/*! 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 6 When an ionic bond is formed, wh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 6

When an ionic bond is formed, what type of atom (in terms of relative electronegativity) is likely to: a) lose one or more electrons? b) gain one or more electrons? Explain your reasoning.

Short Answer

Expert verified
In an ionic bond, a less electronegative atom (usually a metal) is likely to lose one or more electrons, while a more electronegative atom (usually a non-metal) is likely to gain one or more electrons. This process forms cations and anions respectively.

Step by step solution

01

Understanding Electronegativity

Electronegativity is a measure of how strongly an atom attracts a bonding pair of electrons towards itself. In the periodic table, electronegativity increases from bottom to top and from left to right.
02

Identify the Atom That Loses Electron(s)

An atom that loses one or more electrons is likely to be less electronegative. Electronegative atoms tend to attract and retain more electrons. In comparison, less electronegative atoms (usually metals) find it energetically favorable to lose electrons and become positively charged ions (cations). Examples include sodium (Na) and potassium (K).
03

Identify the Atom That Gains Electron(s)

An atom that gains one or more electrons during an ionic bond formation is likely to have higher electronegativity. Highly electronegative atoms (usually non-metals) attract electrons and form negatively charged ions (anions). Examples include fluorine (F) and oxygen (O).

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

Describe the trend in EN as one moves down a group of the periodic table.

For \(\mathrm{HF}\) and \(\mathrm{HBr}, \delta_{\mathrm{H}}=0.29\) and \(0.09\), respectively. Use electronegativities to explain why the partial charge on \(\mathrm{H}\) in \(\mathrm{HF}\) is more positive than the partial charge on \(\mathrm{H}\) in \(\mathrm{HBr}\).

Why do homonuclear molecules \(\left(\mathrm{H}_{2}, \mathrm{Cl}_{2}, \mathrm{~N}_{2}\right.\), and so on) have nonpolar bonds?

Describe the trend in EN as one moves from left to right across a period of the periodic table.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

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.