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.13E

(a)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for I3 molecule 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 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.13E

The shape for I3 molecule is linear, corresponding VSEPR formula is X2E3 and bond angle is 180°.

Explanation of Solution

I3 has I as central atom. I possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in I3 calculated as follows:

  Total valence electrons=7(3)+1=22 (11 pairs)

The skeleton structure in I3 has two bond pairs that comprise 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=224=18

These 9 electron pairs are allotted as lone pairs of each of the iodine atoms to satisfy their octet. Hence, the Lewis structure in I3 is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.13E , additional homework tip  1

It is evident that in I3, the central iodine atom has two bond pairs and three lone pairs. Lone pairs tend to occupy the equatorial locations of trigonal plane in trigonal bi-pyramidal arrangement so as to have minimum repulsions in accordance with VSPER model. This results in linear I3 molecule. The shape is linear so bond angle is 180°.

If lone pairs are represented by E, and other attached bond pairs by X, then for any linear species the VSEPR formula is predicted as X2E3.

(b)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for molecule has to be predicted.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 2E.13E

The shape for POCl3 is tetrahedral, corresponding VSEPR formula is AX4 and bond angle is 109.5°.

Explanation of Solution

POCl3 has P as central atom. P possesses five valence electrons and O and Cl have 6 and 7 electrons respectively.

Total valence electrons are sum of the valence electrons on atom in POCl3 calculated as follows:

  Total valence electrons=5(1)+6(1)+7(3)=32 (16 pairs)

The skeleton structure in POCl3 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 or multiple bonds to satisfy respective octets. Hence, the Lewis structure in POCl3 and corresponding VSEPR geometry and bond angle is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.13E , additional homework tip  2

It is evident that in POCl3 the central phosphorus atom has four bond pairs because the double bond is also regarded as single unit. To minimize repulsion all bond pairs can easily be accommodated in a tetrahedral shape in accordance with VSPER model.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any tetrahedral species the VSEPR formula is predicted to be AX4.

(c)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for IO3 molecule has to be predicted.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 2E.13E

The shape for IO3 is trigonal pyramidal, corresponding VSEPR formula is AX4E2 and bond angle is less than 109.5°.

Explanation of Solution

IO3 has I as central atom. I has seven valence electrons and oxygen possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on atom in IO3 calculated as follows:

  Total valence electrons=7(1)+6(3)+1=26 (13 pairs)

The skeleton structure in IO3 has three bond pairs that comprise 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 to each of the oxygen to satisfy its octet. Hence, the Lewis structure in IO3 along with corresponding VSEPR geometry is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.13E , additional homework tip  3

It is evident that in IO3 the central iodine atom has three bond pairs and one lone pairs. Such four electron pairs correspond to tetrahedral arrangement.

Lone pairs tend to be localized on apical position while 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 IO3.

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.

(d)

Interpretation Introduction

Interpretation:

The VSEPR formula, shape for the Lewis structure of N2O molecule along with bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 2E.13E

The shape for N2O molecule is linear with bond angle as 180° and corresponding VSEPR formula is X2E3.

Explanation of Solution

N2O has N as central atom that possesses 5 valence electrons. Oxygen possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in N2O calculated as follows:

  Total valence electrons=5(2)+6=16

The skeleton structure in N2O has two bond pairs that comprise 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  Remaining electrons=164=12

These 6 electron pairs are allotted as lone pairs or double bonds on each of the atom in N2O to satisfy the octet. Hence, the Lewis structure in N2O and corresponding VSPER geometry is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2E.13E , additional homework tip  4

It is evident that in N2O, the central iodine atom has two bond pairs (since each triple or double bond is treated as one super pair). This results in linear N2O in accordance with VSPER model and consequently, bond angle must be 180°.

If central atom is represented by A, and other attached bond pairs by X, then for any linear species the VSEPR formula is predicted as AX2.

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Chapter 2 Solutions

Chemical Principles: The Quest for Insight

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