/*! 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 46 Explain which of these anions is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 46

Explain which of these anions is larger, and why: \(\mathrm{Se}^{2-}\) or \(\mathrm{Te}^{2-}\)

Short Answer

Expert verified
The \( \mathrm{Te}^{2-} \) anion is larger. This is due to the fact that atomic size increases as we move down the periodic table. Tellurium (Te) is located below Selenium (Se) on the table, meaning it has more electron energy levels and therefore, a larger atomic size.

Step by step solution

01

Understanding the Periodic Table

The first thing to remember is that in the Periodic Table, as we move down a group (vertical column), the atomic size or atomic radius increases. This is because each period represents a new energy level in the electron cloud around the nucleus. As such, \( \mathrm{Se}^{2-} \) and \( \mathrm{Te}^{2-} \) are in the same group, with selenium (Se) above tellurium (Te).
02

Analyzing the Atomic Number

Tellurium (Te) has a greater atomic number than Selenium (Se), indicating that there are more protons in the nucleus of Tellurium and also more electrons in its electron cloud. This increased number of energy levels in the electron cloud means that the distance from the nucleus to the outermost energy level of Te will be greater than that of Se.
03

Comparing the size of anions

Considering the above points, we can see that \( \mathrm{Te}^{2-} \) will be larger than \( \mathrm{Se}^{2-} \). Because as we add more energy levels (each additional period on the table), the atomic size increases.

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.

Atomic Radius
The atomic radius is a key concept in understanding how elements compare in size. It refers to the distance from an atom's nucleus to the outermost electron shell. It plays a crucial role in determining an element's size. As you move down the Periodic Table, each element has additional electron shells. This increases the atomic radius. For example, moving from selenium to tellurium down the group allows tellurium to have a larger atomic radius due to more electron shells. Despite having more protons which can pull electrons closer, the overall size increases due to these added layers.
Periodic Table
The Periodic Table organizes elements in a way that reveals recurring chemical properties. It arranges elements in order of increasing atomic number, which corresponds to the number of protons in an element. Elements in the same column or 'group' have similar valence electron configurations. Moving down a group, elements have more electron shells, causing them to have larger atomic radii. This is because, although nuclear charge increases down a group, the onset of new shells has a dominating effect. This principle is evident in the comparison between selenium and tellurium, where tellurium is in a period below selenium, resulting in a larger overall size.
Selenium
Selenium, symbolized as \( ext{Se}\), is an element found in the 16th group of the Periodic Table. It is situated above tellurium. Selenium in its anionic form, \(\text{Se}^{2-}\), gains two additional electrons, increasing its electron cloud. However, selenium's smaller number of energy levels compared to tellurium keeps it smaller in size. Despite gaining electrons, which generally increases atomic size, its less number of electron shells constrains its overall size. This is why when comparing \(\text{Se}^{2-}\) and \(\text{Te}^{2-}\), selenium's reduced energy level count makes its ionic form smaller than tellurium's.
Tellurium
Tellurium (Te), another member of the 16th group of the Periodic Table, is known for its substantial atomic radius due to its position below selenium. In the formation of \(\text{Te}^{2-}\), tellurium gains electrons just like selenium, becoming an anion. The ionic form retains tellurium's characteristic of having more electron shells than selenium. This expanded electron cloud means that even though \(\text{Te}^{2-}\) and \(\text{Se}^{2-}\) share the same charge state, tellurium's additional energy level gives it a larger radius. This is a prime example of how the structure of the Periodic Table affects atomic size.

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

Why do elements that have high ionization energies usually have more positive electron affinities?

Acidic oxides are those that react with water to produce acid solutions, whereas the reactions of basic oxides with water produce basic solutions. Nonmetallic oxides are usually acidic, whereas those of metals are basic. Predict the products of the reactions of these oxides with water: \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{BaO}, \mathrm{CO}_{2}\), \(\mathrm{N}_{2} \mathrm{O}_{5}, \mathrm{P}_{4} \mathrm{O}_{10}, \mathrm{SO}_{3} .\) Write an equation for each of the reactions.

What is the characteristic of the electron configuration of stable ions derived from representative elements?

Name the element that forms compounds, under appropriate conditions, with every other element in the periodic table except He and Ne.

Two atoms have the electron configurations \(1 s^{2} 2 s^{2} 2 p^{6}\) and \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\). The first ionization en ergy of one is \(2080 \mathrm{~kJ} / \mathrm{mol},\) and that of the other is \(496 \mathrm{~kJ} / \mathrm{mol}\). Pair each ionization energy with one of the given electron configurations. Justify your choice.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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.