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.14E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of SiCl4 has to be drawn and VSEPR formula, molecular shape, and bond angle have 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.14E

The shape for SiCl4 molecule is tetrahedral, VSEPR formula is AX4 and corresponding bond angle is 109.5 °.

Explanation of Solution

SiCl4 has Si as central atom. Si has four valence electrons while Cl possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each atom in SiCl4 calculated as follows:

  Total valence electrons=4+7(4)=32

The skeleton structure in SiCl4 has four bond pairs that comprise of 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=328=24

These 12 electron pairs are assigned as lone pairs of each of the Cl atoms to satisfy its octet. Hence, the Lewis structure of SiCl4 along with VSEPR molecular geometry and corresponding bond angle is illustrated below.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.14E , additional homework tip  1

If lone pairs are represented by E, central atom with A and other attached bon pairs by X, then for any see-saw species the VSEPR formula is predicted as AX4E.

It is evident that in SiCl4,the central silicon atom has four bond pairs and no lone pairs. Thus four bonds are separated at an angle of 109.5 ° so as to have minimum bond pair- bond pair repulsions in accordance with VSPER model. This results in tetrahedral shape in SiCl4 molecule.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of PF5 has to be drawn and VSEPR formula, molecular shape, and bond angle have to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 2E.14E

The shape for PF5 is trigonal pyramidal, corresponding VSEPR formula is AX5 and the bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

PF5 has P as central atom. P has five valence electrons while F possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each atom in PF5 calculated as follows:

  Total valence electrons=5+7(5)=40

The skeleton structure in PF5 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 F atoms to satisfy respective octets. Hence, the Lewis structure PF5 that has shape corresponding trigonal bi-pyramidal arrangement is illustrated below.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.14E , additional homework tip  2

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 PF5, the central phosphorus atom has five bond pairs and no lone pairs. to have minimum repulsions three fluorine form bonds in trigonal equatorial plane while two other fluorine atoms form bonds in axial plane to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape in PF5 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.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of SBr2 has to be drawn and VSEPR formula, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 2E.14E

The shape for SBr2 molecule is angular or bent, VSEPR formula is AX2E and corresponding bond angle is less than 105 °.

Explanation of Solution

SBr2 has S as central atom. S has six valence electrons and Br also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in SBr2 calculated as follows:

  Total valence electrons=6+7(2)=20

The skeleton structure in SBr2 has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=204=16

These 8 electron pairs are allotted as lone pairs to each of the bromine atoms and central sulfur to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in SBr2 is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.14E , additional homework tip  3

It is evident that in SBr2, the central sulfur atom has two bond pairs and two lone pairs.  This corresponds to tetrahedral arrangement. However, as the two localized lone pairs tend to occupy maximum space the bond angle reduces slightly from usual tetrahedral bond angle to nearly 105 ° so as to have minimum repulsions in accordance with VSPER model. This results in angular or bent shaped for SBr2.

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 AX2E.

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of ICl2+ has to be drawn and VSEPR formula, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 2E.14E

The shape for ICl2+ molecule is linear, VSEPR formula is AX2E3 and corresponding bond angles are 120 ° , 180 ° and 90 °.

Explanation of Solution

ICl2+ has I as central atom. I has seven valence electrons and chlorine also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in ICl2+ calculated as follows:

  Total valence electrons=7+7(2)1=20

The skeleton structure in ICl2+ has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=204=16

These 8 electron pairs are allotted as lone pairs of each of the chlorine atoms and central iodine to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ICl2+ is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.14E , additional homework tip  4

It is evident that in ICl2+ the central iodine atom has two bond pairs to two chlorine atoms and three 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 AX2E3.

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 two chlorine atoms take up axial position so that they are 180 ° apart from each other and are at right angles to equatorial lone pairs. This results in linear shape for ICl2+.

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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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