/*! 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 75 Give the maximum number of elect... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 75

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. \(n=2, \ell=1\)

Short Answer

Expert verified
a. 32 electrons b. 2 electrons c. 25 electrons d. 10 electrons e. 6 electrons

Step by step solution

01

Calculate total number of orbitals

For n = 4, we can have: 鈩� = 0, 1, 2, 3 For 鈩� = 0, m_鈩� = 0, m_s = +1/2, -1/2 (one s-orbital) For 鈩� = 1, m_鈩� = -1, 0, 1, m_s = +1/2, -1/2 (three p-orbitals) For 鈩� = 2, m_鈩� = -2, -1, 0, 1, 2, m_s = +1/2, -1/2 (five d-orbitals) For 鈩� = 3, m_鈩� = -3, -2, -1, 0, 1, 2, 3, m_s = +1/2, -1/2 (seven f-orbitals)
02

Count maximum electrons

Each orbital can accommodate two electrons (m_s = +1/2 and -1/2). So: 1s: 2 electrons 3p: 6 electrons 5d: 10 electrons 7f: 14 electrons Total number of electrons = 2 + 6 + 10 + 14 = 32 electrons #b. n = 5, m_鈩� = +1#
03

Determine possible values of 鈩� and m_s

Given m_鈩� = 1, 鈩� can only be 1 (p-orbital), as values for m_鈩� are from -鈩� to +鈩�. m_s has two possible values: +1/2 and -1/2
04

Count maximum electrons

As 鈩� = 1, it means that we are referring to a p-orbital. A p-orbital can hold a maximum of 2 electrons with different spin quantum numbers (m_s = +1/2 and -1/2). Total number of electrons = 2 electrons #c. n = 5, m_s = +1/2#
05

Determine total number of orbitals

For n = 5, we can have: 鈩� = 0, 1, 2, 3, 4 Since m_s value is constant, all orbitals can accommodate only one electron with a given m_s value.
06

Count maximum electrons

For each value of 鈩�, we can count the number of electrons: 鈩� = 0 (s): 1 electron 鈩� = 1 (p): 3 electrons 鈩� = 2 (d): 5 electrons 鈩� = 3 (f): 7 electrons 鈩� = 4 (g): 9 electrons Total number of electrons: 1 + 3 + 5 + 7 + 9 = 25 electrons #d. n = 3, 鈩� = 2#
07

Determine possible values of m_鈩� and m_s

Since 鈩� = 2, we'll have m_鈩� values from -2 to 2. m_鈩� = -2, -1, 0, 1, 2 m_s has two possible values: +1/2 and -1/2
08

Count maximum electrons

As 鈩� = 2, it means that we are referring to a d-orbital. A d-orbital can hold a maximum of 10 electrons (2 electrons for each m_鈩� value). Total number of electrons = 10 electrons #e. n = 2, 鈩� = 1#
09

Determine possible values of m_鈩� and m_s

Since 鈩� = 1, we'll have m_鈩� values from -1 to 1. m_鈩� = -1, 0, 1 m_s has two possible values: +1/2 and -1/2
10

Count maximum electrons

As 鈩� = 1, it means that we are referring to a p-orbital. A p-orbital can hold a maximum of 6 electrons (2 electrons for each m_鈩� value). Total number of electrons = 6 electrons

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!

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

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is \(890.1 \mathrm{~kJ} / \mathrm{mol}\). Is light with a wavelength of \(225 \mathrm{~nm}\) capable of ionizing a gold atom (removing an electron) in the gas phase?

Calculate the velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^{2} \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\), respectively.

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).

The electron affinities of the elements from aluminum to chlorine are \(-44,-120,-74,-200.4\), and \(-384.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Rationalize the trend in these values.

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell\), determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell}\), determines the direction that the atomic orbitals point in space.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

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.