Skip to main content
Ch.8 - Basic Concepts of Chemical Bonding

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 Solution
Video duration:
2m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Video transcript

hey everyone, we're shown the connectivity of atoms in the oxydol molecule and were asked to complete the lewiS structure and to determine the number of sigma and pI bonds. Now we've learned that single bonds will contain sigma bonds while double bonds will contain a sigma plus a pi bond, a triple bond will contain a sigma plus two pi bonds. Now that we have this in mind, we can use this as reference when we determine the number of sigma and pi bonds. So let's go ahead and complete our structure first. We need to complete the octet of each atom in order to do so we need to add a double bond between our two carbons and a double bond between our carbon and nitrogen. We also have to add a lone pair on our nitrogen and two lone pairs on our oxygen. Now we have the complete lewis structure of oxydol. Now to answer the second part of our question, I'm going to go ahead and highlight our single bonds as pink are double bonds as blue. And our triple bonds as yellow counting are single bonds. First We have 1, 2, 3, 4, 5 and six. So we have a total of six single bonds. Now let's go ahead and count our double bonds. We have one, two. We only have two double buttons now to determine the number of sigma and pi bonds, we know that our six single bonds are going to give us six sigma bonds while our two double bonds are going to give us two sigma bonds Plus two Pi bonds in total, we have six plus two sigma bonds, which will get us to eight sigma bonds, And we have two pi bonds. And this is going to be our final answer since we also didn't have any trouble bonds. Now, I hope that made sense and let us know if you have any questions.
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?

365
views
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
views
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?

338
views
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.


2
views
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?

921
views
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?

1177
views