Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 9, Problem 67AE
Vitamin B6 is an organic compound whose deficiency in the human body can cause apathy, irritability, and an increased susceptibility to infections. Below is an incomplete Lewis structure for vitamin B6. Complete the Lewis structure and answer the following questions. Hint: Vitamin B6 can be classified as an organic compound (a compound based on carbon atoms). The majority of Lewis structures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure below to give each atom a formal charge of zero.
a. How many σ bonds and π bonds exist in vitamin B6?
b. Give approximate values for the bond angles marked a through g in the structure.
c. How many carbon atoms are sp2 hybridized?
d. How many carbon, oxygen, and nitrogen atoms are sp3 hybridized?
e. Does vitamin B6 exhibit delocalized π bonding? Explain.
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Unshared, or lone, electron pairs play an important role in determining the chemical and physical properties of organic compounds. Thus, it is important
to know which atoms carry unshared pairs.
Use the structural formulas below to determine the number of unshared pairs at each designated atom.
Be sure your answers are consistent with the formal charges on the formulas.
The number of unshared pairs at atom a is
4
The number of unshared pairs at atom b is 2
H
The number of unshared pairs at atom c is
3
The number of unshared pairs at atom a is
The number of unshared pairs at atom b is
C-NEN
а b
H
The number of unshared pairs at atom c is
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Use this condensed chemical structure to complete the table below.
OH
The condensed chemical structure of phenol.
Some facts about the phenol molecule:
molecular formula:
number of major resonance structures:
number of carbon-hydrogen single (C-H)
bonds:
number of oxygen-hydrogen single (O-H)
bonds:
number of lone pairs:
0
0
0
0
0
Unshared, or lone, electron pairs play an important role in determining the chemical and physical properties of organic compounds.
Thus, it is important to know which atoms carry unshared pairs.
Use the structural formulas below to determine the number of unshared pairs at each designated atom.
Be sure your answers are consistent with the formal charges on the formulas.
HC=CH
aN
b
H
The number of unshared pairs at atom a is
The number of unshared pairs at atom b is
The number of unshared pairs at atom c is
The number of unshared pairs at atom a is
The number of unshared pairs at atom b is
The number of unshared pairs at atom c is
Chapter 9 Solutions
Chemistry
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Molecule A...Ch. 9 - Draw the Lewis structures for TeCl4, ICl5, PCl5,...Ch. 9 - A variety of chlorine oxide fluorides and related...Ch. 9 - Pelargondin is the molecule responsible for the...Ch. 9 - Complete a Lewis structure for the compound shown...Ch. 9 - Which of the following statements concerning SO2...Ch. 9 - Consider the molecular orbital electron...Ch. 9 - Place the species B2+ , B2, and B2 in order of...Ch. 9 - Consider the following computer-generated model of...Ch. 9 - Cholesterol (C27liu;O) has the following...Ch. 9 - Cyanamide (H2NCN), an important industrial...Ch. 9 - A flask containing gaseous N2 is irradiated with...Ch. 9 - Prob. 92CPCh. 9 - Values of measured bond energies may vary greatly...Ch. 9 - Use the MO model to explain the bonding in BeH2....Ch. 9 - Prob. 95CPCh. 9 - Arrange the following from lowest to highest...Ch. 9 - Use the MO model to determine which of the...Ch. 9 - Given that the ionization energy of F2 is 290...Ch. 9 - Carbon monoxide (CO) forms bonds to a variety of...Ch. 9 - Prob. 100CPCh. 9 - As the bead engineer of your starship in charge of...Ch. 9 - Determine the molecular structure and...
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- Write Lewis structures for these ions. Show all valence electrons and all formal charges. (a) Amide ion, NH2 (b) Bicarbonate ion, HCO3 (c) Carbonate ion, CO32 (d) Nitrate ion, NO3 (e) Formate ion, HCOO (f) Acetate ion, CH3COOarrow_forwardThe study of carbon-containing compounds and their properties is called organic chemistry. Besides carbon atoms, organic compounds also can contain hydrogen, oxygen, and nitrogen atoms (as well as other types of atoms). A common trait of simple organic compounds is to have Lewis structures where all atoms have a formal charge of zero. Consider the following incomplete Lewis structure for an organic compound called histidine (an amino acid), which is one of the building blocks of proteins found in our bodies: Draw a complete Lewis structure for histidine in which all atoms have a formal charge of zero.arrow_forwardAspartame is an artificial sweetener marketed under the name Nutra-Sweet. A partial Lewis structure for aspartame is shown below. Aspartame can be classified as an organic compound (a compound based on carbon atoms). The majority of Lewis structures for simple organic compounds have all atoms with a formal charge of zero. Therefore, add lone pairs and multiple bonds to the structure above to give each atom a formal charge of zero when drawing the Lewis structure. Also note that the six-sided ring is shorthand notation for a benzene ring (C6H5). Benzene is discussed in Section 4-7. Complete the Lewis structure for aspartame. How many C and N atoms exhibit sp1 hybridization? How many C and O atoms exhibit sp3 hybridization? How many and bonds are in aspartame?arrow_forward
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- Unshared, or lone, electron pairs play an important role in determining the chemical and physical properties of organic compounds. Thus, it is important to know which atoms carry unshared pairs. Use the structural formulas below to determine the number of unshared pairs at each designated atom. Be sure your answers are consistent with the formal charges on the formulas. The number of unshared pairs at atom a is CH3 The number of unshared pairs at atom b is The number of unshared pairs at atom c is The number of unshared pairs at atom a is v. The number of unshared pairs at atom b is The number of unshared pairs at atom c is HC CNH₂ a Nb Varrow_forwardA resonance hybrid is a structure that can be depicted by more than one valid Lewis structure. part1: Draw the major resonance form of fulminic acid, HCNO, with the atoms connected as indicated in the formula. Your structure should have nonzero formal charges minimized, and it should include all nonzero formal charges and all nonbonding electrons. part2: Draw the second most important resonance form of fulminic acid, HCNO, with the atoms connected as indicated in the formula. Your structure should have nonzero formal charges minimized, and it should include all nonzero formal charges and all nonbonding electrons. part3: Draw the least important resonance contributor for fulminic acid, HCNO, with the atoms connected as indicated in the formula. Your structure should have nonzero formal charges minimized and should include all nonzero formal charges and all nonbonding electrons.arrow_forwardA resonance hybrid is a structure that can be depicted by more than one valid Lewis structure. part1: Draw the major resonance form of fulminic acid, HCNO, with the atoms connected as indicated in the formula. Your structure should have nonzero formal charges minimized, and it should include all nonzero formal charges and all nonbonding electrons. part2: Draw the second most important resonance form of fulminic acid, HCNO, with the atoms connected as indicated in the formula. Your structure should have nonzero formal charges minimized, and it should include all nonzero formal charges and all nonbonding electrons.arrow_forward
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