Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 7, Problem 7.143QP

Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to Fe2+ in the hemoglobin molecule. The molecular orbitals of CO have the same energy order as those of the N2 molecule, (a) Draw a Lewis structure of CO and assign formal charges. Explain why CO has a rather small dipole moment of 0.12 D. (b) Compare the bond order of CO with that from molecular orbital theory, (c) Which of the atoms (C or O) is more likely to form bonds with the Fe2+ ion in hemoglobin?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The Lewis structure of the carbon monoxide and its corresponding formal charges should be drawn.  The bond order of carbon monoxide with that of molecular orbital theory should be compared.  Also to found out which atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

Concept Introduction:

  • Lewis structures which also known as Lewis dot structures represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
  • A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.
  • Formal charge of an atom can be determined by the given formula.

Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Dipole moment occurs when there is a difference in the electronegativity of atoms in a bond which results in charge separation.

Answer to Problem 7.143QP

  1. (a) Lewis structure of the carbon monoxide and its corresponding formal charges is Chemistry: Atoms First, Chapter 7, Problem 7.143QP , additional homework tip  1
  2. (c) It was found out that carbon atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

Explanation of Solution

Given molecule is CO .

Lewis structure of carbon monoxide is drawn below.

Chemistry: Atoms First, Chapter 7, Problem 7.143QP , additional homework tip  2

  • The total number of valence electrons is found to be 10, where carbon and oxygen atom contribute 4 and 6 electrons respectively.

The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on atom to complete the octet. After the distribution of electrons, atoms does not obey octet rule so triple bond is formed carbon and oxygen atoms.

The formal charge carbon monoxide

Chemistry: Atoms First, Chapter 7, Problem 7.143QP , additional homework tip  3

The formal charge carbon monoxide is calculated,

  • Carbon atom

Number of valenceelectron=4Number of bondingelectron=6Number of non-bondingelectron=2

Substituting these values to the equation,

FC=4-(12×6)-2=-1

  • Oxygen atom (a)

  Number of valenceelectron=6Number of bondingelectron=6Number of non-bondingelectron=2

Substituting these values to the equation,

FC=6-(12×6)-2=+1

Due to the difference in the electronegativity of carbon and oxygen atoms, the C-O bond is polar. So carbon atom gets a partial positive charge whereas oxygen atoms gets partial negative charge which means concentration of electron density is expected to have on oxygen atom. After calculating the formal charges, the negative charge is assigned on the carbon atom. Thus carbon monoxide is expected to have low dipole moment.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The Lewis structure of the carbon monoxide and its corresponding formal charges should be drawn.  The bond order of carbon monoxide with that of molecular orbital theory should be compared.  Also to found out which atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

Concept Introduction:

  • Lewis structures which also known as Lewis dot structures represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
  • A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.
  • Formal charge of an atom can be determined by the given formula.

Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Dipole moment occurs when there is a difference in the electronegativity of atoms in a bond which results in charge separation.

In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals and only contain a maximum of two electrons. The number of newly formed molecular orbital is equal to the number of atomic orbitals involved in the bonding.

There are two types of molecular orbitals,

  1. a) Bonding molecular orbitals: sharing of electron density is between the nuclei and has comparatively lower energy and fills first.
  2. b) Antibonding molecular orbitals: Two nuclei are pulled by the electrons density in opposite direction and have higher energy comparing to bonding molecular orbital.

Explanation of Solution

Total number of electrons in carbon monoxide is 14 where carbon and oxygen atom have 6 and 8 electrons respectively.

Electronic configuration of CO is (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2px)2(π2py)2(σpz)2 .

The bond order of CO is determined.

The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.

Bond order =(No.ofe-inbondingorbitals)-(No.ofe-inantibondingorbitals)2

No.ofe-inbondingorbitals=8No.ofe-inantibondingorbitals=2

Bondorder=12(8-2)=3

The bond order of CO is 3 which is similar to the triple bond that drawn in Lewis structure. CO and N2 is isoelectronic which means they  have same number of valence electrons and atoms.

Conclusion

The bond order of carbon monoxide with that of molecular orbital theory was compared.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The Lewis structure of the carbon monoxide and its corresponding formal charges should be drawn.  The bond order of carbon monoxide with that of molecular orbital theory should be compared.  Also to found out which atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

Concept Introduction:

  • Lewis structures which also known as Lewis dot structures represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
  • A formal charge (FC) is the charge assigned to an atom in a molecule, irrespective of relative electronegativity by thinking that electrons in all chemical bonds are shared equally among atoms.
  • Formal charge of an atom can be determined by the given formula.

Formalcharge(FC)=(no.ofvalenceelectroninatom)12(no.ofbondingelectrons)(no.ofnon-bondingelectrons)

  • Dipole moment occurs when there is a difference in the electronegativity of atoms in a bond which results in charge separation.

