Use VSEPR theory to predict the probable geometric structures of (a) \(\mathrm{XeO}_{3} ;\) (b) \(\mathrm{XeO}_{4} ;\) (c) \(\mathrm{XeF}_{5}^{+}\).

Short Answer

Expert verified
Using the VSEPR theory: (a) \(\mathrm{XeO}_{3}\) predicts a trigonal pyramidal structure. (b) \(\mathrm{XeO}_{4}\) predicts a tetrahedral structure. (c) \(\mathrm{XeF}_{5}^{+}\) predicts a square pyramidal structure.

Step by step solution

01

Determine the Total Number of Electrons

Count the total number of valence electrons around the central atom (Xe). For \(\mathrm{XeO}_{3}\), the count is 8 (from Xe) + 3(6) (from each O) = 26 electrons. For \(\mathrm{XeO}_{4}\), the count is 8 (from Xe) + 4(6) (from each O) = 32 electrons. For \(\mathrm{XeF}_{5}^{+}\), the count is 8 (from Xe) + 5(7) (from each F) = 43 electrons, but we subtract one electron due to the +1 charge for a total of 42 electrons.
02

Place Electrons and Determine the Steric Number

For \(\mathrm{XeO}_{3}\), placing the electrons, we have 1 bond with each O (6 electrons), and one lone pair on Xe (2 electrons). This gives us a steric number of 4. For \(\mathrm{XeO}_{4}\), we have one bond with each O (8 electrons), and no lone pairs giving us a steric number of 4. For \(\mathrm{XeF}_{5}^{+}\), we have one bond with each F (10 electrons), and one lone pair on Xe (2 electrons). This gives us a steric number of 6.
03

Predict the Shape

Using the steric number, we can predict the shape. For \(\mathrm{XeO}_{3}\) with a steric number of 4 and 1 lone pair, the shape is trigonal pyramidal. For \(\mathrm{XeO}_{4}\) with a steric number of 4 and no lone pairs, the shape is tetrahedral. For \(\mathrm{XeF}_{5}^{+}\) with a steric number of 6 and 1 lone pair, the shape is square pyramidal.

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 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

Give an appropriate formula for each of the following compounds: (a) calcium sulfate dihydrate; (b) hydrosulfuric acid; (c) sodium hydrogen sulfate; (d) disulfuric acid.

In water, \(\mathrm{O}^{2-}\) is a strong base. If \(50.0 \mathrm{mg}\) of \(\mathrm{Li}_{2} \mathrm{O}\) is dissolved in \(750.0 \mathrm{mL}\) of aqueous solution, what will be the pH of the solution?

Give a practical laboratory method that you might use to produce small quantities of the following gases and comment on any difficulties that might arise: (a) \(\mathrm{O}_{2} ;\) (b) \(\mathrm{NO} ;\) (c) \(\mathrm{H}_{2} ;\) (d) \(\mathrm{NH}_{3} ;\) (e) \(\mathrm{CO}_{2}\).

Use the following electrode potential diagram for basic solutions to classify each of the statements below as true or false. Assume standard conditions. $$\begin{aligned}\mathrm{SO}_{4}^{2-} \stackrel{-0.936 \mathrm{V}}{\longrightarrow} \mathrm{SO}_{3}^{2-} & \stackrel{-0.576 \mathrm{V}}{\longrightarrow} \\\& \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \stackrel{-0.74 \mathrm{V}}{\longrightarrow} \mathrm{S} \stackrel{-0.476 \mathrm{V}}{\longrightarrow} \mathrm{S}^{2-}\end{aligned}$$ (a) Sulfate \(\left(\mathrm{SO}_{4}^{2-}\right)\) is a stronger oxidant than thiosulfate \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\right)\) in basic solution. (b) \(S^{2-}\) can be used as a reducing agent in basic solutions. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution.

What mass of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) was present in a sample that required \(26.50 \mathrm{mL}\) of \(0.0510 \mathrm{M} \mathrm{KMnO}_{4}\) for its oxidation to \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in an acidic solution? \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

See all solutions

Recommended explanations on Chemistry Textbooks

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