Use the localized electron model to describe the bonding in \(\mathrm{CCl}_{4}\).

Short Answer

Expert verified
The bonding in CCl4 can be described using the localized electron model as follows: CCl4 has a tetrahedral electron-pair geometry and molecular geometry. The bonds between the carbon atom and each of the four chlorine atoms are sigma (σ) bonds, formed by the overlap of an sp3 hybrid orbital from the carbon atom and a singly occupied atomic p-orbital from each chlorine atom. There are no pi (π) bonds in CCl4.

Step by step solution

01

Drawing the Lewis structure of CCl4

First, draw the Lewis structure of CCl4. The Lewis structure is a representation of the molecule showing the arrangement of atoms and the distribution of valence electrons. To draw the Lewis structure for CCl4: 1. Find the number of valence electrons for each atom. Carbon has 4 valence electrons, and each Chlorine atom has 7 valence electrons. 2. Combine all valence electrons (4 for carbon + 7x4 for four chlorines = 32 electrons). 3. Place the carbon atom in the center and arrange the four chlorine atoms around it. 4. Distribute the valence electrons following the octet rule: 8 electrons for each chlorine atom (forming single bonds with the carbon atom) and 8 electrons for the carbon atom (forming one bond with each chlorine atom). 5. Check that each atom has a complete octet.
02

Determining the electron-pair geometry

Next, determine the electron-pair geometry of CCl4. The electron-pair geometry is the arrangement of electron pairs around the central atom. In this case, the central atom is carbon, and it is surrounded by four electron pairs, each of them shared with a chlorine atom. According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, these electron pairs will repel each other and be arranged as far away from each other as possible, resulting in a tetrahedral geometry.
03

Determining the molecular geometry

Now, identify the molecular geometry. Molecular geometry is the arrangement of atoms in a molecule. Since the carbon atom in CCl4 is surrounded by four chlorine atoms and has no lone pairs, the molecular geometry is similar to the electron-pair geometry, also tetrahedral.
04

Describing the sigma and pi bonding

Finally, describe the sigma (σ) and pi (π) bonding in CCl4 using the localized electron model. In the localized electron model, bonds are described as either sigma (head-on overlap of atomic orbitals) or pi (side-on overlap of atomic orbitals). In CCl4, the carbon atom has four single bonds with chlorine atoms. In each C-Cl bond, an sp3 hybrid orbital from the carbon atom and a singly occupied atomic p-orbital from the chlorine atom overlap, resulting in the formation of four sigma (σ) bonds. There are no pi (π) bonds in CCl4. In conclusion, the localized electron model helps describe the bonding in CCl4 as follows: The bonds are formed by four sigma (σ) bonds, with the electron-pair geometry and molecular geometry both being tetrahedral.

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!

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

The three NO bonds in \(\mathrm{NO}_{3}^{-}\) are all equivalent in length and strength. How is this explained even though any valid Lewis structure for \(\mathrm{NO}_{3}^{-}\) has one double bond and two single bonds to nitrogen?

A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. \(\mathrm{FClO}_{3}\) is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for \(\mathrm{F}_{3} \mathrm{ClO}\) \(\mathrm{F}_{2} \mathrm{ClO}_{2}^{+},\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\). What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?

Predict the molecular structure for each of the following. (See Exercises 25 and \(26 .\) ) a. \(\mathrm{BrFI}_{2}\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) c. \(\operatorname{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Consider three molecules: \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\). Molecule A has a hybridization of \(s p^{3} .\) Molecule \(\mathrm{B}\) has two more effective pairs (electron pairs around the central atom) than molecule A. Molecule C consists of two \(\sigma\) bonds and two \(\pi\) bonds. Give the molecular structure, hybridization, bond angles, and an example for each molecule.

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

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