The molecular geometry of PBr 3 is to be determined. Concept introduction: In order to find the geometry of a molecule, follow the steps: First, draw the Lewis dot structure of the molecule. Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom. Use the VSEPR model to find the electron domain geometry. According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space. Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry. The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

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Answer 1

Question

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

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Students have asked these similar questions

The Concept of Aromaticity 21.15 State the number of 2p orbital electrons in each molecule or ion. (a) (b) (e) (f) (c) (d) (h) (i) DA (k) 21.16 Which of the molecules and ions given in Problem 21.15 are aromatic according to the Hückel criteria? Which, if planar, would be antiaromatic? 21.17 Which of the following structures are considered aromatic according to the Hückel criteria? ---0-0 (a) (b) (c) (d) (e) (h) H -H .8.0- 21.18 Which of the molecules and ions from Problem 21.17 have electrons donated by a heteroatom?

1. Show the steps necessary to make 2-methyl-4-nonene using a Wittig reaction. Start with triphenylphosphine and an alkyl halide. After that you may use any other organic or inorganic reagents. 2. Write in the product of this reaction: CH3 CH₂ (C6H5)₂CuLi H₂O+

3. Name this compound properly, including stereochemistry. H₂C H3C CH3 OH 4. Show the step(s) necessary to transform the compound on the left into the acid on the right. Bri CH2 5. Write in the product of this LiAlH4 Br H₂C OH

Answer 2

Question

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

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Answer 3

Question

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

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Answer 4

Question

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

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Answer 5

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

Answer 6

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

Answer 7

Chapter 9, Problem 1KSP

Interpretation Introduction

Interpretation:

The molecular geometry of PBr3 is to be determined.

Concept introduction: In order to find the geometry of a molecule, follow the steps:

First, draw the Lewis dot structure of the molecule.

Find the number of electron pairs or domains around the central atom. Electron domain is the bond and lone pair of electrons around the central atom.

Use the VSEPR model to find the electron domain geometry.

According to VSEPR, the electron domains repel each other and arrange themselves as apart as possible in space.

Molecular geometry is the arrangement of bonded atoms. If no lone pairs are present, the molecular geometry is the same as the electron domain geometry.

The repulsion between two lone pairs is greater than that in case of lone pair–bond pair, which, in turn, is greater than the bond pair–bond pair repulsion.

Answer 8

Expert Solution & Answer

Answer to Problem 1KSP

Correct answer: Option (c)

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Reason for the correct option:

Lewis dot structure of PBr3 is shown below:

EBK CHEMISTRY, Chapter 9, Problem 1KSP

In this molecule, the phosphorus atom is bonded to three bromine atoms by single bonds and there is one lone pair on the phosphorus atom.

Thus, the electron domains around the phosphorus atom are four.

According to the VSEPR model, the four electron domains repel each other and arrange themselves in space in a tetrahedral shape.

However, in this case, one lone pair is present on the phosphorus atom.

Thus, the molecular geometry is trigonal pyramidal.

Hence, option (c) is correct.

Reasons for the incorrect options:

Option (a) is incorrect because a trigonal planar shape is obtained when there is no lone pair on the central atom.

Option (b) is incorrect because a tetrahedral shape is obtained when four bonds are present.

Option (d) is incorrect because a bent shape is obtained when there are two bonds present.

Option (e) is incorrect because a T-shape is obtained from the trigonal bipyramidal geometry with two lone pairs.

Hence, options (a), (b), (d), and (e) are incorrect.

Answer 12

Ch. 9.1 - Practice Problem ATTEMPT Determine the shapes of...Ch. 9.1 - Prob. 1PPB Ch. 9.1 - Prob. 1PPC Ch. 9.1 - 9.1.1 What are the electron-domain geometry and...Ch. 9.1 - What are the electron-domain geometry and...Ch. 9.1 - Prob. 3CP Ch. 9.1 - Prob. 4CP Ch. 9.1 - Prob. 5CP Ch. 9.2 - Practice Problem ATTEMPT Ethanolamine has a...Ch. 9.2 - Practice Problem BUILD The bond angle in is...

Answer to Problem 1KSP

Explanation of Solution

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