CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
7th Edition
ISBN: 9781319420994
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 2, Problem 2C.4E

(a)

Interpretation Introduction

Interpretation:

Lewis structure of formate ion (HCO2)− has to be drawn.

Concept Introduction:

Lewis structure represents covalent bonds and describes valence electrons configuration of atoms. The covalent bonds are depicted by lines and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

  • The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
  • Total valence electrons are estimated for each atom.
  • single bond is first placed between each atom pair.
  • The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
  • Add charge on the overall structure in case of polyatomic cation or anion.

(a)

Expert Solution
Check Mark

Explanation of Solution

The molecule (HCO2)− consists of one H atom, one C atom and two O atoms.

The symbol for carbon is C with atomic number 6. Its electronic configuration is [He]2s22p2. Thus, it possesses 4 valence electrons.

The symbol for hydrogen is H with atomic number 1. Its electronic configuration is 1s1 Thus, it possesses 1 valence electron.

The symbol for oxygen is O; with atomic number 8. Its electronic configuration is [He]2s22p4. Thus, it possesses 6 valence electrons.

One negative charge on molecule is added up as one valence electron in the total count.

Thus total valence electrons are sum of the valence electrons for each atom in (HCO2)− It is calculated as follows:

  Total valence electrons=4+6(2)+1+1=18(9 pairs)

The skeleton structure (HCO2)− has three bonds that comprise 6 electrons. The electrons left to be allocated are determined as follows:

  Remaining electrons=18−6=12(6 pairs)

To complete the valence electrons of carbon, it forms a double bond with one oxygen atom.

  Total remaining electrons=12−2=10(5 pairs)

Hence, 10 electrons are allocated as 3 lone pairs on singly bonded oxygen atom and two lone pairs on doubly bonded oxygen atom. The Lewis structure of (HCO2)− is as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 1

The negative charge on molecule is delocalized on each oxygen atom and since there are two oxygen atoms, therefore, two resonance structures are formed and possible resonance structures are as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 2

(b)

Interpretation Introduction

Interpretation:

Lewis structure of hydrogen phosphite ion (HPO32−) has to be drawn.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The molecule (HPO32−) consists of one H and three O atoms

The symbol for oxygen is O; with atomic number 8. Its electronic configuration is [He]2s22p4. Thus, it possesses 6 valence electrons.

The symbol for hydrogen is H; with atomic number 1. Its electronic configuration is 1s1. Thus it possesses 1 valence electron.

The symbol for phosphorus is P; with atomic number 15. Its electronic configuration is [Ne]3s23p3. Thus, it possesses 5 valence electrons.

Two negative charges on molecule are added up as two valence electrons in the total count.

Thus total valence electrons are sum of the valence electrons for each atom in (HPO32−). It is calculated as follows:

  Total valence electrons=1+6(3)+5+2=26(13 pairs)

The skeleton structure (HPO32−) has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  Remaining electrons=26−8=18(9 pairs)

Hence, 18 electrons are allocated as 3 lone pairs on two singly bonded oxygen atoms and 2 lone pairs on doubly bonded oxygen atom to complete their octet. The Lewis structure is as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 3

The negative charge on molecule is delocalized on three oxygen atoms therefore three resonance structures will be formed and possible resonance structures are as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 4

(c)

Interpretation Introduction

Interpretation:

Lewis structure of bromate ion (BrO3)− has to be drawn:

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

The molecule (BrO3)− consists of one Br atom and three O atoms.

The symbol for oxygen is O; with atomic number 8. Its electronic configuration is [He]2s22p4. Thus, it possesses 6 valence electrons.

The symbol for bromine is Br; with atomic number 35. Its electronic configuration is [Ar]3d104s24p5. Thus it possesses 7 valence electrons.

One negative charge on molecule is added up to the total valence count.

Thus total valence electrons are sum of the valence electrons for each atom in (BrO3)−. It is calculated as follows:

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

The skeleton structure (BrO3)− has three bonds that comprise 6 electrons. The electrons left to be allocated are determined as follows:

  Remaining electrons=26−6=20(10 pairs)

To complete the valence electrons of bromine it forms two double bonds with two oxygen atoms.

  Total remaining electrons=20−4=16(8 pairs)

Hence, 16 electrons are allocated as 3 lone pairs on the singly bonded oxygen atom, 2 lone pairs on doubly bonded oxygen atom and 1 lone pair on bromine to complete their respective octet. The Lewis structure is as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 5

The negative charge on molecule participates in resonance by resonating on each oxygen atom. Since there are three oxygen atoms, therefore three resonating structures are drawn as:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 6

(d)

Interpretation Introduction

Interpretation:

Lewis structure of selenate ion (SeO4)2− and its resonance structures have to be drawn:

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

The molecule (SeO4)2− consists of one Se atom and four O atoms.

The symbol for oxygen is O; with atomic number 8. Its electronic configuration is [He]2s22p4. Thus, it possesses 6 valence electrons.

The symbol for selenium is Se; with atomic number 34. Its electronic configuration is [Ar]3d104s24p4. Thus it possesses 6 valence electrons.

Two negative charges on molecule are added up as two valence electrons in the total count

Thus total valence electrons are sum of the valence electrons for each atom in (SeO4)2−. It is calculated as follows:

  Total valence electrons=6+6(4)+2=32(16 pairs)

The skeleton structure (SeO4)2− has four bonds that comprise 8 electrons. The electrons left to be allocated are determined as follows:

  Remaining electrons=32−8=24(12 pairs)

To complete the valence electrons of selenium, it forms double bond with two oxygen atom.

  Total remaining electrons=24−4=20(10 pairs)

Hence, 20 electrons are allocated as 3 lone pairs on singly bonded oxygen atoms and 3 lone pairs on doubly bonded oxygen atoms to complete their octet. The Lewis structure is as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 7

The two negative charges on molecule are delocalized on two oxygen atoms and since there are four oxygen atoms, therefore, four resonance structures will be formed and possible resonance structures are as follows:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2C.4E , additional homework tip 8

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Chapter 2 Solutions

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM

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 - 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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