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Chapter 22: Problem 8

Use VSEPR theory to predict the probable geometric structures of the molecules (a) \(\mathrm{O}_{2} \mathrm{XeF}_{2} ;\) (b) \(\mathrm{O}_{3} \mathrm{XeF}_{2}\) (c) OXeF \(_{4}\).

Short Answer

Expert verified
\(\mathrm{O}_{2} \mathrm{XeF}_{2}\) has a square planar geometry, \(\mathrm{O}_{3} \mathrm{XeF}_{2}\) has a square pyramidal geometry, and OXeF \(_{4}\) is octahedral.

Step by step solution

01

Identifying the Central Atom

Identify the central atom in each molecule. This is usually the atom with the least electronegativity or the single unique atom in the molecules. In \(\mathrm{O}_{2} \mathrm{XeF}_{2}\), \(\mathrm{O}_{3} \mathrm{XeF}_{2}\), and OXeF \(_{4}\), the central atom is Xenon (Xe).
02

Determine the Electron Pair Geometry

Count the number of bonding pairs (BP) and non-bonding pairs (lone pairs, LP) of electrons around the central atom. In \(\mathrm{O}_{2} \mathrm{XeF}_{2}\), we have 4 BP and 2 LP; in \(\mathrm{O}_{3} \mathrm{XeF}_{2}\), we have 5 BP and 1 LP; and in OXeF \(_{4}\), we have 5 BP and no LP.
03

Predict the Molecular Geometry

Using VSEPR theory and the number of BP and LP, we can predict the molecular geometry: \(\mathrm{O}_{2} \mathrm{XeF}_{2}\) has a square planar geometry due to 4 BP and 2 LP around the central atom. \(\mathrm{O}_{3} \mathrm{XeF}_{2}\) has a square pyramidal geometry due to 5 BP and 1 LP around the central atom. OXeF \(_{4}\) is octahedral because it has 6 BP (5 bonded with F and 1 bonded with O) around the central atom and no LP.

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Most popular questions from this chapter

Without per frming detailed calculations, determine which of the following compounds has the greatest percent oxygen by mass: dinitrogen tetroxide, aluminum oxide, tetraphosphorus hexoxide, or carbon dioxide.

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