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Chapter 6: Problem 38

Name the following compounds, draw their Lewis structures, and determine their shapes. a. \(\mathrm{SiCl}_{4}\) b. \(\mathrm{BCl}_{3}\) c. \(\mathrm{NBr}_{3}\)

Short Answer

Expert verified
a) Silicon Tetrachloride (\(\mathrm{SiCl}_{4}\)), Lewis structure has Si at center with Cl atoms surrounding it, the shape is tetrahedral. b) Boron Trichloride (\(\mathrm{BCl}_{3}\)), Lewis structure has B at center with Cl atoms surrounding it, the shape is trigonal planar. c) Nitrogen Tribromide (\(\mathrm{NBr}_{3}\)), Lewis structure has N at center and Br atoms surrounding it, the shape is pyramidal.

Step by step solution

01

Naming Compounds

Naming each compound is quite straightforward. a. Silicon Tetrachloride (\(\mathrm{SiCl}_{4}\)) b. Boron Trichloride (\(\mathrm{BCl}_{3}\)) c. Nitrogen Tribromide (\(\mathrm{NBr}_{3}\))
02

Drawing Lewis Structures

a. For \(\mathrm{SiCl}_{4}\), Silicon (Si) is the central atom surrounded by four Chlorine atoms (Cl). Si has 4 valence electrons and each Cl has 7, thus making a complete octet for each atom after sharing. b. For \(\mathrm{BCl}_{3}\), Boron (B) is the central atom surrounded by three Chlorine atoms (Cl). B has 3 valence electrons and each Cl has 7, allowing for a full octet for each Cl after sharing. B has satisfied the octet rule as it can hold up to 6 electrons. c. For \(\mathrm{NBr}_{3}\), Nitrogen (N) is the central atom surrounded by three Bromine atoms (Br). N has 5 valence electrons and each Br has 7, making a complete octet for each atom after sharing.
03

Determining Molecular Shapes

a. The shape of \(\mathrm{SiCl}_{4}\) is tetrahedral since it has four bonded groups and no lone pairs. b. The shape of \(\mathrm{BCl}_{3}\) is trigonal planar as it has three bonded groups and no lone pairs. c. The shape of \(\mathrm{NBr}_{3}\) is pyramidal given it has three bonded groups and one lone pair on the Nitrogen atom.

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