/*! 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 38 (a) What is the trend in electro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 38

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) True or false: The most easily ionizable elements are the most electronegative.

Short Answer

Expert verified
(a) Electronegativity increases from left to right across a row (period) in the periodic table due to decreasing atomic size and increasing effective nuclear charge experienced by valence electrons. (b) Electronegativity generally decreases going down a column (group) in the periodic table, as atomic size increases and valence electrons are added to higher energy levels further from the nucleus. (c) False. The most easily ionizable elements (i.e., group 1 alkali metals) are not the most electronegative elements, as they have low ionization energies and weak electrostatic attraction between the nucleus and their valence electrons.

Step by step solution

01

(a) Electronegativity Trend Across a Row

Electronegativity is the tendency of an atom to attract shared electrons in a chemical bond. As we move from left to right across a row (period) in the periodic table, the electronegativity of the elements increases. This is due to the decrease in atomic size and an increase in the effective nuclear charge experienced by the valence electrons, which increases the electrostatic attraction between the nucleus and the electrons.
02

(b) Electronegativity Trend Down a Column

Going down a column (group) in the periodic table, the electronegativity of the elements generally decreases. This is because the atomic size increases, and electrons are being added to higher energy levels which are further from the nucleus. The increasing distance between the nucleus and the valence electrons reduces the attractive force between them, resulting in lower electronegativity values.
03

(c) Ionization and Electronegativity Statement

"The most easily ionizable elements are the most electronegative." This statement is false. The most easily ionizable elements are found in group 1 (alkali metals), as they have low ionization energies due to their large atomic size and weak electrostatic attraction between the nucleus and their single valence electron. On the other hand, the electronegative elements such as fluorine are not easily ionizable, as they have high electronegativity values and strong attraction between their nucleus and valence electrons. Therefore, the statement is false.

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

Which of the following trends in lattice energy is due to differences in ionic radii? (a) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr}\), (b) \(\mathrm{BaO}>\mathrm{KF}\), (c) \(\mathrm{SrO}>\mathrm{SrCl}_{2}\).

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) How many bonding electrons are between the two oxygen atoms? (c) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ? Explain.

(a) Describe the molecule xenon trioxide, \(\mathrm{XeO}_{3}\), using four possible Lewis structures, one each with zero, one, two, or three Xe-O double bonds. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? (c) Do any of the four Lewis structures have multiple resonance structures? If so, how many resonance structures do you find? (d) Which of the Lewis structures in (a) yields the most favorable formal charges for the molecule?

(a) Draw the dominant Lewis structure for the phosphorus trifluoride molecule, \(\mathrm{PF}_{\mathfrak{3} .}\) (b) Determine the oxidation numbers of the \(P\) and \(F\) atoms. (c) Determine the formal charges of the \(\mathrm{P}\) and \(\mathrm{F}\) atoms.

(a) State whether the bonding in each compound is likely to be covalent or not: (i) iron, (ii) sodium chloride, (iii) water, (iv) oxygen, (v) argon. (b) A substance XY, formed from two different elements, boils at \(-33{ }^{\circ} \mathrm{C}\). Is XY likely to be a covalent or an ionic substance?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear 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.