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

Confirm that the octet rule is obeyed by each of the atoms in the following molecules: (a) \(\mathrm{CF}_{4},\) (b) \(\mathrm{O}_{2}\) , (c) AsBr \(_{3}\) (d) \(\mathrm{SF}_{2}\)

Short Answer

Expert verified
All atoms in CF₄, O₂, AsBr₃, and SF₂ obey the octet rule.

Step by step solution

01

Understand the Octet Rule

The octet rule states that atoms tend to form molecules in such a way that each atom achieves eight electrons in its valence shell, representing a full stable electron configuration similar to noble gases. Exceptions do exist, mainly for elements beyond neon.
02

Analyze CF_4

In CF extsubscript{4}, carbon has 4 valence electrons and needs 4 more to complete an octet. Each fluorine atom shares one electron with carbon. Since fluorine has 7 valence electrons, sharing one electron allows each fluorine to have 8 electrons. Thus, both carbon and fluorine obey the octet rule.
03

Analyze O_2

The O extsubscript{2} molecule has two oxygen atoms. Each oxygen atom has 6 valence electrons and can share 2 electrons with the other oxygen atom to achieve an octet. The double bond formed indicates that both oxygen atoms satisfy the octet rule.
04

Analyze AsBr_3

In AsBr extsubscript{3}, arsenic has 5 valence electrons and forms 3 bonds with bromine atoms. Each bromine atom, which requires one more electron, shares an electron with arsenic, completing an octet for each bromine. Arsenic also achieves an octet by having 8 electrons in its valence shell after bonding.
05

Analyze SF_2

In SF extsubscript{2}, sulfur has 6 valence electrons and needs 2 more electrons to complete an octet. Each fluorine atom shares one electron with sulfur. Fluorine atoms have 7 valence electrons and achieve an octet by sharing one electron with sulfur. Each atom in SF extsubscript{2} obeys the octet rule.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the electrons present in the outermost shell of an atom. They play a crucial role in chemical bonding. For many elements, these are the electrons involved when atoms interact to form molecules. The octet rule is based on the behavior of valence electrons. For example, in the molecule \(\mathrm{CF}_{4}\), carbon contributes its four valence electrons to bond with four fluorine atoms. Meanwhile, each fluorine atom brings its seven valence electrons into play. In forming bonds, these atoms arrange their valence electrons to achieve a full outer shell.

Valence electrons determine the types of bonds atoms can form and their reactivity. To understand molecules fully, knowing the number of valence electrons in atoms is essential.
  • For carbon (C), it's four valence electrons.
  • Oxygen (O) has six valence electrons.
  • Arsenic (As) has five.
  • Fluorine (F) and Bromine (Br) have seven each.
Each element's unique properties are determined by the behavior of these outer electrons.
Molecule Stability
Molecule stability is achieved when an atom obtains a full outer shell of electrons—most often conforming to the octet rule. This stability results in the reduction of energy within the molecule, making it less reactive and more energy-efficient. Consider the example of \(\mathrm{O}_2\), where two oxygen atoms share two electrons each to form a double bond.
By sharing these electrons, each oxygen atom fills its valence shell with eight electrons, thereby reaching a stable state. Additionally, the concept of stable molecules is reflected in their low reactivity under normal conditions.

Stable molecules, such as those following the octet rule, do not readily react with other elements, providing natural balancing.
  • A stable molecule distribution, like in \(\mathrm{AsBr}_{3}\), showcases each bromine atom achieving the octet through bonding with arsenic.
  • The low energy state in such configurations is what holds the molecule intact until an external force acts upon it.
Electron Sharing
Electron sharing is a fundamental aspect of covalent bonding. Atoms share electrons to fill their valence orbitals, often resulting in a stable configuration. For example, the \(\mathrm{SF}_{2}\) molecule shows sulfur sharing electrons with two fluorine atoms. Each fluorine atom shares its electrons back to complete the formation.
This sharing process is not only essential for the fulfillment of the octet rule but also for the formation of covalent bonds that hold molecules together. In \(\mathrm{O}_2\), a double bond forms when each oxygen atom shares two electrons equally with the other.
  • Such sharing allows atoms to reach a stable electronic state similar to noble gases, highlighting the efficiency of electron sharing in chemical bonding.
  • Electron sharing can be seen in single, double, or triple bonds, depending on the number of electron pairs shared.
This level of sharing contributes greatly to the diversity in chemical compounds we see.
Noble Gas Configuration
A noble gas configuration is the end goal for many atoms involved in chemical reactions. These configurations refer to the fully filled valence shells that noble gases naturally possess. For example, neon has a full 8-electron outer shell, providing the template for achieving the octet.
To emulate the noble gas configuration, atoms undergo bonding processes. In \(\mathrm{CF}_{4}\), carbon bonds with four fluorine atoms, effectively sharing electrons to match the stable configuration seen in neon, a noble gas.
  • When atoms reach this noble gas configuration, their chemical reactivity decreases significantly.
  • This level of stability extends across molecules fulfilling the octet rule, like \(\mathrm{O}_2\) and \(\mathrm{AsBr}_{3}\).
Achieving this configuration is the driving force behind chemical bonds, prompting atoms to participate in electron sharing until stability, akin to noble gases, is achieved.

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

Use the VSEPR model to predict the structures of (a) \(\mathrm{H}_{2} \mathrm{Se},\) (b) \(\left[\mathrm{BH}_{4}\right]^{-},(\mathrm{c}) \mathrm{NF}_{3},(\mathrm{d}) \mathrm{SbF}_{5}\) \((\mathrm{e})\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}\) (f) \(\mathrm{IF}_{7},(\mathrm{g})\left[\mathrm{I}_{3}\right]^{-},(\mathrm{h})\left[\mathrm{I}_{3}\right]^{+},\) (i) \(\mathrm{SO}_{3}\)

Draw Lewis structures for (a) \(\mathrm{CO}_{2},(\mathrm{b}) \mathrm{SO}_{2},(\mathrm{c}) \mathrm{OF}_{2}\) and (d) \(\mathrm{H}_{2} \mathrm{CO}\)

(a) On steric grounds, should cis- or \(\operatorname{trans}=\left[\mathrm{PtCl}_{2}\left(\mathrm{PPh}_{3}\right)_{2}\right]\) be favoured? (b) Use the VSEPR model to rationalize why SNF \(_{3}\) is tetrahedral but \(\mathrm{SF}_{4}\) is disphenoidal (c) Suggest why \(\mathrm{KrF}_{2}\) is a linear rather than bent molecule.

(a) Construct an MO diagram for the formation of \(\mathrm{O}_{2}\) show only the participation of the valence orbitals of the oxygen atoms. (b) Use the diagram to rationalize the following trend in \(\mathrm{O}-\mathrm{O}\) bond distances: \(\mathrm{O}_{2}, 121 \mathrm{pm}\) \(\left[\mathrm{O}_{2}\right]^{+}, 112 \mathrm{pm} ;\left[\mathrm{O}_{2}\right]^{-}, 134 \mathrm{pm} ;\left[\mathrm{O}_{2}\right]^{2-}, 149 \mathrm{pm} .\) (c) Which of these species are paramagnetic?

Suggest reasons for the following observations. (a) The second ionization energy of He is higher than the first despite the fact that both electrons are removed from the \(1 s\) atomic orbital. (b) Heating \(\mathrm{N}_{2} \mathrm{F}_{2}\) at \(373 \mathrm{K}\) results in a change from a non-polar to polar molecule. (c) \(\mathrm{S}_{2}\) is paramagnetic.

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