/*! 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 105 One bit of evidence that the qua... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 105

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for Li, N, Ni, Te, Ba, and Hg. Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

Short Answer

Expert verified
The paramagnetic atoms among the given elements are Li, N, and Ni. Li has one unpaired electron (1s虏 2s鹿), N has three unpaired electrons (1s虏 2s虏 2p鲁), and Ni has two unpaired electrons (1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鈦�). The other elements (Te, Ba, and Hg) have all their electrons paired.

Step by step solution

01

Determine electron configurations of given elements.

To do this, use the periodic table and the Aufbau principle. The order in which electron subshells are filled is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, and so on. Consider the electron configurations of the given elements: Li (Atomic number: 3): 1s^2 2s^1 N (Atomic number: 7): 1s^2 2s^2 2p^3 Ni (Atomic number: 28): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 Te (Atomic number: 52): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^4 Ba (Atomic number: 56): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 Hg (Atomic number: 80): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10
02

Identify paramagnetic atoms and count unpaired electrons.

An atom is paramagnetic if it has unpaired electrons in its electron configuration. For Li: 1s^2 2s^1 -> One unpaired electron For N: 1s^2 2s^2 2p^3 -> Three unpaired electrons For Ni: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 -> Two unpaired electrons For Te: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^4 -> All electrons are paired. For Ba: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 -> All electrons are paired. For Hg: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 -> All electrons are paired. From the above analysis, Li, N, and Ni are paramagnetic, with 1, 3, and 2 unpaired electrons, respectively.

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.

Electron Configuration
Electron configuration refers to the distribution of electrons in an atom's orbitals. These orbitals belong to various energy levels and subshells, designated by the principal quantum number, designated as \(n\), and the subshells \(s, p, d,\) and \(f\). The configuration follows a specific order, filling from lower to higher energy levels. This is depicted with notations like \(1s^2 2s^2 2p^6\), where the superscript indicates the number of electrons in that subshell.
Understanding electron configurations helps in predicting an atom's chemical behavior. It explains why elements exhibit specific bonding patterns, and their reactivity among other properties. For instance, nitrogen (N) with the configuration \(1s^2 2s^2 2p^3\) has three electrons in its 2p subshell, making it likely to form bonds with three other atoms to complete its octet. In contrast, mercury (Hg) has all of its subshells filled, making it less reactive.
Aufbau Principle
The Aufbau principle is crucial in determining electron configurations. The German word "Aufbau" means "building up." It suggests that electrons fill atomic orbitals in order from the lowest energy level to higher ones. This order is visually represented by the periodic table and depicted by the sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, etc.

The principle relies on the energy hierarchy, where electrons seek the least energy configuration possible. For example, lithium (Li), requiring only three electrons, follows the configuration \(1s^2 2s^1\). In this case, the first electron fills the 1s orbital, while the remaining electron occupies the 2s orbital, aligning with this principle.
  • The sequence is sometimes adapted due to exceptions involving d and f orbitals. But, for most elements, the Aufbau principle remains applicable.
Unpaired Electrons
Unpaired electrons are those valence electrons occupying an atomic orbital alone. Unlike paired electrons that fill an orbital with their opposite spin counterparts, unpaired electrons contribute to magnetism and reactivity.
In atoms like nitrogen (N), with the configuration \(1s^2 2s^2 2p^3\), the three electrons in the p-shell remain unpaired. These unpaired electrons cause a paramagnetic behavior, meaning the atom is attracted to magnetic fields.

Here's why unpaired electrons matter:
  • They determine whether an atom is paramagnetic or diamagnetic. Paramagnetic atoms have at least one unpaired electron while diamagnetic atoms have all paired up.
  • They play a vital role in chemical bonding. Unpaired electrons often participate in covalent bonding, as they seek stability by pairing up with electrons from other atoms.
Understanding unpaired electrons is essential in chemistry, as they provide insights into the magnetic and reactive properties of an element.

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

Calculate, to four significant figures, the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=5\) state and then fall to states with smaller values of \(n\).

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=4\) b. \(n=5, m_{\ell}=+1\) c. \(n=5, m_{s}=+\frac{1}{2}\) d. \(n=3, \ell=2\) e. \(n=2, \ell=1\)

An electron is excited from the \(n=1\) ground state to the \(n=\) 3 state in a hydrogen atom. Which of the following statements is/are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\) d. The wavelcngth of light cmittcd when the clectron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\) e. For \(n=3,\) the electron is in the first excited state.

The wave function for the \(2 p_{z}\) orbital in the hydrogen atom is $$ \psi_{2 p_{i}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\alpha / 2} \cos \theta $$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p^{2}}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis) and for } \theta=90^{\circ}\right.\) (xy plane).

In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.