Chapter 7, Problem 104QRT

Piperine, the active ingredient in black pepper, has this Lewis structure.

1. (a) Give the values for the indicated bond angles.
2. (b) What is the hybridization of the nitrogen?
3. (c) What is the hybridization of the oxygens?

Interpretation Introduction

Interpretation:

The indicated bond angles has to be given.

Concept Introduction:

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$.

Explanation of Solution

The given structure of Piperine is given below:

  .

The bond angles to be determined are given below:

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Consider the $\text{N-C-O}$(1) bond angle. The electron-region geometry of carbon atom bonded to three other atom is trigonal planar. It is a type of ${\text{AX}}_3{\text{E}}_0$ molecule and the bond angle is ${120}^{\circ }$ for this type of molecule.

Consider the $\text{C-C-C}$(2) bond angle. The electron-region geometry of central carbon atom bonded to three other atom is trigonal planar. It is a type of ${\text{AX}}_3{\text{E}}_0$ molecule and the bond angle is ${120}^{\circ }$ for this type of molecule.

Consider the $\text{O-C-O}$ (3) bond angle. The electron-region geometry of central carbon atom bonded to four other atoms is tetrahedral. It is a type of ${\text{AX}}_4{\text{E}}_0$ molecule and the bond angle is ${109.5}^{\circ }$ for this type of molecule.

Interpretation Introduction

Interpretation:

The hybridization for nitrogen atom in the molecule has to be designated.

Concept Introduction:

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

Geometry of different types of molecule with respect to the hybridizations 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

The given structure of Piperine is shown below:

  .

Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of ${\text{AX}}_3{\text{E}}_1$ molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^{\text{3}}$.

Consider the right carbon. There will be three electron regions in the molecule and hence the electron-region geometry will be triangular planar. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^2$.

Interpretation Introduction

Interpretation:

The hybridization for oxygen atoms in the molecule has to be designated.

Concept Introduction:

Refer to (b).

Explanation of Solution

The given structure of Piperine is shown below:

  .

Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of ${\text{AX}}_3{\text{E}}_1$ molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^{\text{3}}$.

Consider the oxygen atoms with single bonds. There will be four electron regions around both oxygen atoms. They are a type of ${\text{AX}}_2{\text{E}}_2$ molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^{\text{3}}$.

Consider the double bonded oxygen atom. There will be three electron regions around oxygen atom. It is a type of ${\text{AX}}_1{\text{E}}_2$ molecule and hence the electron-region geometry will be triangular planar. For a molecule having triangular planar geometry, the hybridization will be ${\text{sp}}^2$.