Answer to Problem 7.143QP

  1. (d) It was found out that carbon atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

Explanation of Solution

Fe2+ is the metal presence in hemoglobin. When carbon monoxide enters the blood stream Fe2+ cation forms bond with a negatively charged atom. Carbon monoxide is a neutral compound. While calculating the formal charge of CO , it is found that carbon has negative charge. So Fe2+ tend to make bond with carbon atom (OC-Fe2+) instead of oxygen atom (CO-Fe2+) .

Conclusion

The Lewis structure of the carbon monoxide and its corresponding formal charges was drawn. Also it was found out that carbon atom of carbon monoxide forms bond with Fe2+ of hemoglobin.

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Chapter 7 Solutions

Chemistry: Atoms First

Ch. 7.1 - Prob. 7.1.3SRCh. 7.1 - Prob. 7.1.4SRCh. 7.2 - Prob. 7.3WECh. 7.2 - Prob. 3PPACh. 7.2 - For each of the following hypothetical molecules,...Ch. 7.2 - Which of these models could represent a polar...Ch. 7.2 - Prob. 7.2.1SRCh. 7.2 - Prob. 7.2.2SRCh. 7.3 - Prob. 7.4WECh. 7.3 - Prob. 4PPACh. 7.3 - Prob. 4PPBCh. 7.3 - Prob. 4PPCCh. 7.3 - Prob. 7.3.1SRCh. 7.3 - Which of the following exhibits significant...Ch. 7.4 - Hydrogen selenide (H2Se) is a foul-smelling gas...Ch. 7.4 - Prob. 5PPACh. 7.4 - For which molecule(s) can we not use valence bond...Ch. 7.4 - Which of these models could represent a species...Ch. 7.4 - Prob. 7.4.1SRCh. 7.4 - Prob. 7.4.2SRCh. 7.5 - Prob. 7.6WECh. 7.5 - Use hybrid orbital theory to describe the bonding...Ch. 7.5 - Prob. 6PPBCh. 7.5 - Prob. 6PPCCh. 7.5 - Prob. 7.5.1SRCh. 7.5 - Prob. 7.5.2SRCh. 7.6 - Thalidomide (C13H10N2O4) is a sedative and...Ch. 7.6 - The active ingredient in Tylenol and a host of...Ch. 7.6 - Determine the total number of sigma and pi bonds...Ch. 7.6 - In terms of valence bond theory and hybrid...Ch. 7.6 - In addition to its rise in aqueous solution as a...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Use valence bond theory and hybrid orbitals to...Ch. 7.6 - Explain why hybrid orbitals are necessary to...Ch. 7.6 - Prob. 7.6.1SRCh. 7.6 - Prob. 7.6.2SRCh. 7.6 - Prob. 7.6.3SRCh. 7.6 - Prob. 7.6.4SRCh. 7.7 - Prob. 7.9WECh. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - Use molecular orbital theory to determine whether...Ch. 7.7 - For most of the homonuclear diatomic species shown...Ch. 7.7 - Calculate the bond order of N22+, and determine...Ch. 7.7 - Which of the following species is paramagnetic?...Ch. 7.7 - Prob. 7.7.3SRCh. 7.7 - Prob. 7.7.4SRCh. 7.8 - It takes three resonance structures to represent...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Use a combination of valence bond theory and...Ch. 7.8 - Which of the following contain one or more...Ch. 7.8 - Which of the atoms in BCl3 need hybrid orbitals to...Ch. 7.8 - Which of the following can hybrid orbitals be used...Ch. 7.8 - Which of the following enables us to explain the...Ch. 7 - Prob. 7.1KSPCh. 7 - Which of the following species does not have...Ch. 7 - Prob. 7.3KSPCh. 7 - Prob. 7.4KSPCh. 7 - Prob. 7.1QPCh. 7 - Sketch the shape of a linear triatomic molecule, a...Ch. 7 - Prob. 7.3QPCh. 7 - Prob. 7.4QPCh. 7 - In the trigonal bipyramidal arrangement, why does...Ch. 7 - Prob. 7.6QPCh. 7 - Predict the geometry of the following molecules...Ch. 7 - Prob. 7.8QPCh. 7 - Predict the geometries of the following species...Ch. 7 - Predict the geometries of the following ions: (a)...Ch. 7 - Prob. 7.11QPCh. 7 - Prob. 7.12QPCh. 7 - Prob. 7.13QPCh. 7 - Describe the geometry about each of the central...Ch. 7 - Prob. 7.15QPCh. 7 - Prob. 7.16QPCh. 7 - Prob. 7.17QPCh. 7 - Prob. 7.18QPCh. 7 - Prob. 7.19QPCh. 7 - Prob. 7.20QPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Explain the term polarizability. 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