CHEMISITRY W/OWL PKG LOOSELEAF
CHEMISITRY W/OWL PKG LOOSELEAF
9th Edition
ISBN: 9781285903859
Author: ZUMDAHL
Publisher: CENGAGE L
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 9, Problem 23E

Give the expected hybridization of the central atom for the molecules or ions in Exercises 81 and 87 from Chapter 3.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in CCl4.

Answer to Problem 23E

Answer

The given compound is sp3 hybridized.

Explanation of Solution

The central atom in CCl4 is carbon (C). The electronic configuration of carbon is,

1s22s22p2

The valence electrons in carbon are four.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=4+42=4

This means that the central atom shows sp3 hybridization and should have a tetrahedral geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2 find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in NCl3.

Answer to Problem 23E

Answer

The given compound is sp3 hybridized.

Explanation of Solution

The central atom in NCl3 is nitrogen (N). The electronic configuration of nitrogen is,

1s22s22p3

The valence electrons in carbon are five.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=5+32=4

This means that the central atom shows sp3 hybridization and should have a tetrahedral geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in SeCl2.

Answer to Problem 23E

The given compound is sp3 hybridized.

Explanation of Solution

The central atom in SeCl2 is selenium (Se). The electronic configuration of nitrogen is,

1s22s22p63s23p63d104s24p4

The valence electrons in selenium are six.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=6+22=4

This means that the central atom shows sp3 hybridization and should have a tetrahedral geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in ICl.

Answer to Problem 23E

The given compound is sp3 hybridized.

Explanation of Solution

The central atom in ICl is Iodine (I). The electronic configuration of iodine is,

1s22s22p63s23p63d104s24p64d104p45s25p5

The valence electrons in selenium are seven.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=7+12=4

This means that the central atom shows sp3 hybridization and should have a tetrahedral geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(a-I)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in NO2.

Answer to Problem 23E

Answer The given compound is sp2 hybridized.

Explanation of Solution

The central atom in NO2 is nitrogen (N). The electronic configuration of nitrogen is,

1s22s22p3

The valence electrons in nitrogen are five.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=5+12=3

This means that the central atom shows sp2 hybridization and should have a triangular planar geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(a-II)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in NO3.

Answer to Problem 23E

The given compound is sp2 hybridized.

Explanation of Solution

The central atom in NO3 is nitrogen (N). The electronic configuration of nitrogen is,

1s22s22p3

The valence electrons in nitrogen are five.

The formula of number of electron pairs is,

X=V+M±C2

Where,

  • X is number of electron pairs.
  • V is valence electrons of central atom.
  • M is number of monovalent atoms.
  • C is charge on compound.

The number of electron pairs is,

X=V+M±C2X=5+12=3

This means that the central atom shows sp2 hybridization and should have a triangular planar geometry.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present.

(a-III)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in N2O4.

Answer to Problem 23E

The given compound is sp2 hybridized.

Explanation of Solution

The central atom in N2O4 is nitrogen (N). The electronic configuration of nitrogen is,

1s22s22p3

The valence electrons in nitrogen are five.

The structure of this molecule is,

CHEMISITRY W/OWL PKG LOOSELEAF, Chapter 9, Problem 23E , additional homework tip  1

Figure 1

For the nitrogen atom, the number of bonded atoms is two and the number of lone pairs is one. Hence, the steric number for the nitrogen atom (sum of the number of bonded atoms and the number of lone pairs present) is 3 . This corresponds to the sp2 hybridization.

This means that the central atom shows sp2 hybridization and should have a triangular planar geometry.

Conclusion

The geometry of a given compound can be predicted using the steric number value for the atom. The steric number 3 corresponds to the sp2 hybridization.

(b-I)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in OCN.

Answer to Problem 23E

Answer The given compound is sp hybridized.

Explanation of Solution

The central atom in OCN is carbon (C). The electronic configuration of carbon is,

1s22s22p2

The valence electrons in carbon are four.

