
Concept explainers
Draw Lewis structures for the following four isoelectronic species: (a) CO, (b) NO+, (c) CN−, (d) N2. Show formal charges. (See Problem 6.69.)
(a)

Interpretation: The Lewis structures for the given isoelectronic species should be shown.
Concept Introduction:
- Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
- It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
- The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.
To find: The Lewis structure for the given set of isoelectronic species.
Answer to Problem 6.83QP
Explanation of Solution
Given isoelectronic species is below.
Lewis structure of the above isoelectronic species is drawn below.
The total number of valence electrons is found to be 10, where carbon and oxygen has 4 and 6 valence electrons respectively.
The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.
Since the octets of carbon atoms are not filled, a triple bond was made between carbon and oxygen atoms in expense of two electrons where the remaining four electrons are distributed equally over two atoms present in the given species.
(b)

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.
Concept Introduction:
- Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
- It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
- The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.
To find: The Lewis structure for the given set of isoelectronic species.
Answer to Problem 6.83QP
Explanation of Solution
Given isoelectronic species is below.
Lewis structure of the above isoelectronic species is drawn below.
The total number of valence electrons is found to be 11, where nitrogen and oxygen contains 5 and 6 valence electrons respectively. The whole charge of the molecule is +1 that results in the total number of valence electrons as 10.
The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.
Since the octets of nitrogen atoms are not filled, a triple bond is made between nitrogen and oxygen atoms in expense of two electrons where the remaining four electrons are distributed over the 2 atoms present in the given molecule.
(c)

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.
Concept Introduction:
- Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
- It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
- The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.
To find: The Lewis structure for the given set of isoelectronic species.
Answer to Problem 6.83QP
Explanation of Solution
Given isoelectronic molecule is below.
Lewis structure of the above isoelectronic species is drawn below.
The total number of valence electrons is found to be 9, where nitrogen and carbon contributes 5 and 4 electrons respectively. The whole charge of the molecule is -1 making the total number of valence electrons 10.
The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.
Since the octets of carbon atoms are not filled, a triple bond is made between carbon and nitrogen atoms in expense of two electrons where the remaining four electrons are distributed over the atoms present in the given molecule.
(d)

Interpretation: The Lewis structures and the formal charges for the given isoelectronic species should be shown.
Concept Introduction:
- Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
- It is also known as Lewis dot structures which represents the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
- The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.
To find: The Lewis structure for the given set of isoelectronic species.
Answer to Problem 6.83QP
Explanation of Solution
Given isoelectronic species is below.
Lewis structure of above isoelectronic species is drawn below.
The total number of valence electrons is found to be 10, where both nitrogen atoms contribute 5 electrons.
The 8 electrons getting after reducing two electrons for each bond from the total valence electron are distributed on nitrogen atom to complete the octet.
Since the octets of nitrogen atoms are not filled, a triple bond is made between both nitrogen atoms.
(a)

Interpretation: The formal charges for the given isoelectronic species should be shown.
Concept Introduction
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.
This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.
Formal charge of an atom can be determined by the given formula.
Answer to Problem 6.83QP
Explanation of Solution
Formal charge of the given isoelectronic species is given below
The formal charge of the given isoelectronic species is calculated,
- Carbon atom
Substituting these values to the equation,
- Oxygen atom
Substituting these values to the equation,
(b)

Interpretation: The formal charges for the given isoelectronic species should be shown.
Concept Introduction
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.
This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.
Formal charge of an atom can be determined by the given formula.
Answer to Problem 6.83QP
Explanation of Solution
Formal charge of the given isoelectronic species is given below
The formal charge of the given isoelectronic species is calculated,
- Nitrogen atom
Substituting these values to the equation,
- Oxygen atom
Substituting these values to the equation,
(c)

Interpretation: The formal charges for the given isoelectronic species should be shown.
Concept Introduction
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.
This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.
Formal charge of an atom can be determined by the given formula.
Answer to Problem 6.83QP
Explanation of Solution
Formal charge of the given isoelectronic species is given below
The formal charge of the given isoelectronic species is calculated,
- Carbon atom
Substituting these values to the equation,
- Nitrogen atom
Substituting these values to the equation,
(d)

