/*! 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 20 For the following pairs of orbit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 20

For the following pairs of orbitals, indicate which is higher in energy in a many-electron atom. (a) \(3 \mathrm{~s}\) or \(2 \mathrm{p}\) (b) 4 s or \(4 \mathrm{~d}\) (c) \(4 \mathrm{f}\) or \(6 \mathrm{~s}\) (d) 1 s or \(2 \mathrm{~s}\)

Short Answer

Expert verified
Answer: (a) 3s, (b) 4d, (c) 4f, (d) 2s

Step by step solution

01

Identify the principal and azimuthal quantum numbers of each orbital

We have the following orbitals: (a) 3s: Principal quantum number (n) = 3, azimuthal quantum number (l) = 0 2p: Principal quantum number (n) = 2, azimuthal quantum number (l) = 1 (b) 4s: Principal quantum number (n) = 4, azimuthal quantum number (l) = 0 4d: Principal quantum number (n) = 4, azimuthal quantum number (l) = 2 (c) 4f: Principal quantum number (n) = 4, azimuthal quantum number (l) = 3 6s: Principal quantum number (n) = 6, azimuthal quantum number (l) = 0 (d) 1s: Principal quantum number (n) = 1, azimuthal quantum number (l) = 0 2s: Principal quantum number (n) = 2, azimuthal quantum number (l) = 0
02

Apply the n + l rule to compare the energies of the orbitals

Now we'll apply the n + l rule to determine which orbital in each pair has higher energy: (a) 3s: n + l = 3 + 0 = 3 2p: n + l = 2 + 1 = 3 Since both orbitals have the same n + l value, we compare their n values. The 2p orbital has a lower n value, so it has lower energy. Therefore, the 3s orbital has higher energy. (b) 4s: n + l = 4 + 0 = 4 4d: n + l = 4 + 2 = 6 The 4s orbital has a lower n + l value, so it has lower energy. Therefore, the 4d orbital has higher energy. (c) 4f: n + l = 4 + 3 = 7 6s: n + l = 6 + 0 = 6 The 6s orbital has a lower n + l value, so it has lower energy. Therefore, the 4f orbital has higher energy. (d) 1s: n + l = 1 + 0 = 1 2s: n + l = 2 + 0 = 2 The 1s orbital has a lower n + l value, so it has lower energy. Therefore, the 2s orbital has higher energy.
03

Report the higher-energy orbital for each pair

Based on our comparisons, the higher-energy orbitals in the given pairs are as follows: (a) 3s (b) 4d (c) 4f (d) 2s

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.

Principal Quantum Number
The principal quantum number, denoted as \( n \), is one of the key concepts in quantum mechanics, particularly in the study of atomic structure. It is a vital number that determines the overall size and energy level of an electron orbital in an atom. The value of \( n \) is always a positive integer, namely 1, 2, 3, and so forth. Here鈥檚 how it influences the electron orbital:
  • It determines the "shell" or energy level in which the electron is located. Higher \( n \) values correspond to orbitals that are farther from the nucleus and have higher energy.
  • As the value of the principal quantum number increases, the electron is likely to be found at a greater average distance from the nucleus, resulting in larger atomic radii.
  • The principal quantum number is crucial in quantifying the energy levels of electrons in multi-electron atoms, where the complexity increases due to electron-electron interactions.
In comparing orbitals, knowing the \( n \) value is a starting point for comparing their energy levels, largely because electrons further from the nucleus are less tightly bound and often have greater energy.
Azimuthal Quantum Number
The azimuthal quantum number, designated as \( l \), adds more depth to our understanding of an atom鈥檚 electron configuration and its energy level. Also referred to as the angular momentum quantum number, \( l \) determines the shape of the electron鈥檚 orbital and reflects subshells within a given principal energy level. Here's how \( l \) is defined and what it affects:
  • The azimuthal quantum number \( l \) can take on any integer value from 0 to \( n-1 \), where \( n \) is the principal quantum number. This means if \( n = 3 \), \( l \) can be 0, 1, or 2.
  • Each value of \( l \) corresponds to a unique subshell, commonly denoted by letters: \( l = 0 \) is an "s" subshell, \( l = 1 \) is a "p" subshell, \( l = 2 \) is a "d" subshell, and \( l = 3 \) is an "f" subshell.
  • It also impacts the energy levels within a principal quantum number. Generally, within the same \( n \), a higher \( l \) value implies a higher energy subshell due to increased angular momentum.
Understanding the azimuthal quantum number is essential when comparing orbitals' energy levels, such as in the exercises where one analyzes the impact of \( l \) on energy.
n + l Rule
The \( n + l \) rule is a useful guideline in quantum chemistry to predict the relative energy levels of different orbitals within atoms, particularly when dealing with multi-electron systems. This rule helps in determining which orbitals will be filled first with electrons as atoms build up their electron configurations. Here's how it works:
  • Calculate the sum of the principal quantum number \( n \) and the azimuthal quantum number \( l \) for each orbital. This total, \( n + l \), helps in ranking them by energy level.
  • The lower the value of \( n + l \), the lower the energy of the orbital. Orbital filling follows this sequence from lower to higher \( n + l \) values.
  • If two orbitals have the same \( n + l \) value, the one with the lower \( n \) will have lower energy.
This rule is especially valuable when analyzing different orbitals of a multi-electron atom, as it was applied in analyzing exercises for energy comparisons. It intuitively explains why, for example, 3s is of higher energy than 2p due to closer proximity of electrons and available space in higher shells.

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

In what main group(s) of the periodic table do element(s) have the following number of filled p orbitals in the outermost principal level? (a) 0 (b) 1 (c) 2 (d) 3

Which of the following electron configurations (a-e) are for atoms in the ground state? In the excited state? Which are impossible? (a) \(1 s^{2} 2 p^{1}\) (b) \(1 s^{2} 2 s^{2} 2 p^{4}\) (c) \(1 s^{2} 2 s^{2} 2 p^{5} 3 d^{1}\) (d) \(1 s^{2} 2 s^{2} 2 p^{7} 3 s^{2}\) (e) \(1 s^{2} 2 s^{2} 2 p^{6} 4 s^{1} 3 d^{11}\)

For the following pairs of orbitals, indicate which is lower in energy in a many-electron atom. (a) \(3 \mathrm{~d}\) or \(4 \mathrm{~s}\) (b) \(4 \mathrm{f}\) or \(3 \mathrm{~d}\) (c) \(2 \mathrm{~s}\) or \(2 \mathrm{p}\) (d) \(4 \mathrm{f}\) or \(4 \mathrm{~d}\)

In the Paschen series, \(\mathbf{n}_{\text {lo }}=3\). Calculate the longest wavelength possible for a transition in this series.

Indicate whether each of the following is true or false. (a) Effective nuclear charge stays about the same when one goes down a group. (b) Group 17 elements have seven electrons in their outer level. (c) Energy is given off when an electron is removed from an atom.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

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.