Chapter 5, Problem 5.10P

(a)

Interpretation Introduction

Interpretation:

Atomic orbitals which are used to form each $\sigma \text{ and }\pi$ bond have to be identified for the highlighted carbon atom of the given compound.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared unhybridized orbital’s (p, d, etc) electron density are concentrated in above and below of the plane of the molecule.

  $\begin{array}{ccc}\text{Type}\text{of}\text{Bond}& \text{No}\text{.}\text{of}\text{σ}\text{bonds}& \text{No}\text{.}\text{of}\text{π}\text{bonds}\\ \text{Single}& \text{1}& \text{0}\\ \text{Double}& \text{1}& \text{1}\\ \text{Triple}& \text{1}& \text{2}\end{array}$

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

In the marked carbon atom, one s and three p orbital hybridize forming four $s{p}^3$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

In the marked carbon atom, one s and two p orbital hybridize forming three $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

(b)

Interpretation Introduction

Interpretation:

Atomic orbitals which are used to form each $\sigma \text{ and }\pi$ bond have to be identified for the highlighted carbon atom of the given compound.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared unhybridized orbital’s (p, d, etc) electron density are concentrated in above and below of the plane of the molecule.

  $\begin{array}{ccc}\text{Type}\text{of}\text{Bond}& \text{No}\text{.}\text{of}\text{σ}\text{bonds}& \text{No}\text{.}\text{of}\text{π}\text{bonds}\\ \text{Single}& \text{1}& \text{0}\\ \text{Double}& \text{1}& \text{1}\\ \text{Triple}& \text{1}& \text{2}\end{array}$

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

In the marked carbon atom, one s and two p orbital hybridize forming three $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

(c)

Interpretation Introduction

Interpretation:

Atomic orbitals which are used to form each $\sigma \text{ and }\pi$ bond has to be identified for the highlighted carbon atom of the given compound.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared unhybridized orbital’s (p, d, etc) electron density are concentrated in above and below of the plane of the molecule.

  $\begin{array}{ccc}\text{Type}\text{of}\text{Bond}& \text{No}\text{.}\text{of}\text{σ}\text{bonds}& \text{No}\text{.}\text{of}\text{π}\text{bonds}\\ \text{Single}& \text{1}& \text{0}\\ \text{Double}& \text{1}& \text{1}\\ \text{Triple}& \text{1}& \text{2}\end{array}$

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

In the marked carbon atom, one s and two p orbital hybridize forming three $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

(d)

Interpretation Introduction

Interpretation:

Atomic orbitals which are used to form each $\sigma \text{ and }\pi$ bond has to be identified for the highlighted carbon atom of the given compound.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared unhybridized orbital’s (p, d, etc) electron density are concentrated in above and below of the plane of the molecule.

  $\begin{array}{ccc}\text{Type}\text{of}\text{Bond}& \text{No}\text{.}\text{of}\text{σ}\text{bonds}& \text{No}\text{.}\text{of}\text{π}\text{bonds}\\ \text{Single}& \text{1}& \text{0}\\ \text{Double}& \text{1}& \text{1}\\ \text{Triple}& \text{1}& \text{2}\end{array}$

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

In the marked carbon atom, one s and two p orbital hybridize forming three $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

(e)

Interpretation Introduction

Interpretation:

Atomic orbitals which are used to form each $\sigma \text{ and }\pi$ bond has to be identified for the highlighted carbon atom of the given compound.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Sigma (σ) bonds are the bonds in which shared hybrid orbital’s electron density are concentrated along the internuclear axis.

Pi (π) bonds are the bonds in which shared unhybridized orbital’s (p, d, etc) electron density are concentrated in above and below of the plane of the molecule.

  $\begin{array}{ccc}\text{Type}\text{of}\text{Bond}& \text{No}\text{.}\text{of}\text{σ}\text{bonds}& \text{No}\text{.}\text{of}\text{π}\text{bonds}\\ \text{Single}& \text{1}& \text{0}\\ \text{Double}& \text{1}& \text{1}\\ \text{Triple}& \text{1}& \text{2}\end{array}$

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

In the marked carbon atom, one s and one p orbital hybridize forming two $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,

In the marked carbon atom, one s and two p orbital hybridize forming three $s{p}^2$ orbitals. Atomic orbitals are used to form each $\sigma \text{ and }\pi$ bond for the highlighted carbon atom can be given as,