Interpretation: The formal charges for the given isoelectronic species should be shown.
Concept Introduction
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.
This method is used to identify the most probable Lewis structures if more than one possibility exists for a compound.
The Lewis structure with formal charge on each of the atoms close to zero is taken as the most plausible structure.
Formal charge of an atom can be determined by the given formula.
Answer to Problem 6.83QP
Explanation of Solution
Formal charge of the given isoelectronic species is given below
The formal charge of the given isoelectronic species is calculated,
- Nitrogen atom
Substituting these values to the equation,
- Since both the nitrogen atoms are similar, the formal charge of the nitrogen atoms is zero.
Want to see more full solutions like this?
Chapter 6 Solutions
Chemistry Atoms First, Second Edition
- Identifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: ☐ 1.8 mol of HCl is added to 1.0 L of a 1.0M NH3 bases: ☐ solution. other: ☐ 0.18 mol of HNO3 is added to 1.0 L of a solution that is 1.4M in both NH3 and NH₁Br. acids: bases: ☐ other: ☐ 0,0,... ? 000 18 Ar B 1arrow_forwardUsing reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NH3 (g) = N2 (g) +3H₂ —N2 (g) AGº = 34. kJ Now suppose a reaction vessel is filled with 4.19 atm of ammonia (NH3) and 9.94 atm of nitrogen (N2) at 378. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NH 3 tend to rise or fall? ☐ x10 fall Х Is it possible to reverse this tendency by adding H₂? In other words, if you said the pressure of NH 3 will tend to rise, can that be changed to a tendency to fall by adding H₂? Similarly, if you said the pressure of NH3 will tend to fall, can that be changed to a tendency to rise by adding H₂? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H₂ needed to reverse it. Round your answer to 2 significant digits. yes no atm 00. 18 Ar 무ㅎ ?arrow_forwardIdentifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HF is a weak acid. 2.2 mol of NaOH is added to 1.0 L of a 1.4M HF solution. acids: П bases: Х other: ☐ ப acids: 0.51 mol of KOH is added to 1.0 L of a solution that is bases: 1.3M in both HF and NaF. other: ☐ 00. 18 Ararrow_forward
- Using reaction free energy to predict equilibrium composition Consider the following equilibrium: N2O4 (g) 2NO2 (g) AG⁰ = 5.4 kJ Now suppose a reaction vessel is filled with 1.68 atm of dinitrogen tetroxide (N204) at 148. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of N2O4 tend to rise or fall? x10 fall Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to rise by adding NO2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO 2 needed to reverse it. Round your answer to 2 significant digits. yes no 0.42 atm ☑ 5 0/5 ? مله Ararrow_forwardHomework 13 (Ch17) Question 4 of 4 (1 point) | Question Attempt: 2 of 2 ✓ 1 ✓ 2 = 3 4 Time Remaining: 4:25:54 Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction free energy of the following chemical reaction: 2CH3OH (g)+302 (g) → 2CO2 (g) + 4H₂O (g) Round your answer to zero decimal places. ☐ kJ x10 ☐ Subm Check 2020 Hill LLC. All Rights Reserved. Terms of Use | Privacy Cearrow_forwardIdentifying the major species in weak acid or weak base equilibria Your answer is incorrect. • Row 2: Your answer is incorrect. • Row 3: Your answer is incorrect. • Row 6: Your answer is incorrect. 0/5 The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HF is a weak acid. acids: HF 0.1 mol of NaOH is added to 1.0 L of a 0.7M HF solution. bases: 0.13 mol of HCl is added to 1.0 L of a solution that is 1.0M in both HF and KF. Exponent other: F acids: HF bases: F other: K 1 0,0,... ? 000 18 Ararrow_forward
- Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NOCI (g) 2NO (g) + Cl2 (g) AGº =41. kJ Now suppose a reaction vessel is filled with 4.50 atm of nitrosyl chloride (NOCI) and 6.38 atm of chlorine (C12) at 212. °C. Answer the following questions about this system: ? rise Under these conditions, will the pressure of NOCI tend to rise or fall? x10 fall Is it possible to reverse this tendency by adding NO? In other words, if you said the pressure of NOCI will tend to rise, can that be changed to a tendency to fall by adding NO? Similarly, if you said the pressure of NOCI will tend to fall, can that be changed to a tendency to rise by adding NO? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO needed to reverse it. Round your answer to 2 significant digits. 0.035 atm ✓ G 00. 18 Ararrow_forwardHighlight each glycosidic bond in the molecule below. Then answer the questions in the table under the drawing area. HO- HO- -0 OH OH HO NG HO- HO- OH OH OH OH NG OHarrow_forward€ + Suppose the molecule in the drawing area below were reacted with H₂ over a platinum catalyst. Edit the molecule to show what would happen to it. That is, turn it into the product of the reaction. Also, write the name of the product molecule under the drawing area. Name: ☐ H C=0 X H- OH HO- H HO- -H CH₂OH ×arrow_forward
- Draw the Haworth projection of the disaccharide made by joining D-glucose and D-mannose with a ẞ(1-4) glycosidic bond. If the disaccharide has more than one anomer, you can draw any of them. Click and drag to start drawing a structure. Xarrow_forwardEpoxides can be opened in aqueous acid or aqueous base to produce diols (molecules with two OH groups). In this question, you'll explore the mechanism of epoxide opening in aqueous acid. 2nd attempt Be sure to show all four bonds at stereocenters using hash and wedge lines. 0 0 Draw curved arrows to show how the epoxide reacts with hydronium ion. 100 +1: 1st attempt Feedback Be sure to show all four bonds at stereocenters using hash and wedge lines. See Periodic Table See Hint H A 5 F F Hr See Periodic Table See Hintarrow_forward03 Question (1 point) For the reaction below, draw both of the major organic products. Be sure to consider stereochemistry. > 1. CH₂CH₂MgBr 2. H₂O 3rd attempt Draw all four bonds at chiral centers. Draw all stereoisomers formed. Draw the structures here. e 130 AN H See Periodic Table See Hint P C Brarrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY





