Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 9, Problem 33E

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals of the central atom, and predict the overall polarity.

a. CF4

b. NF3

c. OF2

d. BF3

e. BeH2

f. TeF4

g. AsF5

h. KrF2

i. KrF4

j. SeF6

k. IF5

l. IF3

(a)

Expert Solution
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Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for CF4 .

Explanation of Solution

The atomic number of carbon is 6 and its electronic configuration is,

1s22s22p4

The valence electrons present in carbon are four.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7

The molecule CF4 is made of four fluorine atoms and one carbon atom. Hence, the total number of valence electrons is,

C+4F=(4+(4×7))e=32e

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.

The Lewis structure of CF4 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  1

Figure 1

The Lewis structure of the given molecule corresponds to a tetrahedral molecular structure.

The resultant of any three CF bonds is equal in magnitude but opposite in direction to the fourth CF bond. Hence, the molecule is non-polar.

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for NF3 .

Explanation of Solution

The atomic number of nitrogen is 7 and its electronic configuration is,

1s22s22p3

The valence electrons present in nitrogen are five.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule NF3 is made of three fluorine atoms and one nitrogen atom. Hence, the total number of valence electrons is,

N+3F=(5+(3×7))e=26e

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.

The Lewis structure of NF3 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  2

Figure 2

The Lewis structure of the given molecule corresponds to a triangular pyramidal molecular structure.

The given molecule is polar and the bond angle present is 109.5° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for OF2 .

Explanation of Solution

The atomic number of oxygen is 8 and its electronic configuration is,

1s22s22p4

The valence electrons present in nitrogen are six.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule OF2 is made of two fluorine atoms and one oxygen atom. Hence, the total number of valence electrons is,

O+2F=(6+(2×7))e=20e

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.

The Lewis structure of OF2 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  3

Figure 3

The Lewis structure of the given molecule corresponds to a V-shaped molecular structure.

The given molecule is polar and the bond angle present is 109.5° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for BF3 .

Explanation of Solution

The atomic number of boron is 5 and its electronic configuration is,

1s22s22p1

The valence electrons present in boron are three.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule BF3 is made of three fluorine atoms and one boron atom. Hence, the total number of valence electrons is,

B+3F=(3+(3×7))e=24e

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=3+32=3

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

The Lewis structure of BF3 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  4

Figure 4

The Lewis structure of the given molecule corresponds to a triangular planar molecular structure.

The given molecule is non-polar and the bond angle present is 120° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for BeH2 .

Explanation of Solution

The atomic number of beryllium is 4 and its electronic configuration is,

1s22s2

The valence electrons present in beryllium are two.

The atomic number of hydrogen is 1 and its electronic configuration is,

1s1

Hydrogen has only one electron.

The molecule BeH2 is made of two hydrogen atoms and one beryllium atom. Hence, the total number of valence electrons is,

Be+2H=(2+(2×1))e=4e

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=2+22=2

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

The Lewis structure of BeH2 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  5

Figure 5

The Lewis structure of the given molecule corresponds to a triangular planar molecular structure.

The given molecule is non-polar (because of its linear structure) and the bond angle present is 180° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for TeF4 .

Explanation of Solution

The atomic number of tellurium is 52 and its electronic configuration is,

1s22s22p63s23p63d104s24p64d105s25p4

The valence electrons present in tellurium are six.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule TeF4 is made of four fluorine atoms and one tellurium atom. Hence, the total number of valence electrons is,

Te+4F=(6+(4×7))e=34e

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+42=5

This means that the central atom shows sp3d hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of TeF4 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  6

Figure 6

The Lewis structure of the given molecule corresponds to a see-saw shaped molecular structure.

The given molecule is polar (as the equatorial dipoles do not get cancelled) and the bond angle for FTeF is 90° and 120° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for AsF5 .

Explanation of Solution

The atomic number of arsenic is 33 and its electronic configuration is,

1s22s22p63s23p63d104s24p3

The valence electrons present in arsenic are five.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule AsF5 is made of five fluorine atoms and one arsenic atom. Hence, the total number of valence electrons is,

As+5F=(5+(5×7))e=40e

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+52=5

This means that the central atom shows sp3d hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of AsF5 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  7

Figure 7

The Lewis structure of the given molecule corresponds to a trigonal bipyramidal molecular structure.

The given molecule is non-polar (as the axial dipoles being opposite in direction get cancelled; and the resultant of any two equatorial AsF bond is equal in magnitude but opposite in direction to the third) and the bond angle for FAsF is 90° and 120° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(h)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for KrF2 .

Explanation of Solution

The atomic number of krypton is 36 and its electronic configuration is,

1s22s22p63s23p63d104s24p6

The valence electrons present in krypton are eight.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule KrF2 is made of two fluorine atoms and one krypton atom. Hence, the total number of valence electrons is,

Kr+2F=(8+(2×7))e=22e

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=8+22=5

This means that the central atom shows sp3d hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of KrF2 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  8

Figure 8

The Lewis structure of the given molecule corresponds to a linear molecular structure.

