Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Draw the Lewis structures and describe the geometry for the following:

(a) \({\rm{P}}{{\rm{F}}_4} + \)

(b) \({\rm{P}}{{\rm{F}}_5}\)

(c) \({\rm{P}}{{\rm{F}}_6}^ - \)

(d) \({\rm{PO}}{{\rm{F}}_3}\)

Short Answer

Expert verified

a) \({\rm{P}}{{\rm{F}}_4} + \):

b) \({\rm{P}}{{\rm{F}}_5}\):

c) \({\rm{P}}{{\rm{F}}_6}^-\):

d) \({\rm{PO}}{{\rm{F}}_3}\):

Step by step solution

01

Definition of Lewis structure.

A Lewis structure is a diagram that depicts a molecule's covalent bonds and lone electron pairs.

02

Write the Lewis structure.

Count all of a molecule's valent electrones to draw the Lewis structures:

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 4 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 33\end{array}\)

To fill the octet, each atom should contain \(8\) electrons; each bond counts as two electrons, while the rest are lone electron pairs.

The \(PF_4^ + \) has a form of tetradar.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 5 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 40\end{array}\)

The \(P{F_5}\) has trigonal bipyramidal form.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 6 \cdot 7 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 47\end{array}\)

The \(PF_6^ - \) has octahedral form.

\(\begin{array}{l}N\left( {{e^ - }} \right) = N\left( {F,{e^ - }} \right) + N\left( { + ,{e^ - }} \right) + N\left( {P,{e^ - }} \right)\\N\left( {{e^ - }} \right) = 3 \cdot 7 + 1 \cdot 6 + 1 \cdot 5\\N\left( {{e^ - }} \right) = 32\end{array}\)

The \({\rm{PO}}{{\rm{F}}_3}\) has tetrahedral form.

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

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