![Study Guide for Zumdahl/Zumdahl/DeCoste’s Chemistry, 10th Edition](https://www.bartleby.com/isbn_cover_images/9781305957473/9781305957473_largeCoverImage.gif)
Concept explainers
Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. Carbon is the central atom in CH4, nitrogen is the central atom in NH3, and oxygen is the central atom in H2O.
a. F2
b. O2
c. CO
d. CH4
e. NH3
f. H2O
g. HF
(a)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of fluorine
The valence electron of fluorine is 7
The fluorine molecule
The skeletal structure of
Each fluorine atom requires one electron to complete the octet. Hence, the mutual sharing of two electrons takes place. The 12 valence electrons present are placed as lone pairs in such a way that each atom gets three lone pairs.
The Lewis structure of
Figure 1
(b)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of oxygen
The valence electron of oxygen is 6
The oxygen molecule
The skeletal structure of
Each oxygen atom requires two electrons to complete the octet. Hence, the mutual sharing of four electrons takes place. The 8 valence electrons present are placed as lone pairs in such a way that each atom gets two lone pairs.
The Lewis structure of
Figure 2
(c)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of oxygen is 8 and its electronic configuration is,
The valence electron of oxygen is 6
The atomic number of carbon
The valence electron of carbon is 4
The molecule
The skeletal structure of
Each oxygen atom requires two electrons to complete the octet whereas carbon requires four electrons to complete the octet. Hence, the mutual sharing of six electrons takes place which is represented by triple bond. The 4 valence electrons left are placed as lone pairs in such a way that each atom gets one lone pair.
The Lewis structure of
Figure 3
(d)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of carbon is 6 and its electronic configuration is,
The valence electron of carbon is 4
The atomic number of hydrogen
The valence electron of hydrogen is 1
The molecule
The skeletal structure of
Figure 4
Each hydrogen atom requires one electron to complete the octet whereas carbon requires four electrons to complete the octet. In the molecule
The Lewis structure of
Figure 5
(e)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of nitrogen
The valence electron of nitrogen is 5
The atomic number of hydrogen is 1 and its electronic configuration is,
The valence electron of hydrogen is 1
The molecule
The skeletal structure of
Figure 6
Each hydrogen atom requires one electron to complete the octet whereas nitrogen requires three electrons to complete the octet. In the molecule
The Lewis structure of
Figure 7
(f)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of oxygen is 8 and its electronic configuration is,
The valence electron of oxygen is 6
The atomic number of hydrogen is 1 and its electronic configuration is,
The valence electron of hydrogen is 1
The molecule
The skeletal structure of
Figure 8
Each hydrogen atom requires one electron to complete the octet whereas oxygen requires two electrons to complete the octet. In the molecule
The Lewis structure of
Figure 9
(g)
![Check Mark](/static/check-mark.png)
Interpretation: The Lewis structure is to be drawn for the given molecules.
Concept introduction: The Lewis structure is also known as dot structure. This structure depicts the bonding between atoms and the lone pairs of electrons if exists.
The octet rule states that atoms or molecules gain or lose electrons to get the electronic configuration of nearest noble gas.
To determine: The Lewis structure of the molecule
Answer to Problem 85E
Explanation of Solution
The first step in determining the Lewis structure is to determine the number of valence electrons. The atomic number of fluorine is 9 and its electronic configuration is,
The valence electron of fluorine is 7
The atomic number of hydrogen is 1 and its electronic configuration is,
The valence electron of hydrogen is 1
The molecule
The skeletal structure of
Each hydrogen atom requires one electron to complete the octet whereas fluorine requires one electron to complete the octet. Hence, the mutual sharing of eight electrons takes place.
The Lewis structure of
Figure 10
The Lewis dot structure is drawn to satisfy the octets of atoms. The octet rule states that elements gain or lose electrons to get the nearest noble gas configuration.
Want to see more full solutions like this?
