Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 7, Problem 104QRT

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

Chapter 7, Problem 104QRT, Piperine, the active ingredient in black pepper, has this Lewis structure. (a) Give the values for

  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?

(a)

Expert Solution
Check Mark
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.

TypeofmoleculeHybridaizationGeometryBondangleAX2spLinear180°AX3,AX2Bsp2Trigonalplanar120°AX4,AX3B,AX2B2sp3Tetrahedral109.5°AX5,AX4B,AX3B2,AX2B3sp3dTrigonalbipyramidal120°,90°AX6,AX5B,AX4B2sp3d2Octahedral90°ACentralatomXAtomsbondedtoABNonbondingelectronpairsonA.

Explanation of Solution

The given structure of Piperine is given below:

  Chemistry: The Molecular Science, Chapter 7, Problem 104QRT , additional homework tip  1.

The bond angles to be determined are given below:

  Chemistry: The Molecular Science, Chapter 7, Problem 104QRT , additional homework tip  2.

Consider the 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 AX3E0 molecule and the bond angle is 120 for this type of molecule.

Consider the 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 AX3E0 molecule and the bond angle is 120 for this type of molecule.

Consider the 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 AX4E0 molecule and the bond angle is 109.5 for this type of molecule.

(b)

Expert Solution
Check Mark
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,

TypeofmoleculeHybridaizationAtomicorbitalsusedforhybridaizationGeometryAX2sp1s+1pLinearAX3,AX2Bsp21s+2pTrigonalplanarAX4,AX3B,AX2B2sp31s+3pTetrahedralAX5,AX4B,AX3B2,AX2B3sp3d1s+3p+1dTrigonalbipyramidalAX6,AX5B,AX4B2sp3d21s+3p+2dOctahedralACentralatomXAtomsbondedtoABNonbondingelectronpairsonA

Explanation of Solution

The given structure of Piperine is shown below:

  Chemistry: The Molecular Science, Chapter 7, Problem 104QRT , additional homework tip  3.

Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of AX3E1 molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be sp3 .

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 sp2 .

(c)

Expert Solution
Check Mark
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:

  Chemistry: The Molecular Science, Chapter 7, Problem 104QRT , additional homework tip  4.

Consider the nitrogen atom. There will be four electron regions around nitrogen atom. It is a type of AX3E1 molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be sp3 .

Consider the oxygen atoms with single bonds. There will be four electron regions around both oxygen atoms. They are a type of AX2E2 molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be sp3 .

Consider the double bonded oxygen atom. There will be three electron regions around oxygen atom. It is a type of AX1E2 molecule and hence the electron-region geometry will be triangular planar. For a molecule having triangular planar geometry, the hybridization will be sp2 .

