An orbital picture of H 2 C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn. Concept introduction: The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs. The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

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Answer 1

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Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

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Draw the major product of this reaction. Ignore inorganic byproducts. Assume that the water side product is continuously removed to drive the reaction toward products. (CH3)2NH, TSOH Drawing

Answer 2

Question

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

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Answer 3

Question

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

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Answer 4

Question

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

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Answer 5

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

Answer 6

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

Answer 7

Chapter 3, Problem 3.17P

Interpretation Introduction

Interpretation:

An orbital picture of H2C=NH molecule, similar to the picture shown in Figure 3-21, is to be drawn.

Concept introduction:

The orbital picture of a molecule can be drawn on the basis of the geometries of each atom. The geometry determines the atom’s hybridization and the MOs hybrid orbitals it uses for formation of single and double bonds and for lone pairs.

The end-on overlap of hybrid orbitals leads to the formation of σ bonds. π bonds are formed by the side-by-side overlap of pure p orbitals. Lone pairs are unshared electrons, therefore they reside in orbitals that do not have a significant overlap with orbitals of another atom.

Answer 8

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Answer 11

Ch. 3 - Prob. 3.1P Ch. 3 - Prob. 3.2P Ch. 3 - Prob. 3.3P Ch. 3 - Prob. 3.4P Ch. 3 - Prob. 3.5P Ch. 3 - Prob. 3.6P Ch. 3 - Prob. 3.7P Ch. 3 - Prob. 3.8P Ch. 3 - Prob. 3.9P Ch. 3 - Prob. 3.10P

Solution Summary: The author explains how the orbital picture of a molecule can be drawn on the basis of the geometries of each atom.

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Chapter 3 Solutions