Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 2, Problem 2E.20E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of PBr5 has to be drawn, number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Valence Shell Electron Pair Repulsion model predicts shape by inclusion of bond angles and most distant arrangement of atoms that leads to minimum repulsion.

For molecules that have lone pairs around central atom, lone pairs influence shape, because there are no atoms at the positions occupied by these lone pairs. The key rule that governs the molecular shape, in this case, is the extent of lone pair–lone pair repulsions are far greater than lone bond pair or bond pair-bond pair repulsions. The table that summarized the molecular shapes possible for various combinations of bonded and lone pairs are given as follows:

Steric numberNumber of lone pairsMolecular geometryBond angles20Linear180 °301Trigonal planarBent120 °4012TetrahedralTrigonal pyramidalBent109.5 °50123Trigonal Bi-pyramidalSee-SawT-shapedLinear90 °,120 °,180 °6012OctahedralSquare pyramidalSquare planar90 °,180 °

(a)

Expert Solution
Check Mark

Answer to Problem 2E.20E

The shape for PBr5 is trigonal pyramidal, corresponding VSEPR formula is AX5 and the bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

PBr5 has P as central atom. P has five valence electrons while Br possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each atom in PBr5 calculated as follows:

  Total valence electrons=5+7(5)=40

The skeleton structure in PBr5 has five bond pairs that comprise 10 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=4010=30

These 15 electron pairs are assigned as lone pairs of each of the Br atoms to satisfy respective octets.

Hence, the Lewis structure PBr5 that has shape corresponding trigonal bi-pyramidal arrangement is illustrated below.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.20E , additional homework tip  1

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any trigonal pyramidal geometry the VSEPR formula is predicted as AX5.

It is evident that in PBr5, the central phosphorus atom has five bond pairs and no lone pairs. To have minimum repulsions three bromine form bond in trigonal equatorial plane while two other bromine atoms form bonds in axial plane to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape in PBr5 molecule.

The bond angles are 120 ° in the trigonal equatorial plane and 180 ° for axial fluorine. The equatorial plane is at 90 ° to axial plane.

(b)

Interpretation Introduction

Interpretation:

The Lewis structure of XeOF2 has to be drawn, number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 2E.20E

The shape for XeOF2 molecule T-shaped, number of lone pairs on central xenon is two and bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

XeOF2 has Xe as central atom. Xe has eight valence electrons and XeOF2 fluorine possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in XeOF2 calculated as follows:

  Total valence electrons=8+6+7(2)=28

The skeleton structure in XeOF2 has three bond pairs. This comprises of 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=286=22

These 11 electron pairs are allotted as lone pairs of each of the fluorine, oxygen atoms and central xenon to satisfy respective octets. Thus, the Lewis structure and corresponding VSEPR geometry XeOF2 is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.20E , additional homework tip  2

It is evident that in XeOF2, the central xenon atom has three bond pairs and two lone pair. Lone pairs tend to occupy the two equatorial positions so as to stay at 120 ° apart from each other and at right angles to axial fluorine are 180 ° apart from each other in accordance with VSPER model.  Thus bond angles that correspond to T-shaped XeOF2 are 120 ° , 180 ° and 90 °.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of SF5+ has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 2E.20E

The shape for SF5+ molecule trigonal pyramidal, number of lone pairs on central sulphur is zero and bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

SF5+ has S as central atom. S has six valence electrons and F possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in SF5+ calculated as follows:

  Total valence electrons=6+7(5)1=40

The skeleton structure in SF5+ has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=4010=30

These 15 electron pairs are allotted as lone pairs to each of the F atoms to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in SF5+ is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.20E , additional homework tip  3

It is evident that in SF5+, the central sulfur atom has five bond pairs and zero lone pairs.  This corresponds to trigonal bi-pyramidal arrangement. The equatorial bonds locations in trigonal plane are 120 ° apart so as to have minimum repulsions in accordance with VSPER model while the axial are 180 ° apart from each other and are at right angles to equatorial bonds.

