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Chapter 10: Problem 60

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 mercury(II) bromide (HgBr2) consists of a central Hg atom single bonded to two Br atoms, each surrounded by three lone pairs. The molecule is linear due to its two bonded atoms and no lone pairs around the central atom, according to the VSEPR theory.

Step by step solution

01

Understand Mercury(II) Bromide

Mercury(II) bromide, also known as mercuric bromide, is a compound with the formula HgBr2. In this compound, mercury(II) has an oxidation state of +2, and bromine has an oxidation state of -1.
02

Draw Lewis Structure

A Lewis structure represents molecular structures based on the shared electron concept of covalent bonds. For HgBr2, draw a central Hg atom with single bonds to two Br atoms. Place three lone pairs (dots representing two electrons) around each Br atom. The Lewis structure doesn't depict ionic bonds, but it helps visualize how electrons are distributed around atoms in a molecule.
03

Establish Geometry

The molecular geometry can be established based on the VSEPR (Valence Shell Electron Pair Repulsion) theory. This theory states that electron pairs in a molecule tend to orient themselves as far apart as possible to minimize repulsion. According to this theory, HgBr2 is considered to be linear because it has 2 single bonds and no lone pairs around the central atom, minimizing the electron-electron repulsion.

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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.

For each pair listed here, state which one has a higher first ionization energy and explain your choice: (a) H or \(\mathrm{H}_{2},(\mathrm{~b}) \mathrm{N}\) or \(\mathrm{N}_{2},\) (c) \(\mathrm{O}\) or \(\mathrm{O}_{2},\) (d) \(\mathrm{F}\) or \(\mathrm{F}_{2}\).

The allene molecule \(\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) is linear (the three \(\mathrm{C}\) atoms lie on a straight line). What are the hybridization states of the carbon atoms? Draw diagrams to show the formation of sigma bonds and pi bonds in allene.

What are the hybrid orbitals of the carbon atoms in these molecules? (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}\) (e) \(\mathrm{CH}_{3} \mathrm{COOH}\).

Use molecular orbital theory to explain why the \(\mathrm{Be}_{2}\) molecule does not exist
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