Chapter 1, Problem 1.70P

Interpretation Introduction

(a)

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

Concept introduction: The geometry and hybridisation of an atom is determined by the number of group around it. If the number of groups attached to an atom is $2$, then the geometry will be linear. If the number of groups attached to an atom is $3$, then the geometry will be trigonal planar. If the number of groups attached to an atom is $4$, then the geometry will be tetrahedral.

Answer to Problem 1.70P

Orbitals used in the formation of given bonds are:

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

Explanation of Solution

The given compound is,

Figure $1$

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

Conclusion

Orbitals used in the formation of given bonds are:

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

Interpretation Introduction

(b)

Interpretation: The orbitals used to form the indicated bondsare to be predicted.

Concept introduction: The geometry and hybridisation of an atom is determined by the number of group around it. If the number of groups attached to an atom is $2$, then the geometry will be linear. If the number of groups attached to an atom is $3$, then the geometry will be trigonal planar. If the number of groups attached to an atom is $4$, then the geometry will be tetrahedral.

Answer to Problem 1.70P

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 ${\text{C}}_{s{p}^2}-{\text{C}}_{s{p}^3}$ orbitals.

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

(c)

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

Concept introduction: The geometry and hybridisation of an atom is determined by the number of group around it. If the number of groups attached to an atom is $2$, then the geometry will be linear. If the number of groups attached to an atom is $3$, then the geometry will be trigonal planar. If the number of groups attached to an atom is $4$, then the geometry will be tetrahedral.

Answer to Problem 1.70P

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

(d)

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

Concept introduction: The geometry and hybridisation of an atom is determined by the number of group around it. If the number of groups attached to an atom is $2$, then the geometry will be linear. If the number of groups attached to an atom is $3$, then the geometry will be trigonal planar. If the number of groups attached to an atom is $4$, then the geometry will be tetrahedral.

Answer to Problem 1.70P

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

Orbitals used in the formation of given bonds are:

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