The given molecule is non-polar (because of its linear structure) and the bond angle present is 180° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(i)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for KrF4 .

Explanation of Solution

The atomic number of krypton is 36 and its electronic configuration is,

1s22s22p63s23p63d104s24p6

The valence electrons present in krypton are eight.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule KrF4 is made of two fluorine atoms and one krypton atom. Hence, the total number of valence electrons is,

Kr+4F=(8+(4×7))e=36e

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=8+42=6

This means that the central atom shows sp3d2 hybridization and should have a square bipyramidal geometry.

The Lewis structure of KrF4 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  9

Figure 9

The Lewis structure of the given molecule corresponds to a square planar molecular structure.

The given molecule is non-polar (the dipoles get cancelled) and the bond angle present is 90° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(j)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for SeF6 .

Explanation of Solution

The atomic number of selenium is 34 and its electronic configuration is,

1s22s22p63s23p63d104s24p4

The valence electrons present in selenium are six.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule SeF6 is made of six fluorine atoms and one selenium atom. Hence, the total number of valence electrons is,

Se+6F=(6+(6×7))e=48e

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+62=6

This means that the central atom shows sp3d2 hybridization and should have a square bipyramidal geometry.

The Lewis structure of SeF6 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  10

Figure 10

The Lewis structure of the given molecule corresponds to a square bipyramidal molecular structure.

The given molecule is non-polar (the dipoles get cancelled) and the bond angle present is 90° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(k)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for IF5 .

Explanation of Solution

The atomic number of iodine is 53 and its electronic configuration is,

1s22s22p63s23p63d104s24p64d105s25p5

The valence electrons present in iodine are seven.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule IF5 is made of five fluorine atoms and one iodine atom. Hence, the total number of valence electrons is,

I+5F=(7+(5×7))e=42e

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+52=6

This means that the central atom shows sp3d2 hybridization and should have a square bipyramidal geometry.

The Lewis structure of IF5 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  11

Figure 11

The Lewis structure of the given molecule corresponds to a square pyramidal molecular structure.

The given molecule is polar (the dipoles do not get cancelled, it is polar because of the axial dipole) and the bond angle present is 90° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

(l)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for each species is to be determined.

Concept introduction: Molecular structure is a three-dimensional shape of a molecule. 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.

A Lewis structure depicts the bonding between the atoms, present in a molecule and also the number of lone pairs present in the structure.

To determine: The Lewis structure; the molecular structure; the expected hybridization of the central atom and the overall polarity for IF3 .

Explanation of Solution

The atomic number of iodine is 53 and its electronic configuration is,

1s22s22p63s23p63d104s24p64d105s25p5

The valence electrons present in iodine are seven.

The atomic number of fluorine is 9 and its electronic configuration is,

1s22s22p5

The valence electron of fluorine is 7 .

The molecule IF3 is made of five fluorine atoms and one iodine atom. Hence, the total number of valence electrons is,

I+3F=(7+(3×7))e=28e

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+32=5

This means that the central atom shows sp3d hybridization and should have a triangular bipyramidal geometry.

The Lewis structure of IF3 is,

Chemistry, Chapter 9, Problem 33E , additional homework tip  12

Figure 12

The Lewis structure of the given molecule corresponds to a trigonal bipyramidal molecular structure.

The given molecule is non-polar (as the axial dipoles being opposite in direction get cancelled; and the resultant of any two equatorial IF bond is equal in magnitude but opposite in direction to the third) and the bond angle for FIF is 90° and 120° .

Conclusion

The geometry of a given compound can be predicted using the hybridization of the central atom present. The Lewis structure gives us the final molecular structure for the given molecule.

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Chemistry

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 - The molecules N2 and CO are isoelectronic but...Ch. 9 - Do lone pairs about a central atom affect the...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 - As compared with CO and O2, CS and S2 are very...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. 35ECh. 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 - Minoxidil (C9H15N15O) is a compound produced 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 the molecular orbital model to describe the...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 - The antibiotic thiarubin-A was discovered by...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. 73AECh. 9 - The three most stable oxides of carbon are carbon...Ch. 9 - Complete the following resonance structures for...Ch. 9 - Prob. 77AECh. 9 - The transport of O2 in the blood is carried out by...Ch. 9 - Using molecular orbital theory, explain why the...Ch. 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 - Values of measured bond energies may vary greatly...Ch. 9 - Use the MO model to explain the bonding in BeH2....Ch. 9 - Prob. 101CPCh. 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. 106CPCh. 9 - As the bead engineer of your starship in charge of...Ch. 9 - Determine the molecular structure and...Ch. 9 - Although nitrogen trifluoride (NF3) is a thermally...
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