/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Describe the type of bonding tha... [FREE SOLUTION] | 91影视

91影视

Log out

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 F鈧�(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 F鈧�(g) molecule

In a F鈧�(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 F鈧�(g) and HF(g) molecules, let's compare them. 1. Similarity: Both F鈧�(g) and HF(g) molecules have covalent bonds, which involve the sharing of electron pairs between atoms. 2. Difference: The covalent bond in F鈧�(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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

An ionic compound made from the metal \(M\) and the diatomic gas \(X_{2}\) has the formula \(M_{a} X_{b},\) in which \(a=1\) or 2 and \(b=1\) or \(2 .\) Use the data provided to determine the most likely values for \(a\) and \(b,\) along with the most likely charges for each of the ions in the ionic compound. Data (in units of \(\mathrm{kJ} / \mathrm{mol} )\) Successive ionization energies of \(\mathrm{M} : 480 ., 4750\) . Successive electron affinity values for \(\mathrm{X} :-175,920\) . Enthalpy of sublimation for \(\mathrm{M}(s) \rightarrow \mathrm{M}(g) : 110\) . Lattice energy for MX \(\left(\mathrm{M}^{+} \text { and } \mathrm{X}^{-}\right) :-1200\) . Lattice energy for \(\mathrm{MX}_{2}\left(\mathrm{M}^{2+} \text { and } \mathrm{X}^{-}\right) :-3500\) . Lattice energy for \(\mathrm{M}_{2} \mathrm{X}\left(\mathrm{M}^{+} \text { and } \mathrm{X}^{2-}\right) :-3600\) . Lattice energy for \(\mathrm{MX}\left(\mathrm{M}^{2+} \text { and } \mathrm{X}^{2-}\right) :-4800\) .

The compound \(\mathrm{NF}_{3}\) is quite stable, but \(\mathrm{NCl}_{3}\) is very unstable \(\mathrm{(NCl}_{3}\) was first synthesized in 1811 by \(\mathrm{P}\) . L. Dulong, who lost three fingers and an eye studying its properties). The compounds \(\mathrm{NBr}_{3}\) and \(\mathrm{NI}_{3}\) are unknown, although the explosive compound \(\mathrm{NI}_{3} \cdot \mathrm{NH}_{3}\) is known. Account for the instability of these halides of nitrogen.

The structure of \(\mathrm{TeF}_{5}-\) is Draw a complete Lewis structure for \(\mathrm{TeF}_{5}-\), and explain the distortion from the ideal square pyramidal structure. (See Exercise 116.)

Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being 鈥減ulled鈥� from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

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.