Chapter 10, Problem 10.118QE

(a)

Interpretation Introduction

Interpretation:

Bond angle has to be predicted using VSEPR model for the given structure and also the hybrid orbitals on the central atoms has to be given.  The molecule is polar or not also has to be indicated.

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

Resonance structure:

The given species is shown below;

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

    $\begin{array}{c}\text{Total}\text{No}\text{of}\text{valence}\text{electrons}=(2\times 1)+(1\times 4)+(2\times 5)\\ =2+4+10\\ =16\end{array}$

A total of $8$ electrons are involved in the skeletal structure..  Resonance structures can be drawn as shown below by considering the formal charges also;

Hybrid orbitals of central atoms in structure I:

The resonance structure is shown below;

Hybrid orbitals of central nitrogen atom:

The nitrogen 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 nitrogen atom is $sp$.

Hybrid orbital of carbon atom:

The carbon atom does not have 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 central atoms in structure II:

The resonance structure is shown below;

Hybrid orbitals of central nitrogen atom:

The nitrogen 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 nitrogen atom is $sp$.

Hybrid orbital of carbon atom:

The carbon atom have one lone pair of electrons and it is bonded to three atoms.  Therefore, the steric number is calculated as shown below;

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

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

Two resonance structures do not use the same hybrid orbitals because the hybridization of the carbon atom is different in both.

Polarity of the species:

From the above figure, it is found that there is a permanent dipole moment.  Hence, the molecule will be polar.

(b)

Interpretation Introduction

Interpretation:

Bond angle has to be predicted using VSEPR model for the given structure and also the hybrid orbitals on the central atoms has to be given.  The molecule is polar or not also has to be indicated.

Concept Introduction:

Refer part (a).

Explanation of Solution

Resonance structure:

The given species is shown below;

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

    $\begin{array}{c}\text{Total}\text{No}\text{of}\text{valence}\text{electrons}=(3\times 1)+(2\times 4)+(1\times 5)+(3\times 6)\\ =3+8+5+18\\ =34\end{array}$

A total of $16$ electrons are involved in the skeletal structure..  Resonance structures can be drawn as shown below by considering the formal charges also;

Hybrid orbitals of central atoms in structure I and II:

The resonance structures is shown below;

Hybrid orbitals of nitrogen atom:

The nitrogen 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 nitrogen atom is $s{p}^2$.

Hybrid orbital of first carbon atom:

The first carbon atom does not have 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 orbital of second carbon atom:

The second carbon atom does not have 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 $s{p}^2$.

Two resonance structures use the same hybrid orbitals because the hybridization of the carbon atoms and nitrogen atom are same.

Polarity of the species:

From the above figure, it is found that there is a permanent dipole moment.  Hence, the molecule will be polar.