Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
Question
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Chapter 12, Problem 141AE

(a)

Interpretation Introduction

Interpretation: Atomic number for element that has 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3 should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(a)

Expert Solution
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Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 . Since this ion has resulted from loss of five electrons so expected atomic nmber corresponding to this configuration is 24.

(b)

Interpretation Introduction

Interpretation:Electrons in s -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(b)

Expert Solution
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Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 .The configuration 1s22s22p63s23p64s23d2 indicates 6 electrons in s -orbital.

(c)

Interpretation Introduction

Interpretation:Electrons in p -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(c)

Expert Solution
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Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 .Theconfiguration 1s22s22p63s23p64s23d2 indicates12 electrons in p -orbital.

(d)

Interpretation Introduction

Interpretation:Electrons in d -orbital for this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(d)

Expert Solution
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Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 .The configuration 1s22s22p63s23p64s23d2 indicates twoelectrons in d -orbital.

(e)

Interpretation Introduction

Interpretation:Number of neutrons in this element should be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(e)

Expert Solution
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Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24.

The formula to compute neutrons from mass number is as follows:

  Number of neutrons=Mass numberAtomic number

Atomic number is 24.

Mass number is 52.

Substitute the value in above formula.

  Number of neutrons=Mass numberAtomic number=5224=28

So there are 28 neutrons in chromium.

(f)

Interpretation Introduction

Interpretation:Mass of 3.01×1023 atoms is to be identified.

Concept introduction:To remove the electron situated in outermost shell certain minimum energy must be imparted so as to convert an atom to gaseous species. The energy thus imparted represents ionization energy.

The magnitude of ionization energy is determined by how effectively valence electron is held by the nucleus. If the outermost shell has, for instance, one or two electronsthat require very minimum ionization energy because they can attain the noble gas configuration upon loss of those electrons.

(f)

Expert Solution
Check Mark

Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24. So, element is identified as chromium.

Since molar mass of chromium ion is 51.99 g/mol , so mass of 3.01×1023 atoms is calculated as follows:

  Mass=(3.01×1023 atoms)(1 mol6.022×1023 atoms)(51.99 g1 mol)=25.9 g

Thus, mass of 3.01×1023 atoms is 25.9 g .

(g)

Interpretation Introduction

Interpretation:Ground-state electron configuration of neutral chromium should be written.

Concept introduction:Aufbau rule states that electrons must be filled in lowest energy levels first. For instance, electrons first occupy shells that are lower in energies illustrated as follows:

  Chemical Principles, Chapter 12, Problem 141AE

Pauli’s exclusion principle states thatno two or more than two electrons of a poly electron atom can have same values of 4 quantum numbers that are n , l , m and s .It also states that only two electrons can occupy same orbital with opposite spins.

Hund’s rule of maximum multiplicity states that electrons cannot be allowed to pair until each orbital gets singly filled with one electron. These 3 principles form basis for determination of electronic configuration.However, certain elements that are able to achieve nearest half-filled or fully filed configuration show exceptional configurations.

(g)

Expert Solution
Check Mark

Explanation of Solution

With 2 electrons with n equal to 1, 8 with n equal to 1, and 10 electrons with n equal to 3the configuration can be written as 1s22s22p63s23p64s23d2 . Since this ion has resulted from loss of five electrons so expected atomic number corresponding to this configuration is 24.

With atomic number as 24, expected configuration for Cr is [Ar]4s23d4 while actual ground electronic configuration for Cr is [Ar]4s13d5 . This is because of greater stability that arises in latter case due to half-filled configuration.

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Chapter 12 Solutions

Chemical Principles

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