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.15E

(a)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for CF3Cl molecule 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 the molecules that have no lone pairs around the central atom the bonded-atom unshared -pair arrangement is decided by the table as follows:

Number ofelectron pairsMolecular shape2Linear3Trigonal Planar4Tetrahedral5Trigonal Bipyramidal6Octahedral7Pentagonal Bipyramidal

In order to determine the shape the steps to be followed are indicated as follows:

  1. 1. Lewis structure of molecule should be written.
  2. 2. The type electron arrangement around the central atom should be identified around the central atom. This essentially refers to determination of bond pairs and unshared or lone pairs around central atoms.
  3. 3. Then bonded-atom unshared -pair arrangement that can maximize the distance of electron pairs about central atom determines the shape.

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 –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.15E

The shape for CF3Cl molecule is tetrahedral, bond angle is 109.5 ° and corresponding VSEPR formula is AX4.

Explanation of Solution

CF3Cl has C as central atom. I possesses 7 valence electrons, carbon has four and F has 7 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in CF3Cl calculated as follows:

  Total valence electrons=4+7(3)+7=32

The skeleton structure in CF3Cl has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=328=24

These 12 electron pairs are allotted as lone pairs to satisfy respective octets. Hence, the Lewis structure in CF3Cl is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.15E , additional homework tip  1

It is evident that in CF3Cl the central carbon atom has four bond pairs and zero lone pair. to have minimum repulsions CF3Cl adopts tetrahedral accordance with VSPER model and thus the bond angle is 109.5 °.

If central atom is represented by A, and other attached bond pairs by X, then for any tetrahedral species with no lone pairs the VSEPR formula is predicted as AX4.

(b)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for TeCl4 molecule have to be predicted.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 2E.15E

The shape for TeCl4 molecule is see-saw, bond angles are 90 ° and 120 ° and corresponding VSEPR formula is AX4E.

Explanation of Solution

TeCl4 has Te as central atom. Te possesses 6 valence electrons and Cl have 7 electrons respectively.

Total valence electrons are sum of the valence electrons on atom in TeCl4 calculated as follows:

  Total valence electrons=6+7(4)=34

The skeleton structure in TeCl4 has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=348=26

These 13 electron pairs are allotted as lone pairs to satisfy respective octets. Hence, the Lewis structure in TeCl4 along with corresponding VSEPR geometry and bond angle is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.15E , additional homework tip  2

It is evident that TeCl4 has four bond pairs and one lone pairs. Thus with total 5 pairs around central Te atom, it is predicted to correspond to trigonal bipyramidal arrangement.

One lone pair is localized on equatorial positions so as to minimize lone pair–bond pair repulsions in accordance with VSPER model. This leads see-saw shape for TeCl4. Clearly, the bond angles for such a molecular shape is 90 ° and 120 ° respectively since the trigonal plane will be at right angles to the equatorial plane that has each atom oriented at 120 °.

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 to be AX4E .

(c)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for COF2 molecule have to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 2E.15E

The shape for COF2 molecule is trigonal planar, bond angle is 120 ° and corresponding VSEPR formula is AX3.

Explanation of Solution

COF2 has C as central atom. C has 4 valence electrons, oxygen possesses 6 valence electrons and F has 7 valence electrons.

Total valence electrons are sum of the valence electrons on atom in COF2 calculated as follows:

  Total valence electrons=4+6+7(2)=24(12 pairs)

The skeleton structure in COF2 has three bond pairs that comprise 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=246=18

These 9 electron pairs are allotted as lone pairs or as multiple bonds to oxygen so as to satisfy octets. Hence, the Lewis structure in COF2 along with corresponding VSEPR geometry, bond angle is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.15E , additional homework tip  3

It is evident that in COF2 the central carbon atom has three bond pairs and zero lone pairs. Such four electron pairs correspond to trigonal planar arrangement so as to have minimum repulsions in accordance with VSPER model. This results in trigonal planar shape for COF2 each oriented at 120 °.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any trigonal planar species the VSEPR formula is predicted as AX3 .

(d)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for CH3 molecule have to be predicted.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 2E.15E

The shape for CH3 is trigonal pyramidal, bond angle is approximately less than 109.5 ° and corresponding VSEPR formula is AX3E.

Explanation of Solution

CH3 has C as central atom that has 4 valence electrons.

Total valence electrons are sum of the valence electrons on atom in CH3 along with the uni negative-charge calculated as follows:

  Total valence electrons=4+1(3)+1=8

Thus, Lewis structure in CH3 has three bond pairs to each of the three H that leaves one pair to be assigned as lone pair. Hence, the Lewis structure in CH3 along with VSEPR geometry and corresponding bond angle is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.15E , additional homework tip  4

It is evident that in CH3 the central carbon atom has three bond pairs and one lone pair. Such four electron pairs correspond to tetrahedral arrangement.

Lone pair tend to be localized on apical position and so 109.5 ° bond angle reduces slightly from usual tetrahedral bond angle so as to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape for CH3.

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

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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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