/*! 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 57 For each of the following groups... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 57

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

Short Answer

Expert verified
a. Cu > Cu鈦 > Cu虏鈦 b. Pt虏鈦 > Pd虏鈦 > Ni虏鈦 c. O虏鈦 > O鈦 > O d. La鲁鈦 > Eu鲁鈦 > Gd鲁鈦 > Yb鲁鈦 e. Te虏鈦 > I鈦 > Cs鈦 > Ba虏鈦 > La鲁鈦

Step by step solution

01

a. Cu, Cu鈦, Cu虏鈦

The atomic sizes are affected by ionic charges. The more positive the charge, the smaller the size of the ion due to increased effective nuclear charge and less repulsion between electrons. Here, all species belong to the same element (Copper). Therefore, ionic size decreases with an increase in positive charge. So the order is: Cu > Cu鈦 > Cu虏鈦.
02

b. Ni虏鈦, Pd虏鈦, Pt虏鈦

In this group, we have three ions with the same charge (+2). We can compare their sizes based on atomic radii in the periodic table. The size of the atom increases as we move down a group in the periodic table. Ni, Pd, and Pt belong to the same group and Pt being the lowest, will have the largest radius. Therefore, the order is: Pt虏鈦 > Pd虏鈦 > Ni虏鈦.
03

c. O, O鈦, O虏鈦

In this group, the atomic size increases with an increasing number of negative charges. This is due to decreased effective nuclear charge and more repulsion among the electrons. Thus, the order is: O虏鈦 > O鈦 > O.
04

d. La鲁鈦, Eu鲁鈦, Gd鲁鈦, Yb鲁鈦

These four ions have the same charge (+3). The sizes can be compared based on their positions in the periodic table as they all belong to the same group. As the atomic size increases as we move down a group, and La is the highest and Yb being the lowest, the order is: La鲁鈦 > Eu鲁鈦 > Gd鲁鈦 > Yb鲁鈦.
05

e. Te虏鈦, I鈦, Cs鈦, Ba虏鈦, La鲁鈦

In this group, we have different ions with various charges. First, arrange them by charge: Te虏鈦 > I鈦 > Cs鈦 > Ba虏鈦 > La鲁鈦. Then, consider their positions in the periodic table. Te and I are in the same row, so their size will be affected by the charge and shielding effects: Te虏鈦 > I鈦. Cs鈦, Ba虏鈦, and La鲁鈦 are located in the same group, so the size increases going up the group: Cs鈦 > Ba虏鈦 > La鲁鈦. Combining the orders, we get: Te虏鈦 > I鈦 > Cs鈦 > Ba虏鈦 > La鲁鈦.

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.