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of IF3 has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 2E.20E

The shape for IF3 molecule T-shape, number of lone pairs on central iodine is two pairs and corresponding bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

IF3 has I as central atom. I has seven valence electrons and F also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each F and central iodine in IF3 calculated as follows:

  Total valence electrons=7+7(3)=28

The skeleton structure in IF3 has three bond pairs that comprise 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=286=22

These 11 electron pairs are allotted as lone pairs of each of the fluorine atoms and central iodine to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in IF3 is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.20E , additional homework tip  4

It is evident that in IF3 the central iodine atom has three bond pairs to three chlorine atoms and two lone pairs. If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any see-saw species the VSEPR formula is predicted as AX3E2.

Lone pairs tend to occupy the equatorial locations of trigonal plane so that they are 120 ° apart to have minimum repulsions in accordance with VSPER model while the three chlorine atoms take up two axial positions and one equatorial position so that they are 180 ° apart from each other and are at right angles to equatorial lone pairs. This result in linear shape for IF3.

(e)

Interpretation Introduction

Interpretation:

The Lewis structure of BrO3 has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution
Check Mark

Answer to Problem 2E.20E

The shape for BrO3 molecule trigonal pyramidal, number of lone pairs on central bromine is one and corresponding bond angles are 107 °.

Explanation of Solution

BrO3 has Br as central atom. Br has seven valence electrons and O possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in BrO3 along with uni-negative charge calculated as follows:

  Total valence electrons=7+6(3)+1=26

The skeleton structure in BrO3 has three bond pairs that comprises 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=266=20

These 10 electron pairs are allotted as lone pairs or multiple bonds to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in BrO3 is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.20E , additional homework tip  5

It is evident that in BrO3 the central bromine atom has three bond pairs to three oxygen atoms as the double bond is also regarded as single bond unit. No of lone pair on central bromine is one.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any see-saw species the VSEPR formula is predicted as AX3E.

The bond pairs in BrO3 are less than 109.5° apart to have minimum repulsions in accordance with VSPER model. The three oxygen atoms take up triangular plane in base. This results in trigonal pyramidal for BrO3.

