Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 20

Describe the type of bonding that exists in the \(\mathrm{F}_{2}(g)\) molecule. How does this type of bonding differ from that found in the \(\mathrm{HF}(g)\) molecule? How is it similar?

Short Answer

Expert verified
In the F₂(g) molecule, a nonpolar covalent bond exists as two fluorine atoms share a pair of electrons equally due to their same electronegativity value. In contrast, the HF(g) molecule forms a polar covalent bond as electrons are not shared equally between the hydrogen and fluorine atoms due to the significant electronegativity difference. Both molecules possess covalent bonds but differ in the electron distribution, resulting in nonpolar and polar covalent bonds respectively.

Step by step solution

01

Identify bonding in F₂(g) molecule

In a F₂(g) molecule, there are two fluorine atoms sharing one pair of electrons that creates a bond between them. This type of bond is called a covalent bond. Since both fluorine atoms have the same electronegativity value (3.98), the electrons are shared equally between them. This results in a nonpolar covalent bond.
02

Identify bonding in HF(g) molecule

In an HF(g) molecule, there is one hydrogen atom and one fluorine atom. These two atoms share a pair of electrons, creating a bond between them. This bond is also a covalent bond. However, due to the high electronegativity difference between fluorine (3.98) and hydrogen (2.20), the electrons are not shared equally. Fluorine attracts the electrons more towards itself, resulting in a polar covalent bond.
03

Compare the bonding types

Now that we have identified the bonding types in F₂(g) and HF(g) molecules, let's compare them. 1. Similarity: Both F₂(g) and HF(g) molecules have covalent bonds, which involve the sharing of electron pairs between atoms. 2. Difference: The covalent bond in F₂(g) is a nonpolar covalent bond as the bonding electrons are shared equally between the two fluorine atoms. On the other hand, the covalent bond in HF(g) is a polar covalent bond due to the unequal sharing of electrons with fluorine attracting the bonding electrons more towards itself because of its higher electronegativity value compared to hydrogen.

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

Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: $\mathrm{N}(g), 472.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}(g),\( \)216.0 \mathrm{kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{kJ} / \mathrm{mol}$ . Compare your value to the one in Table \(8.5 .\)

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. PF_. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2},\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order) \(-2631,-5359,\) and \(-14,774 \mathrm{kJ} / \mathrm{mol}\) . Match the appropriate formula to each lattice energy. Explain.

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. \(\mathrm{Ne} \quad\) c. \(\mathrm{Kr}\) b. \(\mathrm{Ar} \quad\) d. \(\mathrm{Xe}\)

Write the Lewis structure for $\mathrm{O}_{2} \mathrm{F}_{2} \quad\left(\mathrm{O}_{2} \mathrm{F}_{2}\right.$ exists as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F} )\). Assign oxidation states and formal charges to the atoms in O2F2. This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\)?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

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