Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 61

Consider the Lewis structure for acetic acid, which is known as vinegar: CCC(=O)O (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygen atoms, and what are the approximate bond angles at the oxygen that is connected to carbon and hydrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

Short Answer

Expert verified
\(a)\) The approximate bond angles around Carbon-1 and Carbon-2 are \(109.5^\circ\) and \(120^\circ\), respectively. The hybridizations of orbitals on Carbon-1 and Carbon-2 are \(sp^3\) and \(sp^2\), respectively. \(b)\) The hybridizations of orbitals on Oxygen-1 and Oxygen-2 are \(sp^3\) and \(sp^2\), respectively. The approximate bond angle at Oxygen-1 (connected to carbon and hydrogen) is \(109.5^\circ\). \(c)\) The total number of \(\sigma\) bonds in the entire molecule is 7, and the total number of \(\pi\) bonds is 1.

Step by step solution

01

Draw the Lewis Structure of Acetic Acid

We start by drawing the Lewis structure for acetic acid: H H | | H - C - C = O | | O O | H
02

Determine Bond Angles and Hybridization of Orbitals on Carbon Atoms

First, let's label the carbon atoms. We can refer to the carbon atom connected to three hydrogen atoms as Carbon-1 (C1) and the other carbon atom as Carbon-2 (C2). For each carbon atom, we'll determine the bonding electron domains and identify the hybridization. Carbon-1 (C1): - It has three bonding electron domains (3 single bonds to 3 hydrogen atoms and 1 single bond to Carbon-2). - Hybridization: \(sp^3\) - Approximate bond angle: \(109.5^\circ\) (tetrahedral) Carbon-2 (C2): - It has three bonding electron domains (1 single bond to Carbon-1, 1 single bond to Oxygen, and 1 double bond to another Oxygen atom). - Hybridization: \(sp^2\) - Approximate bond angle: \(120^\circ\) (trigonal planar)
03

Determine Bond Angles and Hybridization of Orbitals on Oxygen Atoms

Next, let's label the oxygen atoms. We'll refer to the oxygen atom connected to Carbon-2 and hydrogen as Oxygen-1 (O1) and the other oxygen atom as Oxygen-2 (O2). Oxygen-1 (O1): - It has three bonding electron domains (1 single bond to Carbon-2, 1 single bond to Hydrogen, and 1 lone pair). - Hybridization: \(sp^3\) - Approximate bond angle (between C2-O1-H): \(109.5^\circ\) Oxygen-2 (O2): - It has three bonding electron domains (1 double bond to Carbon-2 and 1 lone pair). - Hybridization: \(sp^2\) - The bond angles are not required for this oxygen; only those connected to both carbon and hydrogen are requested.
04

Count Total Number of σ Bonds and π Bonds

Counting the bonds in the Lewis structure for acetic acid: Total number of \(\sigma\) bonds: - 4 single bonds from C1 - 2 single bonds from C2 - 1 single bond from O1 Total: 7 \(\sigma\) bonds Total number of \(\pi\) bonds: - 1 double bond from C2, which contains 1 \(\pi\) bond Total: 1 \(\pi\) bond Now we have all the information to answer the questions in the exercise: (a) The approximate bond angles around Carbon-1 and Carbon-2 are \(109.5^\circ\) and \(120^\circ\), respectively. The hybridizations of orbitals on Carbon-1 and Carbon-2 are \(sp^3\) and \(sp^2\), respectively. (b) The hybridizations of orbitals on Oxygen-1 and Oxygen-2 are \(sp^3\) and \(sp^2\), respectively. The approximate bond angle at Oxygen-1 (connected to carbon and hydrogen) is \(109.5^\circ\). (c) The total number of \(\sigma\) bonds in the entire molecule is 7, and the total number of \(\pi\) bonds is 1.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}:\) The compound on the right is called cisplatin, and the compound on the left is called transplatin. (a) Which compound has a nonzero dipole moment? (b) One of these compounds is an anticancer drug, and one is inactive. The anticancer drug works by its chloride ions undergoing a substitution reaction with nitrogen atoms in DNA that are close together, forming a \(\mathrm{N}-\mathrm{Pt}-\mathrm{N}\) angle of about \(90^{*} .\) Which compound would you predict to be the anticancer drug?

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMOLUMO gap; see Exercise 9.104 ). Suppose you have two samples, one is lycopene which is responsible for the red color in tomato, and the other is curcumin which is responsible for the yellow color in turmeric. Which one has the larger HOMO-LUMO gap?

Explain the following: (a) The peroxide ion, \(\mathrm{O}_{2}^{2-}\), has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 \mathrm{p}}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger O- \(O\) bond than \(\mathrm{O}_{2}\) itself.

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{BeF}_{2}\), (b) \(\mathrm{AsCl}_{5}\), (c) \(\mathrm{NO}_{2}^{-}\), (e) \(\mathrm{SF}_{4},(\mathbf{f}) \mathrm{BrF}_{s-}\) (d) \(\mathrm{CS}_{2}\)

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2}^{*}\), orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write ${ }^{4} \mathrm{M}^{\prime \prime}$ at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(\mathrm{M}\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The CO bond axis should be on the \(x\) -axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of $\mathrm{CO} ?\( (e) What kind of bond is being made with the orbitals between \)\mathrm{M}$ and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation
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