Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 95

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

Short Answer

Expert verified
The Lewis structure of BF3 has boron (B) in the center, connected to three fluorine atoms (F) with single bonds and no lone pairs on boron. Calculating the formal charges of B and F following this structure, we find that they are both 0. Despite having only 6 electrons around it, the central boron atom has a formal charge of 0, indicating that the structure is stable. Similarly, each fluorine atom follows the octet rule and has a formal charge of 0. Therefore, using formal charge arguments, we can rationalize that BF3 does not follow the octet rule because the current structure is the most stable one, with all atoms having a formal charge of 0.

Step by step solution

01

Determine the Lewis structure of BF3

First, draw the Lewis structure of the BF3 molecule. Boron (B) will be in the center, connected to three fluorine (F) atoms. Boron has 3 valence electrons, and each fluorine atom has 7 valence electrons. So, the Lewis structure will have single bonds between boron and each fluorine atom, and three lone pairs of electrons on each fluorine atom.
02

Calculate the formal charges of the atoms

To calculate the formal charges, use the formula: Formal charge = (Valence electrons) - (Non-bonding electrons + 1/2 * Bonding electrons) For boron (B): Valence electrons = 3 Non-bonding electrons = 0 (since boron has no lone pairs) Bonding electrons = 6 (since there are three single bonds) Formal charge of B = 3 - (0 + 1/2 * 6) = 0 For fluorine (F): Valence electrons = 7 Non-bonding electrons = 6 (since each fluorine atom has three lone pairs) Bonding electrons = 2 (since there is one single bond between B and each F) Formal charge of each F = 7 - (6 + 1/2 * 2) = 0
03

Analyze the formal charges to explain why the octet rule is not followed

Despite not having a complete octet (only 6 electrons around it), the central boron atom has a formal charge of 0. Meanwhile, each fluorine atom follows the octet rule and also has a formal charge of 0. Since the most stable structure is the one with the least formal charges, the current structure (single bonds between boron and each fluorine atom) is the most stable one for BF3. Thus, using formal charge arguments, it can be rationalized that BF3 does not follow the octet rule because the current structure, which does not have a complete octet for boron, is the most stable one, as all the atoms have a formal charge of 0.

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

Write Lewis structures for the following. Show all resonance structures where applicable. a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{~N}_{2} \mathrm{O}_{4}\right.\) exists as \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{NO}_{2} .\) ) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\left.\mathrm{SCN}^{-} .\right)\)

Rank the following bonds in order of increasing ionic character: \(\mathrm{N}-\mathrm{O}, \mathrm{Ca}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Br}-\mathrm{Br}, \mathrm{K}-\mathrm{F}\).

Write Lewis structures that obey the octet rule for each of the following molecules. a. \(\mathrm{CCl}_{4}\) c. \(\mathrm{SeCl}_{2}\) b. \(\mathrm{NCl}_{3}\) d. \(\mathrm{ICl}\) In each case, the atom listed first is the central atom.

Order the following species with respect to carbon-oxygen bond length (longest to shortest). $$ \mathrm{CO}, \quad \mathrm{CO}_{2}, \quad \mathrm{CO}_{3}{ }^{2-}, \quad \mathrm{CH}_{3} \mathrm{OH} $$ What is the order from the weakest to the strongest carbonoxygen bond? \(\left(\mathrm{CH}_{3} \mathrm{OH}\right.\) exists as \(\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}\).)

Write electron configurations for a. the cations \(\mathrm{Sr}^{2+}, \mathrm{Cs}^{+}, \mathrm{In}^{+}\), and \(\mathrm{Pb}^{2+}\). b. the anions \(\mathrm{P}^{3-}, \mathrm{S}^{2-}\), and \(\mathrm{Br}^{-}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

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