Chapter 2, Problem 2F.8E

(a)

Interpretation Introduction

Interpretation:

Hybridization of underlined $\text{C}$ in ${\text{CH}}_3\underset{\_}{\text{C}}{\text{CCH}}_3$ has to be determined.

Concept Introduction:

Hybridization is the hypothetical concept of mixing of atomic orbital into hybrid orbitals that are of dissimilar shapes, energies and are appropriate for combination of electrons to form bonds in valence bond theory.

Hybridization is calculated by the hybrid orbitals and to calculate hybrid orbitals we need to know the steric number that is given by,

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{number of atoms bonded to the central atom}\right)+\\ \left(\text{number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The table that relates the steric number with hybridization is as follows:

$\begin{array}{cc}\text{Steric Number}& \text{Type of Hybridisation}\\ 2& \text{sp}\\ 3& {\text{sp}}^2\\ 4& {\text{sp}}^3\\ 5& {\text{sp}}^3\text{d}\\ 6& {\text{sp}}^3{\text{d}}^2\end{array}$

Answer to Problem 2F.8E

The molecule ${\text{CH}}_3\underset{\_}{\text{C}}{\text{CCH}}_3$ is $\text{sp}$ hybridized.

Explanation of Solution

The molecule ${\text{CH}}_3\underset{\_}{\text{C}}{\text{CCH}}_3$ consists of four $\text{C}$ atoms and six $\text{H}$ atoms.

The symbol for carbon is $\text{C}$ with atomic number 6. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^2$. Thus, it possesses 4 valence electrons.

The symbol for hydrogen is $\text{H}$ with atomic number 1. Its electronic configuration is $1{\text{s}}^2$. Thus, it possesses 1 valence electron.

Its Lewis structure is as follows:

As from the structure above, 2 atoms are bonded to the underlined atom and 0 lone pair on it.

Substitute 2 for number of atoms bonded to the central atom and 0 for a number of lone pair on the central atom in equation (1) to find out the steric number.

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

Hence, the hybridization of ${\text{CH}}_3\underset{\_}{\text{C}}{\text{CCH}}_3$ is ${\text{sp}}^{}$.

(b)

Interpretation Introduction

Interpretation:

Hybridization of underlined $\text{N}$ in ${\text{CH}}_3\underset{\_}{\text{N}}{\text{NCH}}_3$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 2F.8E

The molecule ${\text{CH}}_3\underset{\_}{\text{N}}{\text{NCH}}_3$ is ${\text{sp}}^2$ hybridized.

Explanation of Solution

The molecule ${\text{CH}}_3\underset{\_}{\text{N}}{\text{NCH}}_3$ consists of two $\text{C}$ atoms, two $\text{N}$ atoms, and six $\text{H}$ atoms.

The symbol for carbon is $\text{C}$ with atomic number 6. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^2$. Thus, it possesses 4 valence electrons.

The symbol for hydrogen is $\text{H}$ with atomic number 1. Its electronic configuration is $1{\text{s}}^2$. Thus, it possesses 1 valence electron.

The symbol for nitrogen is $\text{N}$ with atomic number 7. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^3$. Thus, it possesses 5 valence electrons.

Its Lewis structure is as follows:

As from the structure above, 2 atoms are bonded to the underlined nitrogen atom and 1 lone pair on it.

Substitute 2 for a number of atoms bonded to the central atom and 1 for a number of lone pairs on the central atom in equation (1) to find out the steric number.

$\begin{array}{c}\text{Steric number}=2+1\\ =3\end{array}$

Hence, the hybridization of ${\text{CH}}_3\underset{\_}{\text{N}}{\text{NCH}}_3$ is ${\text{sp}}^2$.

(c)

Interpretation Introduction

Interpretation:

Hybridization of underlined $\text{C}$ in ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{C}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 2F.8E

The molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{C}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_2$ is ${\text{sp}}^2$ hybridized.

Explanation of Solution

The molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{C}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_2$ consists of six $\text{C}$ atoms and twelve $\text{H}$ atoms.

The symbol for carbon is $\text{C}$ with atomic number 6. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^2$. Thus, it possesses 4 valence electrons.

The symbol for hydrogen is $\text{H}$ with atomic number 1. Its electronic configuration is $1{\text{s}}^2$. Thus, it possesses 1 valence electron.

Its Lewis structure is as follows:

As from the structure above, 3 atoms are bonded to the underlined carbon atom and 0 lone pair on the same atom.

Substitute 3 for a number of atoms bonded to the central atom and 0 for a number of lone pair on the central atom in equation (1) to find out the steric number.

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

Hence, the hybridization of $\text{C}$ in the molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{C}}\text{C}{\left({\text{CH}}_{\text{3}}\right)}_2$ is ${\text{sp}}^2$.

(d)

Interpretation Introduction

Interpretation:

Hybridization of underlined $\text{N}$ in ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{N}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 2F.8E

The molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{N}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_2$ is ${\text{sp}}^3$ hybridized.

Explanation of Solution

The molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{N}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_2$ contains four $\text{C}$ atoms, two $\text{N}$ atoms, and twelve $\text{H}$ atoms.

The symbol for carbon is $\text{C}$ with atomic number 6. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^2$. Thus, it possesses 4 valence electrons.

The symbol for hydrogen is $\text{H}$ with atomic number 1. Its electronic configuration is $1{\text{s}}^2$. Thus, it possesses 1 valence electron.

The symbol for nitrogen is $\text{N}$ with atomic number 7. Its electronic configuration is $\left[\text{He}\right]{\text{2s}}^2{\text{2p}}^3$. Thus, it possesses 5 valence electrons.

Its Lewis structure is as follows:

As from the structure above, 3 atoms are bonded to the underlined nitrogen atom and 1 lone pair on the same atom.

Substitute 3 for a number of atoms bonded to central atom and 1 for a number of lone pair on the central atom in equation (1) to find out the steric number.

$\begin{array}{c}\text{Steric number}=3+1\\ =4\end{array}$

Hence, the hybridization of $\text{N}$ in the molecule ${\left({\text{CH}}_{\text{3}}\right)}_2\underset{\_}{\text{N}}\text{N}{\left({\text{CH}}_{\text{3}}\right)}_2$ is ${\text{sp}}^3$.