Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}{ }^{-}\)localized or delocalized?

Short Answer

Expert verified
(a) A localized π bond has electrons confined between two atoms, while a delocalized π bond has electrons spread across more than two atoms, often in resonance structures. (b) To determine delocalized π bonding, look for resonance structures where lone pairs, π bonds, or positive charges are distributed among multiple atoms. (c) The π bond in NO2- is delocalized due to its resonance structures involving delocalized electron distribution.

Step by step solution

01

(Step 1: Understanding Localized and Delocalized π Bonds)

In a localized π bond, the electrons are confined between two atoms, like in a typical carbon-carbon double bond. However, in a delocalized π bond, the electrons are spread across more than two atoms, which often happens in resonance structures.
02

(Step 2: Determining if a molecule or ion exhibits delocalized π bonding)

To determine if a molecule or ion exhibits delocalized π bonding, look for resonance structures where lone pairs, π bonds, or positive charges can be distributed among multiple atoms in the molecule.
03

(Step 3: Analyzing the NO2- ion)

The NO2- ion has the chemical structure: O=N-O with one lone pair on the negatively charged O atom. There are resonance structures for this ion, where the lone pair on the O atom can form a π bond with the N atom, creating a double bond, and the initial double bond between N and the second O atom becomes a single bond.
04

(Step 4: Determining if the π bond in NO2- ion is localized or delocalized)

Since the NO2- ion has resonance structures where the π bond is involved in the delocalized electron distribution, the π bond in the NO2- ion is considered delocalized.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Localized π Bond
If you've ever seen molecules with double or triple bonds, you're already familiar with π (pi) bonds. These bonds form when parallel orbitals overlap and share electrons. In a localized π bond, the electrons are shared directly between just two atoms. Think of it as a specific 'address' for electrons; they're only hanging out there. This is the standard in many organic compounds where double bonds don't interact with each other or with lone electron pairs.

A prime example is ethylene (C2H4), where a localized π bond exists between the two carbon atoms, with no ability for those electrons to wander off to neighboring atoms.
Resonance Structures
Imagine if electrons could have multiple homes, shifting between them like a family with vacation homes. That's what happens in resonance structures. For molecules that have this property, we can draw different structures, called resonance forms, to show different ways electrons are distributed. These are like snapshots of all the possible places where the electrons could be at any moment in time.

It's essential to realize that electrons don't actually flip back and forth between these structures. Instead, the true form of the molecule is a hybrid, a blend of all the possible resonance forms. A famous case is benzene, where electrons in π bonds are delocalized over six carbon atoms instead of being localized between two.
Electron Distribution
The way electrons are spread out in a molecule can tell us a lot about its reactivity and properties. Electron distribution is about understanding where in a molecule electrons prefer to spend their time. In molecules with localized π bonds, the electron distribution is predictable and tied to specific atom pairs. However, when π bonds are delocalized, electron distribution becomes more spread out among several atoms.

This distribution affects the molecule's color, reactivity, and even stability. Delocalized electrons can lower a molecule's energy, making it more stable, as they roam and enjoy the space in their molecular 'neighborhood'. Analyzing electron distribution is key in predicting how a molecule will interact with others and how it behaves under different conditions.
Chemical Bonding
At the heart of chemistry lies chemical bonding, the force that holds atoms together in molecules. There are several types of chemical bonds - ionic, covalent, metallic, and more. π bonds are a type of covalent bond formed by the side-by-side overlap of atomic orbitals. Understanding the different types of chemical bonds and how they form is fundamental to grasping how atoms combine to make molecules with various shapes and functions.

Everything from the water we drink to the medicines we take relies on chemical bonding. Whether bonds are localized, contributing to a molecule's rigid structure, or delocalized, providing flexibility and stability, they are crucial in determining the properties and behavior of the substances around us.

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

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+}\), and \(\mathrm{AlF}_{4}^{-}\). (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electrondomain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry? (d) Which of the ions will exhibit a see-saw molecular geometry?

(a) What does the term diamagnetism mean? (b) How does a diamagnetic substance respond to a magnetic field? (c) Which of the following ions would you expect to be diamagnetic: \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \mathrm{Be}_{2}{ }^{2+}, \mathrm{C}_{2}{ }^{-}\)?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\), (b) \(\mathrm{CH}_{3}^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\), that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

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.

Sign-up for free