/*! 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 41 (a) What is the general relation... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 41

(a) What is the general relationship between the size of an atom and its first ionization energy? (b) Which element in the periodic table has the largest ionization energy? Which has the smallest?

Short Answer

Expert verified
(a) The general relationship between atomic size and ionization energy is that as atomic size increases, ionization energy decreases, and as atomic size decreases, ionization energy increases. This is due to the varying attraction force between the nucleus and electrons. (b) Helium (He) has the largest ionization energy, while Francium (Fr) has the smallest ionization energy in the periodic table.

Step by step solution

01

Define ionization energy and atomic size

Ionization energy is the amount of energy required to remove an electron from a gaseous atom or ion. The general trend for ionization energy is that it increases from left to right across a period and decreases down a group in the periodic table. Atomic size is the distance between the nucleus of an atom and its outermost electrons. The general trend for atomic size is that it increases down a group and decreases from left to right across a period in the periodic table.
02

Relationship between atomic size and ionization energy

The general relationship between atomic size and ionization energy is that as atomic size increases, ionization energy decreases, and as atomic size decreases, ionization energy increases. This is because, as the atom becomes larger, electrons are further from the nucleus and experience a weaker attraction force from the nucleus. As a result, it's easier to remove an electron, and the ionization energy decreases. Conversely, as the atom becomes smaller, electrons are closer to the nucleus and experience a stronger attraction force from the nucleus, making it more difficult to remove an electron. Hence, the ionization energy increases.
03

Identify the element with the largest ionization energy.

To identify the element with the largest ionization energy, we should look for an element with the smallest atomic size, since the ionization energy and atomic size are inversely related. In the periodic table, ionization energy generally increases from left to right across a period and decreases down a group. Therefore, the element with the largest ionization energy should be located at the top-right corner of the periodic table. In this case, the element is Helium (He), which has the largest ionization energy.
04

Identify the element with the smallest ionization energy.

To identify the element with the smallest ionization energy, we should look for an element with the largest atomic size, since the ionization energy and atomic size are inversely related. In the periodic table, atomic size increases down a group and decreases from left to right across a period. Therefore, the element with the smallest ionization energy should be located at the bottom-left corner of the periodic table. In this case, the element is Francium (Fr), which has the smallest ionization energy.

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 a balanced equation for the reaction that occurs in each of the following cases: (a) Cesium is added to water. (b) Strontium is added to water. (c) Sodium reacts with oxygen. (d) Calcium reacts with iodine.

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of \(\mathrm{Xe}\). Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

(a) One of the alkali metals reacts with oxygen to form a solid white substance. When this substance is dissolved in water, the solution gives a positive test for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\). When the solution is tested in a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal? (b) Write a balanced chemical equation for the reaction of the white substance with water.

Mercury in the environment can exist in oxidation states \(0,+1\), and \(+2\). One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, \(\mathrm{X}\) rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's \(4 f\) orbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the \(1 s\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\). (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground-state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full \(+1\).

Consider \(\mathrm{S}, \mathrm{Cl}\), and \(\mathrm{K}\) and their most common ions. (a) List the atoms in order of increasing size. (b) List the ions in order of increasing size. (c) Explain any differences in the orders of the atomic and ionic sizes.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

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.