array}{c}\text{Steric}\text{number}=0+2\\ =2\end{array}$

As the steric number is two, the hybridization of carbon atom is $sp$.

There are one double bond and one triple bond present in the structure.  As multiple bonds are present, acrylonitrile contains pi bonds.