Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 72

Draw the Lewis structure of mercury(II) bromide. Is this molecule linear or bent? How would you establish its geometry?

Short Answer

Expert verified
The Lewis structure of HgBr2 shows mercury donating one electron each to two bromine atoms, making the Bromine complete their octet and achieving a stable state. There are no lone pairs on the mercury atom, and so, based on VSEPR theory, the molecule is linear.

Step by step solution

01

Determine the electron configuration

Start by determining the electron configuration of the atoms in the molecule. Mercury (Hg) is in the 6th period of the periodic table and has an electron configuration of [Xe] 4f14 5d10 6s2. Bromine (Br), located in the 5th period of the periodic table has an electron configuration of [Ar] 3d10 4s2 4p5. For the Lewis structure, we'll particularly focus on the outermost (valence) electrons.
02

Draw the Lewis structure

Mercury(II) bromide (HgBr2) is an ionic compound. Mercury, a metal, donates its two valence electrons to two bromine atoms, each gaining one electron to complete their octets. The Lewis structure represents this by a single line (representing two electrons) between Hg and each Br atom, and pairs of dots around each Br atom to represent the filled p-orbital. The Hg is left with no dots as it has lost its valence electrons.
03

Determine the molecule geometry

The molecule geometry can be established using the Lewis structure. As there are two Br atoms attached directly to the central Hg atom and no lone pairs on the central atom, the geometry of the molecule is linear.

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

The formation of \(\mathrm{H}_{2}\) from two \(\mathrm{H}\) atoms is an energetically favorable process. Yet statistically there is less than a 100 percent chance that any two \(\mathrm{H}\) atoms will undergo the reaction. Apart from energy considerations, how would you account for this observation based on the electron spins in the two \(\mathrm{H}\) atoms?

The compounds carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) are similar in geometry and hybridization. However, \(\mathrm{CCl}_{4}\) does not react with water but \(\mathrm{SiCl}_{4}\) does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the \(\mathrm{Si}\) atom in \(\left.\mathrm{SiCl}_{4} .\right)\)

Predict the geometries of the following ions: (a) \(\mathrm{NH}_{4}^{+}\) (b) \(\mathrm{NH}_{2}^{-},\) (c) \(\mathrm{CO}_{3}^{2-},\) (d) \(\mathrm{ICl}_{2}^{-}\), (e) \(\mathrm{ICl}_{4}^{-}\) (f) \(\mathrm{AlH}_{4}^{-},(\mathrm{g}) \mathrm{SnCl}_{5}^{-}\) (h) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (i) \(\mathrm{BeF}_{4}^{2-}\).

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.

Compare the Lewis and molecular orbital treatments of the oxygen molecule.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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