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Chapter 7 Solutions

Chemistry: The Molecular Science

Ch. 7.5 - Prob. 7.5ECh. 7.6 - Prob. 7.8PSPCh. 7.6 - Prob. 7.7CECh. 7.6 - Prob. 7.9PSPCh. 7.7 - Prob. 7.8CECh. 7.7 - Prob. 7.9CECh. 7 - Write the Lewis structures and give the...Ch. 7 - The structural formula for the open-chain form of...Ch. 7 - Describe the VSEPR model. How is the model used to...Ch. 7 - What is the difference between the electron-region...Ch. 7 - Prob. 3QRTCh. 7 - Prob. 4QRTCh. 7 - If you have three electron regions around a...Ch. 7 - Prob. 6QRTCh. 7 - Prob. 7QRTCh. 7 - Prob. 8QRTCh. 7 - Prob. 9QRTCh. 7 - Prob. 10QRTCh. 7 - Prob. 11QRTCh. 7 - Prob. 12QRTCh. 7 - Prob. 13QRTCh. 7 - Prob. 14QRTCh. 7 - Prob. 15QRTCh. 7 - Prob. 16QRTCh. 7 - Write Lewis structures for XeOF2 and ClOF3. Use...Ch. 7 - Write Lewis structures for HCP and [IOF4]. Use...Ch. 7 - Prob. 19QRTCh. 7 - Prob. 20QRTCh. 7 - Explain why (I3)+ is bent, but (I3) is linear.Ch. 7 - Prob. 22QRTCh. 7 - Prob. 23QRTCh. 7 - Give approximate values for the indicated bond...Ch. 7 - Give approximate values for the indicated bond...Ch. 7 - Prob. 26QRTCh. 7 - Compare the FClF angles in ClF2+ and ClF2. From...Ch. 7 - Prob. 28QRTCh. 7 - Prob. 29QRTCh. 7 - Prob. 30QRTCh. 7 - Prob. 31QRTCh. 7 - Describe the geometry and hybridization of carbon...Ch. 7 - Describe the geometry and hybridization for each C...Ch. 7 - Describe the hybridization around the central atom...Ch. 7 - The hybridization of the two carbon atoms differs...Ch. 7 - The hybridization of the two nitrogen atoms...Ch. 7 - Identify the type of hybridization, approximate...Ch. 7 - Prob. 38QRTCh. 7 - Prob. 39QRTCh. 7 - Prob. 40QRTCh. 7 - Prob. 41QRTCh. 7 - Methylcyanoacrylate is the active ingredient in...Ch. 7 - Prob. 43QRTCh. 7 - Prob. 44QRTCh. 7 - Prob. 45QRTCh. 7 - Prob. 46QRTCh. 7 - Which of these molecules has a net dipole moment?...Ch. 7 - Prob. 48QRTCh. 7 - Use molecular structures and noncovalent...Ch. 7 - Prob. 50QRTCh. 7 - Explain why water “beads up” on a freshly waxed...Ch. 7 - Explain why water will not remove tar from your...Ch. 7 - Prob. 53QRTCh. 7 - Prob. 54QRTCh. 7 - Prob. 55QRTCh. 7 - Prob. 56QRTCh. 7 - The structural formula for vitamin C is Give a...Ch. 7 - Prob. 58QRTCh. 7 - Prob. 59QRTCh. 7 - Prob. 60QRTCh. 7 - Prob. 61QRTCh. 7 - Prob. 62QRTCh. 7 - Prob. 63QRTCh. 7 - Prob. 64QRTCh. 7 - Prob. 65QRTCh. 7 - Prob. 66QRTCh. 7 - Methylcyanoacrylate is the active ingredient in...Ch. 7 - Prob. 68QRTCh. 7 - Prob. 69QRTCh. 7 - Use Lewis structures and VSEPR theory to predict...Ch. 7 - In addition to CO, CO2, and C3O2, there is another...Ch. 7 - Prob. 72QRTCh. 7 - Prob. 73QRTCh. 7 - Prob. 74QRTCh. 7 - Prob. 75QRTCh. 7 - In the gas phase, positive and negative ions form...Ch. 7 - Prob. 77QRTCh. 7 - Prob. 78QRTCh. 7 - Prob. 79QRTCh. 7 - Prob. 80QRTCh. 7 - Prob. 81QRTCh. 7 - Prob. 82QRTCh. 7 - Prob. 83QRTCh. 7 - Prob. 84QRTCh. 7 - Prob. 85QRTCh. 7 - Prob. 86QRTCh. 7 - Prob. 87QRTCh. 7 - Prob. 88QRTCh. 7 - Prob. 89QRTCh. 7 - Prob. 90QRTCh. 7 - Prob. 91QRTCh. 7 - Prob. 92QRTCh. 7 - Prob. 93QRTCh. 7 - Prob. 94QRTCh. 7 - Which of these are examples of hydrogen bonding?Ch. 7 - Prob. 96QRTCh. 7 - Prob. 97QRTCh. 7 - Prob. 98QRTCh. 7 - Halothane, which had been used as an anesthetic,...Ch. 7 - Ketene, C2H2O, is a reactant for synthesizing...Ch. 7 - Gamma hydroxybutyric acid, GHB, infamous as a date...Ch. 7 - There are two compounds with the molecular formula...Ch. 7 - Piperine, the active ingredient in black pepper,...Ch. 7 - Prob. 105QRTCh. 7 - Two compounds have the molecular formula N3H3. One...Ch. 7 - Prob. 108QRTCh. 7 - Prob. 109QRTCh. 7 - Prob. 110QRTCh. 7 - Prob. 111QRTCh. 7 - Prob. 7.ACPCh. 7 - Prob. 7.BCPCh. 7 - Prob. 7.CCPCh. 7 - Prob. 7.DCP
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