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 2B.24E

(a)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 1

Concept Introduction:

The formal charge on each atom in the Lewis structure can be calculated from the equation written as follows:

    Formal charge=[V−L−12B]

Here,

V denotes valence electrons in free atom.

L denotes electrons present as lone pairs.

B denotes electrons present as bond pairs.

Lowest energy structure is the one that has zero or nearly zero formal charge.

(a)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

  Formal charge=[V−L−12B]        (1)

For structure 1:

Substitute 5 for V, 4 for L and 4  for B in equation (1) to calculate formal charge on both similar nitrogen atoms.

    Formal charge(N)=5−4−12(4)=−1

Substitute 4 for V, 0 for L and 8 for B in equation (1).

    Formal charge(C)=4−0−12(8)=0

Similarly for other Lewis structure, substitute 5 for V, 2 for L and 6  for B in equation (1) to calculate formal charge on triply bonded nitrogen.

    Formal charge(N)=5−2−12(6)=0

Substitute 5 for V, 6 for L and 2 for B in equation (1) to calculate formal charge on singly bonded nitrogen.

    Formal charge(N)=5−6−12(2)=−2

Substitute 4 for V, 0 for L and 8 for B in equation (1).

    Formal charge(C)=4−0−12(8)=0

Hence formal charge on each structure is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 2

Since the former has less formal charges on maximum number of atoms so I is lower in energy than II therefore I represents structure of lowest energy.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 3

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    Formal charge=[V−L−12B]        (1)

For structure I:

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the four singly bonded O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 5 for V, 0 for L and 8  for B in equation (1) to calculate formal charge on central arsenic .

    Formal charge(As)=5−0−12(8)=1

For structure II:

Similarly for second Lewis structure, substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the three singly bonded O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 5 for V, 0 for L and 8  for B in equation (1) to calculate formal charge on central arsenic .

    Formal charge(As)=5−0−12(10)=0

For structure III:

Similarly for third Lewis structure, substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the four doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 5 for V, 0 for L and 8  for B in equation (1) to calculate formal charge on central arsenic .

    Formal charge(As)=5−0−12(16)=−3

Hence formal charge in the two structures is illustrated below.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 4

Since the II has a −1 on electronegative oxygen while zero formal charges on central arsenic, it is regarded as most stable and hence II represents structure of lower energy.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 5

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    Formal charge=[V−L−12B]        (1)

For Lewis structureI, substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the three singly bonded O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 7 for V, 0 for L and 14  for B in equation (1) to calculate formal charge on central iodine .

    Formal charge(I)=7−0−12(14)=0

For Lewis structure II,

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the three singly bonded O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 7 for V, 0 for L and 12 for B in equation (1) to calculate formal charge on central iodine .

    Formal charge(I)=7−0−12(12)=+1 \

For Lewis structure III, substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on each of the three singly bonded O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 7 for V, 0 for L and 14  for B in equation (1) to calculate formal charge on central iodine .

    Formal charge(I)=7−0−12(8)=+3

Hence formal charge in the three structures is illustrated below:

Chemical Principles: The Quest for Insight, Chapter 2, Problem 2B.24E , additional homework tip 6

Since I has zero formal charges on central electronegative iodine atom compared to other cases with +1 and +3 formal charges, I regarded as most stable and hence I represent structure of lower energy.

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Chapter 2 Solutions

Chemical Principles: The Quest for Insight

Ch. 2 - Prob. 2A.1ASTCh. 2 - Prob. 2A.1BSTCh. 2 - Prob. 2A.2ASTCh. 2 - Prob. 2A.2BSTCh. 2 - Prob. 2A.3ASTCh. 2 - Prob. 2A.3BSTCh. 2 - Prob. 2A.4ASTCh. 2 - Prob. 2A.4BSTCh. 2 - Prob. 2A.1ECh. 2 - Prob. 2A.2E
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 - 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