/*! 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 27 Identify each statement as true ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 27

Identify each statement as true or false: (a) Cations are larger than their corresponding neutral atoms. (b) \(\mathrm{Li}^{+}\) is smaller than Li. (c) \(\mathrm{Cl}^{-}\) is bigger than I \(^{-}\).

Short Answer

Expert verified
(a) False, (b) True, (c) False.

Step by step solution

01

Understanding Cations

Cations are positively charged ions formed when an atom loses one or more electrons. Typically, when an atom loses electrons, the reduction in electron repulsion among the remaining electrons allows the atom to decrease in size. Thus, cations are generally smaller than their neutral counterparts.
02

Comparing Li and Li+

When lithium (Li) loses an electron to form \( ext{Li}^+\), the resulting ion has one fewer electron but the same number of protons. The loss of electron results in a decrease in electron-electron repulsion, leading to a smaller atomic radius. Therefore, \( ext{Li}^+\) is smaller than Li.
03

Understanding Anions and Their Size

Anions are negatively charged ions formed when atoms gain electrons. This additional electron increases repulsion among the valence electrons, causing the anion to expand in size. Hence, anions are generally larger than their corresponding neutral atoms.
04

Comparing Cl- and I-

Chlorine (Cl) and iodine (I) are both halogens, but iodine is located below chlorine in the periodic table. Elements down a group are larger due to an increased number of electron shells, despite the charge. Thus, even though \( ext{Cl}^{-}\) and \( ext{I}^{-}\) both gain electrons, \( ext{I}^{-}\) remains larger due to its position on the periodic table.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cations
Cations are formed when an atom relinquishes one or more electrons, resulting in a positively charged ion. When electrons are removed, there is a reduction in the electron repulsion between the remaining electrons in the atom. This change leads to a smaller atomic structure and contributes to a decreased atomic radius.

For example, when lithium, represented as Li, loses an electron, it becomes the cation Li\(^+\). The loss of this electron reduces electron repulsion significantly, resulting in a cation that is smaller than the original neutral atom. In summary, cations are typically smaller than their parent atoms.
Anions
Anions are negatively charged ions. They form when an atom gains additional electrons. Adding these electrons causes an increase in forces of repulsion among all electrons because there are more negative charges. This repulsion forces the electrons to spread out more, expanding the size of the ion.

Because of this increase in size, anions are usually larger than their corresponding neutral atoms. For example, when a chlorine atom (Cl) gains an electron, it results in a chloride ion (Cl\(^-\)). This gained electron introduces more electron-electron repulsion, leading to a larger size compared to the neutral chlorine atom.
Atomic Radius
The atomic radius is a measure of the size of an atom from its nucleus to the outer boundary of the surrounding cloud of electrons. When an atom becomes a cation by losing electrons, the atomic radius decreases because there are fewer electrons causing repulsion in its electron cloud.

In contrast, when an atom gains electrons to form an anion, the increased electron-electron repulsion causes the electron cloud to expand, increasing the atomic radius. Thus, atomic radius is essentially affected by whether the atom has gained or lost electrons. Generally, moving down a group in the periodic table, the atomic radius increases due to additional electron shells.
Electron Repulsion
Electron repulsion occurs due to the negative charges of electrons that naturally repel each other. When considering ions, the concept of electron repulsion plays a central role in determining their size.

In cations, since electrons are removed, there is less repulsion, allowing the electrons to be pulled closer to the nucleus, decreasing the size of the ion. On the other hand, anions experience increased electron repulsion because of the extra electrons, and this additional repulsive force pushes the electron cloud outward, enlarging the ion's size.

Understanding electron repulsion is key to predicting and explaining the changes in atomic and ionic sizes during the process of ion formation.

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

As we move across a period of the periodic table, why do the sizes of the transition elements change more gradually than those of the representative elements?

Use electron configurations to explain the following observations: \((\mathbf{a})\) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

Which will experience the greater effect nuclear charge, the electrons in the \(n=2\) shell in \(\mathrm{F}\) or the \(n=2\) shell in \(\mathrm{B}\) ? Which will be closer to the nucleus?

Moseley established the concept of atomic number by studying \(X\) rays emitted by the elements. The \(X\) rays emitted by some of the elements have the following wavelengths: $$ \begin{array}{cc} \hline \text { Element } & \text { Wavelength (pm) } \\ \hline \text { Ne } & 1461 \\ \text { Ca } & 335.8 \\ \text { Zn } & 143.5 \\ \text { Zr } & 78.6 \\ \text { Sn } & 49.1 \\ \hline \end{array} $$ (a) Calculate the frequency, \(\nu,\) of the \(X\) rays emitted by each of the elements, in Hz. (b) Plot the square root of \(\nu\) versus the atomic number of the element. What do you observe about the plot? (c) Explain how the plot in part (b) allowed Moseley to predict the existence of undiscovered elements. (d) Use the result from part (b) to predict the X-ray wavelength emitted by iron. (e) A particular element emits X rays with a wavelength of \(98.0 \mathrm{pm}\). What element do you think it is?

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.