/*! 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} Problem 6 The geometry of \(\mathrm{CH}_{4... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 6

The geometry of \(\mathrm{CH}_{4}\) could be square planar, with the four \(\mathrm{H}\) atoms at the corners of a square and the \(\mathrm{C}\) atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral \(\mathrm{CH}_{4}\) molecule.

Short Answer

Expert verified
The tetrahedral geometry of \(CH_4\) is more stable than the hypothetically proposed square planar shape due to less repulsion between bonding electron pairs resulting from larger H-C-H bond angles in the tetrahedral shape.

Step by step solution

01

Sketching the Square Planar Structure

Start by drawing the Carbon (C) atom at the center and four Hydrogen (H) atoms at the four corners of the square. Each hydrogen atom should be connected to the carbon atom by a straight line representing the bond.
02

Analyzing the Stability of the Square Planar Structure

In a square planar structure, the angle between bonds (H-C-H bond angle) will be 90 degrees. In terms of stability, there could be higher repulsion between the bonding electron pairs due to the smaller bond angles, leading to a less stable molecule.
03

Sketching and Analyzing the Tetrahedral Structure of \(CH_4\)

Draw a tetrahedron, with the Carbon atom at the center and the hydrogen atoms at the corners. The bond angles in a tetrahedral structure are approximately 109.5 degrees. This larger angle lessens electron pair repulsion, thus providing more stability.
04

Comparing the Square Planar and Tetrahedral Structure

Comparing both structures, the tetrahedral configuration, which has larger H-C-H bond angles and less electron pair repulsion, is more stable than the square planar configuration.

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 91影视!

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

Predict the geometry of sulfur dichloride \(\left(\mathrm{SCl}_{2}\right)\) and the hybridization of the sulfur atom.

The molecules cis-dichloroethylene and transdichloroethylene shown on p. 324 can be interconverted by heating or irradiation. (a) Starting with cis-dichloroethylene, show that rotating the \(\mathrm{C}=\mathrm{C}\) bond by \(180^{\circ}\) will break only the pi bond but will leave the sigma bond intact. Explain the formation of trans- dichloroethylene from this process. (Treat the rotation as two, stepwise \(90^{\circ}\) rotations.) (b) Account for the difference in the bond enthalpies for the pi bond (about \(270 \mathrm{~kJ} / \mathrm{mol}\) ) and the sigma bond (about \(350 \mathrm{~kJ} / \mathrm{mol}\) ). (c) Calculate the longest wavelength of light needed to bring about this conversion.

How would you distinguish between a sigma bond and a pi bond?

The geometries discussed in this chapter all lend themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the \(\mathrm{CCl}_{4}\) molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular \(\mathrm{C}-\mathrm{Cl}\) bond to the resultant bond moments of the other three \(\mathrm{C}-\mathrm{Cl}\) bonds in opposite directions, show that the bond angles are all equal to \(109.5^{\circ}\)

The compound 1,2 -dichloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\right)\) is nonpolar, while cis-dichloroethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\right)\) has a dipole moment: The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in \(1,2-\) dichloroethane but not in \(c i s\) -dichloroethylene.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

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.