Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 74

Draw a plausible Lewis structure for the following series of molecules and ions: \((a) \operatorname{SiF}_{6}^{2-} ;\) (b) \(\mathrm{PF}_{5} ;\) (c) \(\mathrm{SF}_{4}\); (d) \(\mathrm{XeF}_{4}\). Describe the electron group geometry and molecular structure of these species.

Short Answer

Expert verified
The Lewis structures are drawn for \(\operatorname{SiF}_{6}^{2-}\), \(PF_{5}\), \(SF_{4}\), and \(XeF_{4}\) according to the octet rule while considering the formal charge. The electron density around the central atoms gives the following geometries: octahedral for \(\operatorname{SiF}_{6}^{2-}\), trigonal bipyramidal for \(PF_{5}\), see-saw for \(SF_{4}\), and square planar for \(XeF_{4}\). As we progress down in the series, the electron group geometry remains the same - octahedral, but the molecular structure changes due to the presence of lone pairs of electrons.

Step by step solution

01

Draw Lewis Structure for \(\operatorname{SiF}_{6}^{2-}\)

Count total number of valence electrons. Silicon has 4 valence electrons and each Fluorine has 7. Add 2 for the 2- charge. So the total is 4+(6*7)+2=46. Draw Silicon in the center and six Fluorine atoms around it. Single bond each Fluorine atom to the Silicon and fill the rest electrons on Fluorine till it completes octet rule. Electron pairs are arranged around silicon in an octahedral.
02

Draw Lewis Structure for \(\mathrm{PF}_{5}\)

Count total number of valence electrons. Phosphorus has 5 and each Fluorine has 7. So the total is 5+(5*7)=40. Place the P in the center and five Fluorine atoms around it. Single bond each Fluorine atom to the Phosphorus. Fill rest of electrons on Fluorine till it completes octet rule. Electron pairs are arranged around phosphorous in a trigonal bipyramidal structure.
03

Draw Lewis Structure for \(\mathrm{SF}_{4}\)

Count total number of valence electrons. Sulfur has 6 and each Fluorine has 7. So the total is 6+(4*7)=34. Place the Sulfur in center and four Fluorine atoms around it. Single bond each Fluorine atom to the Sulfur. Electron pairs are arranged around sulfur in a see-saw structure.
04

Draw Lewis Structure for \( \mathrm{XeF}_{4}\)

Count total number of valence electrons. Xenon has 8 and each Fluorine has 7. So the total is 8+(4*7)=36. Place the Xenon in center and four Fluorine atoms around it. Single bond each Fluorine atom to the Xenon. Two remaining pairs of electrons on Xenon atom. Electron pairs are arranged around Xenon in a square planar structure.

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

Each of the following molecules contains at least one multiple (double or triple) covalent bond. Give a plausible Lewis structure for (a) \(\mathrm{OCS} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{F}_{2} \mathrm{CO} ;\) (d) \(\mathrm{Cl}_{2} \mathrm{SO} ;\) (e) \(\mathrm{C}_{2} \mathrm{H}_{2}\).

Carbon suboxide has the formula \(\mathrm{C}_{3} \mathrm{O}_{2} .\) The carbon- to-carbon bond lengths are \(130 \mathrm{pm}\) and carbon-to-oxygen, \(120 \mathrm{pm} .\) Propose a plausible Lewis structure to account for these bond lengths, and predict the shape of the molecule.

A \(0.325 \mathrm{g}\) sample of a gaseous hydrocarbon occupies a volume of \(193 \mathrm{mL}\) at \(749 \mathrm{mmHg}\) and \(26.1^{\circ} \mathrm{C}\). Determine the molecular mass, and write a plausible condensed structural formula for this hydrocarbon.

Alternative strategies to the one used in this chapter have been proposed for applying the VSEPR theory to molecules or ions with a single central atom. In general, these strategies do not require writing Lewis structures. In one strategy, we write (1) the total number of electron pairs \(=[\) (number of valence electrons) \(\pm\) (electrons required for ionic charge) \(] / 2\) (2) the number of bonding electron pairs \(=\) (number of atoms) -1 (3) the number of electron pairs around central atom \(=\) total number of electron pairs \(-3 \times[\) number of terminal atoms (excluding \(\mathrm{H}\) )] (4) the number of lone-pair electrons = number of central atom pairs - number of bonding pairs After evaluating items \(2,3,\) and \(4,\) establish the VSEPR notation and determine the molecular shape. Use this method to predict the geometrical shapes of the following: (a) \(\mathrm{PCl}_{5} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{ClF}_{3} ;\) (d) \(\mathrm{SO}_{2} ;\) (e) \(\mathrm{ClF}_{4}^{-}\); (f) \(\mathrm{PCl}_{4}^{+}\). Justify each of the steps in the strategy, and explain why it yields the same results as the VSEPR method based on Lewis structures. How does the strategy deal with multiple bonds?

The molecular shape of \(\mathrm{BF}_{3}\) is planar (see Table 10.1 ). If a fluoride ion is attached to the \(B\) atom of \(B F_{3}\) through a coordinate covalent bond, the ion \(\mathrm{BF}_{4}^{-}\) results. What is the shape of this ion?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

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