/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q51. Question: Draw a second resonanc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question: Draw a second resonance structure for each ion. Then, draw the resonance hybrid.

a.

b.

c.

Short Answer

Expert verified

Answer

a.

b.

c.

Step by step solution

01

Step-by-Step Solution Step 1: Detection of multiple bonds, charges, and lone pairs on atoms

The position of multiple bonds, charge and nonbonded electrons determine the site of delocalization of the electrons and charges.

02

Movement of charges and multiple bonds

The curved arrows illustrate the multiple bonds and charge movement among the atoms under resonance. The resulting structure is the second resonance structure.

03

Making of resonance hybrid structures

Several resonance structures are combined to show one hybrid structure.

The delocalized multiple bonds are represented as dotted lines in a resonance hybrid structure. The delocalizing charges on two atoms (if any) are represented as partial negative or partial positive charges on the two concerning atoms.

04

Resonance structures and resonance hybrid of compounds

a. In the given structure, the charge on both the oxygen atoms and multiple bonds in between oxygen-carbon-oxygen atoms are delocalized to give the second resonance structure.

Second resonance structure of a

The resonance hybrid structure is drawn by the combination of two resonance structures, giving a partial negative charge on both the oxygen-atoms and the delocalized pair of electrons distributed between the oxygen-carbon-oxygen atoms.

Resonance hybrid structure of a

b. The nonbonded electron pair is moved onto the ring, making a double bond between the oxygen-carbon,and the double bond in between carbon and oxygen is moved toward oxygen to give an extra electron pair to oxygen, resulting in a negative charge. This gives the second resonance structure.

Second resonance structure of b

The resonance hybrid structure is drawn from the combination of two resonance structures.

Resonance hybrid structure of b

c. Multiple bonds and site of charge are associated with two atoms only, that is, oxygen and carbon. The multiple bonds are moved toward the oxygen atom. Hence, removing the positive charge on oxygen-atom.

The positive charge is generated on the carbon atom, which is bonded with oxygen with a double bond in the given structure. The resulting structure is the second resonance structure.

Second resonance structure of c

The combination of the two resonating structures gives the resonance hybrid structure.

Resonance hybrid structure of c

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: The curved arrow notation introduced in Section 1.6 is a powerful method used by organic chemists to show the movement of electrons not only in resonance structures but also in chemical reactions. Since each curved arrow shows the movement of two electrons, following the curved arrows illustrates what bonds are broken and formed in a reaction. Consider the following three-step process: (a) Add curved arrows in Step [1] to show the movement of electrons. (b) Use the curved arrows drawn in Step [2] to identify the structure of X; X is converted in Step [3] to phenol and HCl.

Question: Answer the following questions about compound A:

  1. Label the shortest single bond.
  2. Label the longest single bond.
  3. Considering all the bonds, label the shortest bond.
  4. Label the weakest bond.
  5. Label the strongest bond.
  6. Explain why bond [1] and bond [2] are different in length, even though they are both single bonds.

Question: Predict the geometry around each highlighted atom

a.

b.

c.

d.

e.

Question: Considering structures, A–D, classify each pair of compounds as isomers, resonance structures, or neither: (a) A and B; (b) A and C; (c) A and D; (d) B and D.

Question: While the most common isotope of nitrogen has a mass number of 14 (nitrogen-14), a radioactive isotope of nitrogen has a mass number of 13 (nitrogen-13). Nitrogen-13 is used in PET (positron emission tomography) scans by physicians to monitor brain activity and diagnose dementia. For each isotope, give the following information: (a) the number of protons; (b) the number of neutrons; (c) the number of electrons in the neutral atom; (d) the group number; and (e) the number of valence electrons.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.