Ionic Radius
The ionic radius of an atom or ion is a crucial aspect when discussing periodic trends. It refers to the effective distance from the center of the nucleus to the outermost shell of an ion. The size of the ionic radius can be influenced by several factors:
  • Charge of the ion: Positive ions (cations) are typically smaller than their neutral atoms because they lose electrons, resulting in fewer electron-electron repulsions. Conversely, negative ions (anions) are larger than their neutral atoms as gaining electrons increases repulsion among them.
  • Effective nuclear charge: Higher positive charges on ions increase the attraction between the nucleus and electrons, reducing the size.
In the case of copper ions, for example, as Cu becomes Cu鈦 and then Cu虏鈦, the removal of electrons leads to a decrease in ionic radius: Cu > Cu鈦 > Cu虏鈦. This trend is consistently observed across similar groups of elements.
Effective Nuclear Charge
The effective nuclear charge (Z_eff) is the net positive charge experienced by an electron in a multi-electron atom. It is crucial for understanding the behavior of atomic and ionic sizes. This concept considers both the positive charge of the nucleus and the shielding effect, where inner electrons partially block the attraction between the nucleus and outer electrons.
Z_eff can be calculated using:\[ Z_{ ext{eff}} = Z - S \]where \( Z \) is the actual nuclear charge (number of protons), and \( S \) is the shielding constant.
  • Increased Z_eff typically leads to a decrease in atomic or ionic size because the nucleus pulls the electron cloud closer.
  • As ions lose electrons, such as Cu鈦 to Cu虏鈦, the Z_eff increases, making the ion smaller.
  • A higher Z_eff also explains why certain positive ions are significantly smaller than their corresponding neutral atoms or anions.
Understanding Z_eff is essential in explaining why the ion sizes vary as we see in periodic trends.
Periodic Table Groups
Periodic Table groups are vertical columns that consist of elements with similar chemical properties. These are essential to understanding periodic trends as elements in the same group behave predictably when it comes to size and charge.
Several characteristics of these groups include:
  • As you move down a group, elements have additional electron shells, which generally increases their atomic size.
  • Valence electrons in each group experience similar effective nuclear charge despite increasing nuclear charge, as additional electrons shield the valence electrons from the nucleus.
  • In groups with charged ions of the same type (e.g., Ni虏鈦, Pd虏鈦, Pt虏鈦), the atomic size increases as you move down due to increased electron shell count.
The importance of groups is evident in exercises like comparing Ni虏鈦, Pd虏鈦, and Pt虏鈦, where the size order follows periodic trends dictated by their group placement.
Atomic Size
Atomic size is a fundamental concept, often referred to as atomic radius, indicating the size of an atom. Understanding this helps explain how and why atoms bond and interact.
Key points about atomic size include:
  • The atomic size is influenced by the number of electron shells and the effective nuclear charge.
  • As you move across a period from left to right, atomic size generally decreases. This is due to increasing nuclear charge pulling the electron cloud closer to the nucleus.
  • Moving down a group, atomic size typically increases because of added electron shells despite increased nuclear charge.
Understanding atomic size is vital, particularly when predicting and rationalizing behaviors in exercises involving atoms and ions like O, O鈦, and O虏鈦, where the increased electron count in anions leads to larger atomic sizes due to added electron-electron repulsion.

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

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{H}, \mathrm{Si}-\mathrm{H}, \mathrm{Sn}-\mathrm{H}\) b. \(\mathrm{Al}-\mathrm{Br}, \mathrm{Ga}-\mathrm{Br}, \mathrm{In}-\mathrm{Br}, \mathrm{Tl}-\mathrm{Br}\) c.\(\mathrm{C}-\mathrm{O}\) or \(\mathrm{Si}-\mathrm{O}\) d. \(\mathrm{O}-\mathrm{F}\) or \(\mathrm{O}-\mathrm{Cl}\)

Which of the following statements is(are) true? Correct the false statements. a. It is impossible to satisfy the octet rule for all atoms in \(\mathrm{XeF}_{2}\) . b. Because \(\mathrm{SF}_{4}\) exists, OF_ should also exist because oxygen is in the same family as sulfur. c. The bond in NO + should be stronger than the bond in \(\mathrm{NO}^{-}\). d. As predicted from the two Lewis structures for ozone, one oxygen-oxygen bond is stronger than the other oxygen-oxygen bond.

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3} .\) Give a possible explanation for why these two reactions occur.

Calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) shells are used by mollusks, corals, and snails to form protective coverings. Draw the Lewis structure for \(\mathrm{CaCO}_{3} .\) Be sure to include any resonance structures.

Without using Fig. 8.3, predict which bond in each of the following groups will be the most polar. a. \(\mathrm{C}-\mathrm{F}, \mathrm{Si}-\mathrm{F}, \mathrm{Ge}-\mathrm{F} \quad\) c. \(\mathrm{S}-\mathrm{F}, \mathrm{S}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}\) b. \(\mathrm{P}-\mathrm{Cl}\) or \(\mathrm{S}-\mathrm{Cl} \quad\) d. \(\mathrm{Ti}-\mathrm{Cl}, \mathrm{Si}-\mathrm{Cl}, \mathrm{Ge}-\mathrm{Cl}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

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.