Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 14

How would you expect the sizes of the hydrogen ion, \(\mathrm{H}^{+},\) and the hydride ion, \(\mathrm{H}^{-},\) to compare with that of the H atom and the He atom? Explain.

Short Answer

Expert verified
The hydrogen ion (\(H^{+}\)) is much smaller than the hydrogen atom because it has lost its sole electron, while a hydride ion (\(H^{-}\)) is larger than the hydrogen atom due to the additional electron, but smaller than the helium atom thanks to helium's larger positively charged nucleus.

Step by step solution

01

Understanding of Atomic Structure

An atom consists of a nucleus containing protons and neutrons, and electrons that move around the nucleus. The size of an atom is determined by the outermost electronic cloud. An atom becomes an ion when it loses or gains an electron.
02

Compare the Sizes of Hydrogen Ion and Hydrogen Atom

The hydrogen ion (\(H^{+}\)) is formed when a hydrogen atom loses an electron. As such, \(H^{+}\) consists only of one proton without any electron, thereby making it practically point-sized.
03

Compare Sizes of Hydride Ion and Helium Atom

The hydride ion (\(H^{-}\)) is formed when a hydrogen atom gains an electron. This means that it has two electrons orbiting one proton, similar to the helium atom. However, since the helium atom has two protons and two neutrons in its nucleus, it has a greater positive charge attracting the electrons, making it smaller than the hydride ion.
04

Size Comparisons

From Step 2 and Step 3, we can conclude that \(H^{+}\) is smaller than the hydrogen atom, and the hydride ion (\(H^{-}\)) is larger than the hydrogen atom but smaller than the helium atom.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The fourth-period element with the largest atom is (a) \(\mathrm{K} ;\) (b) \(\mathrm{Br} ;\) (c) \(\mathrm{Pb} ;\) (d) \(\mathrm{Kr}\).

Explain why the first ionization energy of \(\mathrm{Mg}\) is greater that of \(\mathrm{Na},\) whereas the second ionization of Na is greater than that of Mg.

Explain why the several periods in the periodic table do not all have the same number of members.

Among the following ions, several pairs are isoelectronic. Identify these pairs. \(\mathrm{Fe}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{F}^{-}\) \(\mathrm{Co}^{2+}, \mathrm{Co}^{3+}, \mathrm{Sr}^{2+}, \mathrm{Cu}^{+}, \mathrm{Zn}^{2+}, \mathrm{Al}^{3+}.\)

Neither \(\mathrm{Co}^{2+}\) nor \(\mathrm{Co}^{3+}\) has \(4 \mathrm{s}\) electrons in its electron configuration. How many unpaired electrons would you expect to find in each of these ions? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free