Chapter 8 Solutions
Study Guide for Zumdahl/Zumdahl/DeCoste’s Chemistry, 10th Edition
- How does the bond energy of HCl(g) differ from the standard enthalpy of formation of HCl(g)?arrow_forwardUsing the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions: (a) Cl2(g)+3F2(g)2ClF3(g) (b) H2C=CH2(g)+H2(g)H3CCH3(g) (c) 2C2H6(g)+7O2(g)4CO2(g)+6H2O(g) .arrow_forwardrite the Lewis structure for each of the following atoms. msp;a.He(Z=2)d.Ne(Z=10)b.Br(Z=35)e.I(Z=53)c.Sr(Z=38)f.Ra(Z=88)arrow_forward
- Compare your answers from parts a and b of Exercise 69 of Chapter 3 with H values calculated for each reaction using standard enthalpies of formation in Appendix 4. Do enthalpy changes calculated from bond energies give a reasonable estimate of the actual values?arrow_forwardWrite the Lewis symbols for each of the following ions: (a) As3 (b) I (c) Be2+ (d) O2 (e) Ga3+ (f) Li+ (g) N3arrow_forwardBond Enthalpy When atoms of the hypothetical element X are placed together, they rapidly undergo reaction to form the X2 molecule: X(g)+X(g)X2(g) a Would you predict that this reaction is exothermic or endothermic? Explain. b Is the bond enthalpy of X2 a positive or a negative quantity? Why? c Suppose H for the reaction is 500 kJ/mol. Estimate the bond enthalpy of the X2 molecule. d Another hypothetical molecular compound, Y2(g), has a bond enthalpy of 750 kJ/mol, and the molecular compound XY(g) has a bond enthalpy of 1500 kJ/mol. Using bond enthalpy information, calculate H for the following reaction. X2(g)+Y2(g)2XY(g) e Given the following information, as well as the information previously presented, predict whether or not the hypothetical ionic compound AX is likely to form. In this compound, A forms the A+ cation, and X forms the X anion. Be sure to justify your answer. Reaction: A(g)+12X2(g)AX(s)The first ionization energy of A(g) is 400 kJ/mol. The electron affinity of X(g) is 525 kJ/mol. The lattice energy of AX(s) is 100 kJ/mol. f If you predicted that no ionic compound would form from the reaction in Part e, what minimum amount of AX(s) lattice energy might lead to compound formation?arrow_forward
- 6. Write the singly bonded Lewis dot structure for BF3. Which of the following statements best describes this structure? A. It obeys the octet rule on all atoms. B. It has less than an octet on at least one atom. C. It has a lone pair of electrons on the boron atom. D. It has less than an octet of electrons on all atoms.arrow_forward1-What is the formal charge of the nitrogen atom in hydrogen cyanide (HCN)? a.5 b.−1 c.0 d.1 e.2 2-Which of the following diatomic molecules has the least number of valence electrons? a.N2 b.F2 c.O2 d.Cl2 e.NOarrow_forwardA. State the Octet Rule.B. Write the Lewis symbol of neon. Is the neon atom stable? Does neon exist as atoms in nature? If not, how does neon exist in nature.C. Write the Lewis symbol of nitrogen. Is the nitrogen atom stable? Does nitrogen exist as atoms in nature? If not, how does nitrogen exist in nature.arrow_forward
- Write the Lewis structure for each molecule.a. CH2O b. C2Cl4c. CH3NH2 d. CFCl3 (C central)arrow_forward4. Draw the Lewis dot structures for: (Note: make sure it satisfies the octet rule) a. NI3 b. CS2 c. N₂F2 d. CH₂Cl2 5. a. Which bond is more polar, H-Cl or H-O? b. Which of the following bonds is the most polar, that is, has the greatest ionic character: H-Br, N-H, N-O, P-Cl? 6. Write the formulas and names for the compounds of the following ions: 0²- F p3- K+ Ga³+arrow_forwardWhich bond is the longest? A. HF B. HCI C. HBr D. HIarrow_forward
- Introductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337399425/9781337399425_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337399692/9781337399692_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133611097/9781133611097_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079243/9781305079243_smallCoverImage.gif)