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Chapter 2 Solutions

Chemical Principles: The Quest for Insight

Ch. 2 - Prob. 2A.3ECh. 2 - Prob. 2A.4ECh. 2 - Prob. 2A.5ECh. 2 - Prob. 2A.6ECh. 2 - Prob. 2A.7ECh. 2 - Prob. 2A.8ECh. 2 - Prob. 2A.9ECh. 2 - Prob. 2A.10ECh. 2 - Prob. 2A.11ECh. 2 - Prob. 2A.12ECh. 2 - Prob. 2A.13ECh. 2 - Prob. 2A.14ECh. 2 - Prob. 2A.15ECh. 2 - Prob. 2A.16ECh. 2 - Prob. 2A.17ECh. 2 - Prob. 2A.18ECh. 2 - Prob. 2A.19ECh. 2 - Prob. 2A.20ECh. 2 - Prob. 2A.21ECh. 2 - Prob. 2A.22ECh. 2 - Prob. 2A.23ECh. 2 - Prob. 2A.24ECh. 2 - Prob. 2A.25ECh. 2 - Prob. 2A.26ECh. 2 - Prob. 2A.27ECh. 2 - Prob. 2A.28ECh. 2 - Prob. 2A.29ECh. 2 - Prob. 2A.30ECh. 2 - Prob. 2B.1ASTCh. 2 - Prob. 2B.1BSTCh. 2 - Prob. 2B.2ASTCh. 2 - Prob. 2B.2BSTCh. 2 - Prob. 2B.3ASTCh. 2 - Prob. 2B.3BSTCh. 2 - Prob. 2B.4ASTCh. 2 - Prob. 2B.4BSTCh. 2 - Prob. 2B.5ASTCh. 2 - Prob. 2B.5BSTCh. 2 - Prob. 2B.1ECh. 2 - Prob. 2B.2ECh. 2 - Prob. 2B.3ECh. 2 - Prob. 2B.4ECh. 2 - Prob. 2B.5ECh. 2 - Prob. 2B.6ECh. 2 - Prob. 2B.7ECh. 2 - Prob. 2B.8ECh. 2 - Prob. 2B.9ECh. 2 - Prob. 2B.10ECh. 2 - Prob. 2B.11ECh. 2 - Prob. 2B.12ECh. 2 - Prob. 2B.13ECh. 2 - Prob. 2B.14ECh. 2 - Prob. 2B.15ECh. 2 - Prob. 2B.16ECh. 2 - Prob. 2B.17ECh. 2 - Prob. 2B.18ECh. 2 - Prob. 2B.19ECh. 2 - Prob. 2B.20ECh. 2 - Prob. 2B.21ECh. 2 - Prob. 2B.22ECh. 2 - Prob. 2B.23ECh. 2 - Prob. 2B.24ECh. 2 - Prob. 2C.1ASTCh. 2 - Prob. 2C.1BSTCh. 2 - Prob. 2C.2ASTCh. 2 - Prob. 2C.2BSTCh. 2 - Prob. 2C.3ASTCh. 2 - Prob. 2C.3BSTCh. 2 - Prob. 2C.1ECh. 2 - Prob. 2C.2ECh. 2 - Prob. 2C.3ECh. 2 - Prob. 2C.4ECh. 2 - Prob. 2C.5ECh. 2 - Prob. 2C.6ECh. 2 - Prob. 2C.7ECh. 2 - Prob. 2C.8ECh. 2 - Prob. 2C.9ECh. 2 - Prob. 2C.10ECh. 2 - Prob. 2C.11ECh. 2 - Prob. 2C.12ECh. 2 - Prob. 2C.13ECh. 2 - Prob. 2C.14ECh. 2 - Prob. 2C.15ECh. 2 - Prob. 2C.16ECh. 2 - Prob. 2C.17ECh. 2 - Prob. 2C.18ECh. 2 - Prob. 2D.1ASTCh. 2 - Prob. 2D.1BSTCh. 2 - Prob. 2D.2ASTCh. 2 - Prob. 2D.2BSTCh. 2 - Prob. 2D.1ECh. 2 - Prob. 2D.2ECh. 2 - Prob. 2D.3ECh. 2 - Prob. 2D.4ECh. 2 - Prob. 2D.5ECh. 2 - Prob. 2D.6ECh. 2 - Prob. 2D.7ECh. 2 - Prob. 2D.8ECh. 2 - Prob. 2D.9ECh. 2 - Prob. 2D.10ECh. 2 - Prob. 2D.11ECh. 2 - Prob. 2D.12ECh. 2 - Prob. 2D.13ECh. 2 - Prob. 2D.14ECh. 2 - Prob. 2D.15ECh. 2 - Prob. 2D.16ECh. 2 - Prob. 2D.17ECh. 2 - Prob. 2D.18ECh. 2 - Prob. 2D.19ECh. 2 - Prob. 2D.20ECh. 2 - Prob. 2E.1ASTCh. 2 - Prob. 2E.1BSTCh. 2 - Prob. 2E.2ASTCh. 2 - Prob. 2E.2BSTCh. 2 - Prob. 2E.3ASTCh. 2 - Prob. 2E.3BSTCh. 2 - Prob. 2E.4ASTCh. 2 - Prob. 2E.4BSTCh. 2 - Prob. 2E.5ASTCh. 2 - Prob. 2E.5BSTCh. 