The structure of this molecule is,

CHEMISITRY W/OWL PKG LOOSELEAF, Chapter 9, Problem 23E , additional homework tip  2

Figure 2

For the carbon atom, the number of bonded atoms is two and the number of lone pairs is zero. Hence, the steric number for the carbon atom (sum of the number of bonded atoms and the number of lone pairs present) is 2. This corresponds to the sp hybridization.

This means that the central atom shows sp hybridization and should have a linear geometry.

Conclusion

The geometry of a given compound can be predicted using the steric number value for the atom. The steric number 2 corresponds to the sp hybridization.

(b-II)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in SCN.

Answer to Problem 23E

Answer The given compound is sp hybridized.

Explanation of Solution

The central atom in SCN- is carbon (C). The electronic configuration of carbon is,

1s22s22p2

The valence electrons in carbon are four.

The structure of this molecule is,

CHEMISITRY W/OWL PKG LOOSELEAF, Chapter 9, Problem 23E , additional homework tip  3

Figure 3

For the carbon atom, the number of bonded atoms is two and the number of lone pairs is zero. Hence, the steric number for the carbon atom (sum of the number of bonded atoms and the number of lone pairs present) is 2. This corresponds to the sp hybridization.

This means that the central atom shows sp hybridization and should have a linear geometry.

Conclusion

The geometry of a given compound can be predicted using the steric number value for the atom. The steric number 2 corresponds to the sp hybridization.

(b-III)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The expected hybridization of the central atom for the given species is to be determined.

Concept introduction: The following steps are to be followed to determine the hybridization and the molecular structure of a given compound,

  • The central atom is identified.
  • Its valence electrons are determined.
  • 1e is added to the total number of valence electrons for each monovalent atom present.
  • The total number obtained is divided by 2, to find the number of electron pairs.
  • This further gives us the hybridization of the given compound.

For an atom, present in the structure of a compound, the sum of the number of bonded atoms and the number of lone pairs present on it gives its steric number. The value of the steric number gives us the hybridization of the atom.

To determine: The expected hybridization of the central atom in N3.

Answer to Problem 23E

Answer The given compound is sp hybridized.

Explanation of Solution

The central atom in N3 is nitrogen (N). The electronic configuration of nitrogen is,

1s22s22p3

The valence electrons in nitrogen are five.

The structure of this molecule is,

CHEMISITRY W/OWL PKG LOOSELEAF, Chapter 9, Problem 23E , additional homework tip  4

Figure 4

For the central nitrogen atom, the number of bonded atoms is two and the number of lone pairs is zero. Hence, the steric number for the central nitrogen atom (sum of the number of bonded atoms and the number of lone pairs present) is 2 . This corresponds to the sp hybridization.

This means that the central atom shows sp hybridization and should have a linear geometry.

Conclusion

The geometry of a given compound can be predicted using the steric number value for the atom. The steric number 2 corresponds to the sp hybridization.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Please correct answer and don't used hand raiting
Please correct answer and don't used hand raiting
Please correct answer and don't used hand raiting

