Ch.8 - Basic Concepts of Chemical Bonding

Problem 91

Chapter 8, Problem 91

Structures A, B, and C show the connectivity of the atoms in three different molecules that are isomers of C3H4O. By completing the Lewis structures of these molecules, complete the information in the following table:
Isomer A Isomer B Isomer C
Number of single bonds
Number of double bonds
Number of triple bonds
Number of nonbonding pairs

Verified step by step guidance

insert step 1: Identify the total number of valence electrons for each isomer. Carbon has 4 valence electrons, hydrogen has 1, and oxygen has 6. For C3H4O, calculate the total valence electrons: 3(4) + 4(1) + 6 = 18 valence electrons.

insert step 2: Draw the skeletal structure for each isomer, ensuring that all atoms are connected in the order given by the structures A, B, and C.

insert step 3: Distribute the valence electrons around the atoms in each isomer, starting by forming single bonds between the atoms. Use the remaining electrons to complete the octets of the atoms, starting with the most electronegative atom, which is oxygen.

insert step 4: Identify and count the number of single, double, and triple bonds in each isomer. Single bonds are represented by a single line, double bonds by two lines, and triple bonds by three lines between atoms.

insert step 5: Determine the number of nonbonding pairs (lone pairs) on each atom, especially focusing on the oxygen atom, as it is likely to have lone pairs. Count the pairs of dots around each atom that are not involved in bonding.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Isomerism

Isomerism refers to the phenomenon where two or more compounds have the same molecular formula but different structural arrangements of atoms. In the case of C3H4O, the isomers can differ in the connectivity of atoms, leading to distinct chemical properties and reactivity. Understanding isomerism is crucial for analyzing the different structures and their corresponding Lewis diagrams.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They provide a visual representation of how atoms are connected and help predict the geometry and reactivity of the molecule. Completing the Lewis structures for the isomers of C3H4O is essential for determining the number of single, double, and triple bonds, as well as nonbonding pairs.

Bond Types

In chemistry, bonds between atoms can be classified as single, double, or triple bonds, depending on the number of shared electron pairs. Single bonds involve one pair of shared electrons, double bonds involve two pairs, and triple bonds involve three pairs. Recognizing these bond types is vital for accurately filling out the table regarding the number of each bond type present in the isomers of C3H4O.

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Related Practice

Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH4 can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (b) Which element in NaAlH4 is the most electronegative? Which one is the least electronegative? Which element in NaAlH4 is the least electronegative?

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Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH₄ can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H₂(g). NaAlH₄ possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion.

424

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Textbook Question

A major challenge in implementing the 'hydrogen economy' is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, NaAlH₄ can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H₂(g). NaAlH₄ possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (d) What is the formal charge on hydrogen in the polyatomic ion?

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Textbook Question

The triiodide ion, I3-, exists, whereas the corresponding ion with fluorine, F3-, does not. The I3- ion has a linear structure in which two outer I atoms are each bonded to a central I atom. Although I3- is a known ion, F3- is not.

c. Which of the following statements about the existence of I3- versus the nonexistence of F3- is or are true?

i. The Lewis structure of I3- shows 12 electrons around the central I atom.

ii. Elements from the second row of the periodic table generally do not form hypervalent molecules and ions.

iii. An I atom can form a hypervalent molecule or ion more readily than an F atom because of the larger size of the I atom.

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Textbook Question

The hypochlorite ion, ClO^-, is the active ingredient in bleach. The perchlorate ion, ClO₄^-, is a main component of rocket propellants. Draw Lewis structures for both ions. (b) What is the formal charge of Cl in the perchlorate ion, assuming the Cl—O bonds are all single bonds?

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Textbook Question

The following three Lewis structures can be drawn for N2O:

(a) Using formal charges, which of these three resonance forms is likely to be the most important?

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