/*! 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 71 Draw the Lewis structure for the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 71

Draw the Lewis structure for the \(\mathrm{BeCl}_{4}^{2-}\) ion. Predict its geometry and describe the hybridization state of the Be atom.

Short Answer

Expert verified
The Lewis structure of \(\mathrm{BeCl}_{4}^{2-}\) has a Be atom in the center surrounded by 4 Cl atoms. Each Be-Cl has a single bond and the Be atom carries two extra electrons. The molecule has a tetrahedral geometry. The Be atom's hybridization state is sp3d.

Step by step solution

01

Count Valence Electrons

The atom Be has 2 valence electrons and each atom Cl has 7 valence electrons. In \(\mathrm{BeCl}_{4}^{2-}\), there are 4 Cl atoms. There are also 2 additional electrons because of the '2-' charge. Thus, we have \(2 + (4 \cdot 7) + 2 = 30 \) valence electrons.
02

Draw the Lewis Structure

Berillium (Be) as the least electronegative atom becomes the central atom. Leaving Be in the center, organize the 4 Cl atoms around it. Be, being in the 2nd period of the periodic table, can expand its octet to bond with the 4 Cl atoms. Draw a single bond between the Be atom and each Cl atom using a pair of electrons. Each Cl atom still has 6 valence electrons left, place them around each Cl atom (3 lone pairs around each). The total electrons used now equals \(4\cdot 2 + 3\cdot 6\cdot 4 = 28\), which is less than 30. Hence, place the two remaining electrons on Be.
03

Predict the Geometry

Apply the VSEPR theory to predict the geometry. Be has 4 bond pairs and no lone pair, hence the geometry will be tetrahedral according to VSEPR.
04

Describe the Hybridization State

Hybridization state can be calculated using the hybridization trigonal formula which is number of bond pairs + number of lone pairs on the central atom. In this case therefore, there are 4 bond pairs and 1 lone pair on the central Be atom. This adds up to 5, which implies sp3d hybridization (s represents one orbital, and each p and d represent three and five orbitals respectively, thus total five orbitals imply sp3d hybridization).

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

Draw the Lewis structure of ketene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}\right)\) and describe the hybridization states of the \(\mathrm{C}\) atoms. The molecule does not contain \(\mathrm{O}-\mathrm{H}\) bonds. On separate diagrams, sketch the formation of sigma and pi bonds.

Describe the bonding scheme of the \(\mathrm{AsH}_{3}\) molecule in terms of hybridization.

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) has a tendency to lose two protons \(\left(\mathrm{H}^{+}\right)\) and form the carbide ion \(\left(\mathrm{C}_{2}^{2-}\right),\) which is present in a number of ionic compounds, such as \(\mathrm{CaC}_{2}\) and \(\mathrm{MgC}_{2}\). Describe the bonding scheme in the \(\mathrm{C}_{2}^{2}-\) ion in terms of molecular orbital theory. Compare the bond order in \(\mathrm{C}_{2}^{2-}\) with that in \(\mathrm{C}_{2}\)

Assume that the third-period element phosphorus forms a diatomic molecule, \(\mathrm{P}_{2}\), in an analogous way as nitrogen does to form \(\mathrm{N}_{2}\). (a) Write the electronic configuration for \(\mathrm{P}_{2}\). Use \(\left[\mathrm{Ne}_{2}\right]\) to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Describe the hybridization state of arsenic in arsenic pentafluoride \(\left(\mathrm{AsF}_{5}\right)\).
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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.