Chapter 9 Solutions

CHEMISITRY W/OWL PKG LOOSELEAF

Ch. 9 - Which is the more correct statement: The methane...Ch. 9 - Compare and contrast the MO model with the local...Ch. 9 - What are the relationships among bond order, bond...Ch. 9 - In the hybrid orbital model, compare and contrast ...Ch. 9 - In the molecular orbital mode l, compare and...Ch. 9 - Why are d orbitals sometimes used to form hybrid...Ch. 9 - The atoms in a single bond can rotate about the...Ch. 9 - Compare and contrast bonding molecular orbitals...Ch. 9 - What modification to the molecular orbital model...Ch. 9 - Why does the molecular orbital model do a better...Ch. 9 - The three NO bonds in NO3 are all equivalent in...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - Use the localized electron model to describe the...Ch. 9 - The space-filling models of ethane and ethanol are...Ch. 9 - The space-filling models of hydrogen cyanide and...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - Give the expected hybridization of the central...Ch. 9 - For each of the following molecules, write the...Ch. 9 - For each of the following molecules or ions that...Ch. 9 - Prob. 31ECh. 9 - The allene molecule has the following Lewis...Ch. 9 - Indigo is the dye used in coloring blue jeans. The...Ch. 9 - Urea, a compound formed in the liver, is one of...Ch. 9 - Biacetyl and acetoin are added to margarine to...Ch. 9 - Many important compounds in the chemical industry...Ch. 9 - Two molecules used in the polymer industry are...Ch. 9 - Hot and spicy foods contain molecules that...Ch. 9 - One of the first drugs to be approved for use in...Ch. 9 - The antibiotic thiarubin-A was discovered by...Ch. 9 - Consider the following molecular orbitals formed...Ch. 9 - Sketch the molecular orbital and label its type (...Ch. 9 - Which of the following are predicted by the...Ch. 9 - Which of the following are predicted by the...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - Consider the following electron configuration:...Ch. 9 - Using molecular orbital theory, explain why the...Ch. 9 - Using the molecular orbital model to describe the...Ch. 9 - The transport of O2 in the blood is carried out by...Ch. 9 - A Lewis structure obeying the octet rule can be...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - Using the molecular orbital model, write electron...Ch. 9 - In which of the following diatomic molecules would...Ch. 9 - In terms of the molecular orbital model, which...Ch. 9 - Show how two 2p atomic orbitals can combine to...Ch. 9 - Show how a hydrogen 1s atomic orbital and a...Ch. 9 - Use Figs. 4-54 and 4-55 to answer the following...Ch. 9 - Acetylene (C2H2) can be produced from the reaction...Ch. 9 - Describe the bonding in NO+, NO, and NO, using...Ch. 9 - Describe the bonding in the O3 molecule and the...Ch. 9 - Describe the bonding in the CO32 ion using the...Ch. 9 - Draw the Lewis structures, predict the molecular...Ch. 9 - FClO2 and F3ClO can both gain a fluoride ion to...Ch. 9 - Two structures can be drawn for cyanuric acid: a....Ch. 9 - Give the expected hybridization for the molecular...Ch. 9 - Vitamin B6 is an organic compound whose deficiency...Ch. 9 - Aspartame is an artificial sweetener marketed...Ch. 9 - Prob. 69AECh. 9 - The three most stable oxides of carbon are carbon...Ch. 9 - Complete the following resonance structures for...Ch. 9 - Prob. 73AECh. 9 - Describe the bonding in the first excited state of...Ch. 9 - Using an MO energy-level diagram, would you expect...Ch. 9 - Show how a dxz. atomic orbital and a pz, atomic...Ch. 9 - What type of molecular orbital would result from...Ch. 9 - Consider three molecules: A, B, and C. Molecule A...Ch. 9 - Draw the Lewis structures for TeCl4, ICl5, PCl5,...Ch. 9 - A variety of chlorine oxide fluorides and related...Ch. 9 - Pelargondin is the molecule responsible for the...Ch. 9 - Complete a Lewis structure for the compound shown...Ch. 9 - Which of the following statements concerning SO2...Ch. 9 - Consider the molecular orbital electron...Ch. 9 - Place the species B2+ , B2, and B2 in order of...Ch. 9 - Consider the following computer-generated model of...Ch. 9 - Cholesterol (C27liu;O) has the following...Ch. 9 - Cyanamide (H2NCN), an important industrial...Ch. 9 - A flask containing gaseous N2 is irradiated with...Ch. 9 - Prob. 92CPCh. 9 - Values of measured bond energies may vary greatly...Ch. 9 - Use the MO model to explain the bonding in BeH2....Ch. 9 - Prob. 95CPCh. 9 - Arrange the following from lowest to highest...Ch. 9 - Use the MO model to determine which of the...Ch. 9 - Given that the ionization energy of F2 is 290...Ch. 9 - Carbon monoxide (CO) forms bonds to a variety of...Ch. 9 - Prob. 100CPCh. 9 - As the bead engineer of your starship in charge of...Ch. 9 - Determine the molecular structure and...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY