Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078131615
Author: Martin Silberberg Dr.
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.22P

(a)

Interpretation Introduction

Interpretation:

The hybrid orbitals used and the types of bonds formed by the central atom in O3 are to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.

(b)

Interpretation Introduction

Interpretation:

The hybrid orbitals used and the types of bonds formed by the central atom in I3 are to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.

(c)

Interpretation Introduction

Interpretation:

The hybrid orbitals used and the types of bonds formed by the central atom in COCl2 are to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

Sigma bond is formed by the end to end overlapping of atomic or hybrid orbitals. All single bonds are sigma bonds. One double bond is made up of one sigma bond and one pi bond. One triple bond is made up of one sigma bond and two pi bonds.

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6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?
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Chapter 11 Solutions

Student Study Guide for Silberberg Chemistry: The Molecular Nature of Matter and Change

Ch. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Give the number and type of hybrid orbital that...Ch. 11 - What is the hybridization of nitrogen in each of...Ch. 11 - What is the hybridization of carbon in each of the...Ch. 11 - Prob. 11.9PCh. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Phosphine (PH3) reacts with borane (BH3) as...Ch. 11 - The illustrations below depict differences in...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Methyl isocyanate, , is an intermediate in the...Ch. 11 - Are these statements true or false? Correct any...Ch. 11 - Prob. 11.21PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.23PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Certain atomic orbitals on two atoms were combined...Ch. 11 - Prob. 11.28PCh. 11 - Antibonding MOs always have at least one node. Can...Ch. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - The molecular orbitals depicted are derived from...Ch. 11 - The molecular orbitals depicted below are derived...Ch. 11 - Prob. 11.34PCh. 11 - Use an MO diagram and the bond order you obtain...Ch. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Epinephrine (or adrenaline; below) is a naturally...Ch. 11 - Prob. 11.41PCh. 11 - Isoniazid (below) is an antibacterial agent that...Ch. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Tryptophan is one of the amino acids found in...Ch. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Sulfur forms oxides, oxoanions, and halides. What...Ch. 11 - Prob. 11.54PCh. 11 - Use an MO diagram to find the bond order and...Ch. 11 - Acetylsalicylic acid (aspirin), the most widely...Ch. 11 - Prob. 11.57P
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