Chapter 1, Problem 1.65P

Interpretation Introduction

(a)

Interpretation: The orbitals used to form the highlighted bonds are to be predicted.

Concept introduction: Hybridization is the combination of two or more atomic orbitals to form the same number of hybrid orbitals, each having the same shape and energy.

Answer to Problem 1.65P

Orbitals used in the formation of given bonds are:

$\left[1\right]\to {\text{C}}_{s{p}^2}-{\text{H}}_{1s}\text{, }\left[\text{2}\right]\to {\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^2}\text{, }\left[\text{3}\right]\to {\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$

Explanation of Solution

The given compound is,

Figure $1$

In bond $\left[1\right]$, there are three groups around $\text{C}$ and no lone pair. Thus, it is $s{p}^2$ hybridized. This bond is formed from ${\text{C}}_{s{p}^2}-{\text{H}}_{1s}$ orbitals. In bond $\left[2\right]$, there are three groups around each $\text{C}$ and no lone pair. Thus, it is $s{p}^2$ hybridized. This bond is formed from ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^2}$. In bond $\left[3\right]$, there are three groups around $\text{C}\left(\text{II}\right)$ and no lone pair. Thus, it is $s{p}^2$ hybridized. And there are four groups around $\text{C}\left(\text{III}\right)$ and no lone pair. Thus, it is $s{p}^3$ hybridized. This bond is formed from ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$.

Conclusion

Orbitals used in the formation of given bonds are:

$\left[1\right]\to {\text{C}}_{s{p}^2}-{\text{H}}_{1s}\text{, }\left[\text{2}\right]\to {\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^2}\text{, }\left[\text{3}\right]\to {\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$

Interpretation Introduction

(b)

Interpretation: The orbitals used to form the highlighted bonds are to be predicted.

Concept introduction: Hybridization is the combination of two or more atomic orbitals to form the same number of hybrid orbitals, each having the same shape and energy.

Answer to Problem 1.65P

Orbitals used in the formation of given bonds are:

$\left[1\right]\to {\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}\text{, }\left[\text{2}\right]\to {\text{C}}_{s{p}^2}-{\text{O}}_{s{p}^2}$

Explanation of Solution

Figure 2

In bond $\left[1\right]$, there are four groups around $\text{C}\left(\text{I}\right)$ including two hydrogen atoms and no lone pair. Thus, it is $s{p}^3$ hybridized. This bond is formed from ${\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}$ orbitals. In bond $\left[2\right]$, there are three groups around $\text{C}$ and no lone pair. Thus, it is $s{p}^2$ hybridized. And there is one group and two lone pairs around $\text{O}$. Thus, it is $s{p}^2$ hybridized. This bond is formed from ${\text{C}}_{s{p}^2}-{\text{O}}_{s{p}^2}$ orbitals.

Conclusion

Orbitals used in the formation of given bonds are:

$\left[1\right]\to {\text{C}}_{s{p}^3}-{\text{C}}_{s{p}^2}\text{, }\left[\text{2}\right]\to {\text{C}}_{s{p}^2}-{\text{O}}_{s{p}^2}$.

Interpretation Introduction

(c)

Interpretation: The orbitals used to form the highlighted bonds are to be predicted.

Concept introduction: Hybridization is the combination of two or more atomic orbitals to form the same number of hybrid orbitals, each having the same shape and energy.

Answer to Problem 1.65P

Orbitals used in the formation of given bonds are:

$\left[1\right]\to {\text{H}}_{1s}-{\text{C}}_{sp}\text{, }\left[\text{2}\right]\to {\text{C}}_{sp}-{\text{C}}_{sp},\left[3\right]\to {\text{C}}_{sp}-{\text{C}}_{s{p}^2}\text{and }\left[4\right]\to {\text{N}}_{s{p}^2}-{\text{C}}_{s{p}^3}$

Explanation of Solution

Figure 3

In bond $\left[1\right]$, there are two groups around $\text{C}\left(\text{I}\right)$, including one hydrogen atom and triply bonded $\text{C}$ atom. Thus, it is $sp$ hybridized. This bond is formed from ${\text{H}}_{1s}{\text{-C}}_{sp}$ orbitals. In bond $\left[2\right]$, there are two groups around each $\text{C}$ atom. Thus, it is $sp$ hybridized. This bond is formed from ${\text{C}}_{sp}-{\text{C}}_{sp}$ orbitals. In bond $\left[3\right]$, there are two groups around $\text{C(II)}$ atom. Thus, it is $sp$ hybridized. And there three groups around $\text{C}\left(\text{III}\right)$ including one hydrogen atom. Thus, it is $s{p}^2$ hybridized This bond is formed from ${\text{C}}_{sp}-{\text{C}}_{s{p}^2}$ orbitals. In bond $\left[4\right]$, there are two groups around $\text{N}$ atom and one lone pair. Thus, it is $s{p}^2$ hybridized. And there are four groups around $\text{C}$ including three hydrogen atoms. Thus, it is $s{p}^3$ hybridized. This bond is formed from ${\text{N}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ orbitals.

Conclusion

(a) $\left[1\right]\to {\text{H}}_{1s}-{\text{C}}_{sp}\text{, }\left[\text{2}\right]\to {\text{C}}_{sp}-{\text{C}}_{sp},\left[3\right]\to {\text{C}}_{sp}-{\text{C}}_{s{p}^2}\text{and }\left[4\right]\to {\text{N}}_{s{p}^2}-{\text{C}}_{s{p}^3}$