CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 11, Problem 11.38P

(a)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in BrO3 are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(a)

Expert Solution
Check Mark

Answer to Problem 11.38P

In BrO3, the hybridization of bromine is sp3. Its shape is trigonal pyramidal, the ideal bond angle is 109.5° and the deviated bond angle is less than 109.5°.

Explanation of Solution

The Lewis structure of BrO3 is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  1

Boron forms three single bonds with three oxygen atoms and one lone pair is present on it so four hybrid orbitals are required and therefore the hybridization of boron in BrO3 is sp3. According to hybridization, its shape should be tetrahedral and the ideal bond angle should be 109.5°. But due to the presence of a lone pair on bromine atom, its shape becomes trigonal pyramidal and the bond angle becomes less than 109.5°.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(b)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in AsCl4 are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(b)

Expert Solution
Check Mark

Answer to Problem 11.38P

In AsCl4, the hybridization of bromine is sp3d. Its shape is seesaw, ideal bond angles are 120° and 90°. But the deviated bond angles are less than 120° and 90°.

Explanation of Solution

The Lewis structure of AsCl4 is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  2

Arsenic forms four single bonds with four chlorine atoms and one lone pair is present on it so five hybrid orbitals are required and therefore the hybridization of arsenic in AsCl4 is sp3d. According to hybridization, its shape should be trigonal bipyramidal and the ideal bond angles should be 120° and 90°. But due to the presence of a lone pair on the arsenic atom, its shape becomes seesaw and the bond angle becomes less than 120° and 90°.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(c)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in SeO42 are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(c)

Expert Solution
Check Mark

Answer to Problem 11.38P

In SeO42, the hybridization of selenium is sp3. Its shape is tetrahedral and the ideal bond angles is 109.5°. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of SeO42 is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  3

Selenium forms four single bonds with four oxygen atoms so four hybrid orbitals are required and therefore the hybridization of selenium in SeO42 is sp3. According to hybridization, its shape is tetrahedral and the ideal bond angle should be 109.5°. There will be no deviation in the bond angle due to the absence of a lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(d)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in BiF52 are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(d)

Expert Solution
Check Mark

Answer to Problem 11.38P

In BiF52, the hybridization of bismuth is sp3d2. Its shape is square pyramidal and ideal bond angle is 90°. But the deviated bond angle is less than 90°.

Explanation of Solution

The Lewis structure of BiF52 is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  4

Bismuth forms five single bonds with five fluorine atoms and one lone pair is present on it so six hybrid orbitals are required and therefore the hybridization of bismuth in BiF52 is sp3d2. According to hybridization, its shape should be octahedral and the ideal bond angle should be 90°. But due to the presence of a lone pair on bismuth atom, its shape becomes square pyramidal and the bond angle becomes less than 90°.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

(e)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in SbF4+ are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(e)

Expert Solution
Check Mark

Answer to Problem 11.38P

In SbF4+, the hybridization of selenium is sp3. Its shape is tetrahedral and ideal bond angle is 109.5°. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of SbF4+ is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  5

Antimony forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of antimony in SbF4+ is sp3. According to hybridization, its shape is tetrahedral and the ideal bond angle is 109.5°. There will be no deviation in the bond angle due to the absence of a lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(f)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in AlF63 are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(f)

Expert Solution
Check Mark

Answer to Problem 11.38P

In AlF63, the hybridization of aluminium is sp3d2. Its shape is octahedral, the ideal bond angle is 90°. There is no deviation in the bond angle.

Explanation of Solution

The Lewis structure of AlF63 is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  6

Aluminium forms six single bonds with six fluorine atoms so six hybrid orbitals are required and therefore the hybridization of aluminium in AlF63 is sp3d2. According to hybridization, its shape is octahedral and the ideal bond angle is 90°. There will be no deviation in the bond angle due to the absence of lone pair of electrons.

Conclusion

The shape of the molecule is determined by the hybridization only if no lone pair is present on the central atom.

(g)

Interpretation Introduction

Interpretation:

The shape, hybridization of the central atom, ideal and deviated bond angle in IF4+ are to be determined.

Concept introduction:

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.

The shape of the molecule is determined by the electron bond pairs and lone pairs that are present around the central atom. The angle between the two bonds is called the bond angle. It is determined by the hybridization of the central atom and the presence of lone pairs around it.

(g)

Expert Solution
Check Mark

Answer to Problem 11.38P

In IF4+, the hybridization of iodine is sp3d. Its shape is seesaw, ideal bond angles are 120° and 90°. But the deviated bond angles are 120° and 90°.

Explanation of Solution

The Lewis structure of IF4+ is as follows:

CHEMISTRY >CUSTOM<                     , Chapter 11, Problem 11.38P , additional homework tip  7

Iodine forms four single bonds with four fluorine atoms and one lone pair is present on it so five hybrid orbitals are required and therefore the hybridization of iodine in IF4+ is sp3d. According to hybridization, its shape should be trigonal bipyramidal and the ideal bond angles should be 120° and 90°. But due to the presence of a lone pair on iodine atom, its shape becomes seesaw and the bond angles become less than 120° and 90°.

Conclusion

The presence of lone pair on the central atom results in the deviation of bond angles from that of the ideal ones.

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Chapter 11 Solutions

CHEMISTRY >CUSTOM<

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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