2 - Prob. 2E.1ECh. 2 - Prob. 2E.2ECh. 2 - Prob. 2E.3ECh. 2 - Prob. 2E.4ECh. 2 - Prob. 2E.5ECh. 2 - Prob. 2E.6ECh. 2 - Prob. 2E.7ECh. 2 - Prob. 2E.8ECh. 2 - Prob. 2E.9ECh. 2 - Prob. 2E.10ECh. 2 - Prob. 2E.11ECh. 2 - Prob. 2E.12ECh. 2 - Prob. 2E.13ECh. 2 - Prob. 2E.14ECh. 2 - Prob. 2E.15ECh. 2 - Prob. 2E.16ECh. 2 - Prob. 2E.17ECh. 2 - Prob. 2E.18ECh. 2 - Prob. 2E.19ECh. 2 - Prob. 2E.20ECh. 2 - Prob. 2E.21ECh. 2 - Prob. 2E.22ECh. 2 - Prob. 2E.23ECh. 2 - Prob. 2E.24ECh. 2 - Prob. 2E.25ECh. 2 - Prob. 2E.26ECh. 2 - Prob. 2E.27ECh. 2 - Prob. 2E.28ECh. 2 - Prob. 2E.29ECh. 2 - Prob. 2E.30ECh. 2 - Prob. 2F.1ASTCh. 2 - Prob. 2F.1BSTCh. 2 - Prob. 2F.2ASTCh. 2 - Prob. 2F.2BSTCh. 2 - Prob. 2F.3ASTCh. 2 - Prob. 2F.3BSTCh. 2 - Prob. 2F.4ASTCh. 2 - Prob. 2F.4BSTCh. 2 - Prob. 2F.1ECh. 2 - Prob. 2F.2ECh. 2 - Prob. 2F.3ECh. 2 - Prob. 2F.4ECh. 2 - Prob. 2F.5ECh. 2 - Prob. 2F.6ECh. 2 - Prob. 2F.7ECh. 2 - Prob. 2F.8ECh. 2 - Prob. 2F.9ECh. 2 - Prob. 2F.10ECh. 2 - Prob. 2F.11ECh. 2 - Prob. 2F.12ECh. 2 - Prob. 2F.13ECh. 2 - Prob. 2F.14ECh. 2 - Prob. 2F.15ECh. 2 - Prob. 2F.16ECh. 2 - Prob. 2F.17ECh. 2 - Prob. 2F.18ECh. 2 - Prob. 2F.19ECh. 2 - Prob. 2F.20ECh. 2 - Prob. 2F.21ECh. 2 - Prob. 2G.1ASTCh. 2 - Prob. 2G.1BSTCh. 2 - Prob. 2G.2ASTCh. 2 - Prob. 2G.2BSTCh. 2 - Prob. 2G.1ECh. 2 - Prob. 2G.2ECh. 2 - Prob. 2G.3ECh. 2 - Prob. 2G.4ECh. 2 - Prob. 2G.5ECh. 2 - Prob. 2G.6ECh. 2 - Prob. 2G.7ECh. 2 - Prob. 2G.8ECh. 2 - Prob. 2G.9ECh. 2 - Prob. 2G.11ECh. 2 - Prob. 2G.12ECh. 2 - Prob. 2G.13ECh. 2 - Prob. 2G.14ECh. 2 - Prob. 2G.15ECh. 2 - Prob. 2G.16ECh. 2 - Prob. 2G.17ECh. 2 - Prob. 2G.18ECh. 2 - Prob. 2G.19ECh. 2 - Prob. 2G.20ECh. 2 - Prob. 2G.21ECh. 2 - Prob. 2G.22ECh. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 2.4ECh. 2 - Prob. 2.5ECh. 2 - Prob. 2.6ECh. 2 - Prob. 2.7ECh. 2 - Prob. 2.8ECh. 2 - Prob. 2.9ECh. 2 - Prob. 2.10ECh. 2 - Prob. 2.11ECh. 2 - Prob. 2.12ECh. 2 - Prob. 2.13ECh. 2 - Prob. 2.14ECh. 2 - Prob. 2.17ECh. 2 - Prob. 2.19ECh. 2 - Prob. 2.22ECh. 2 - Prob. 2.23ECh. 2 - Prob. 2.24ECh. 2 - Prob. 2.25ECh. 2 - Prob. 2.26ECh. 2 - Prob. 2.27ECh. 2 - Prob. 2.28ECh. 2 - Prob. 2.29ECh. 2 - Prob. 2.30ECh. 2 - Prob. 2.31ECh. 2 - Prob. 2.32ECh. 2 - Prob. 2.33ECh. 2 - Prob. 2.34ECh. 2 - Prob. 2.35ECh. 2 - Prob. 2.36ECh. 2 - Prob. 2.37ECh. 2 - Prob. 2.39ECh. 2 - Prob. 2.40ECh. 2 - Prob. 2.41ECh. 2 - Prob. 2.42ECh. 2 - Prob. 2.43ECh. 2 - Prob. 2.44ECh. 2 - Prob. 2.45ECh. 2 - Prob. 2.46ECh. 2 - Prob. 2.47ECh. 2 - Prob. 2.48ECh. 2 - Prob. 2.49ECh. 2 - Prob. 2.50ECh. 2 - Prob. 2.51ECh. 2 - Prob. 2.52ECh. 2 - Prob. 2.53ECh. 2 - Prob. 2.54ECh. 2 - Prob. 2.55ECh. 2 - Prob. 2.56ECh. 2 - Prob. 2.57ECh. 2 - Prob. 2.58ECh. 2 - Prob. 2.59ECh. 2 - Prob. 2.60ECh. 2 - Prob. 2.61ECh. 2 - Prob. 2.62ECh. 2 - Prob. 2.63ECh. 2 - Prob. 2.64E
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