Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 5, Problem 34A

(a)

Interpretation Introduction

Interpretation: The number of sublevels in principal energy level 1 needs to be determined.

Concept introduction: An atomic orbital explains the position of an electron in an atom. An atomic orbital can be occupied by a maximum of two electrons with opposite spins. The four basic atomic orbitals are s, p, d, and f . The number of electrons in each orbital can be explained with the help of quantum numbers. There are four quantum numbers; principal quantum number ( n ), azimuthal quantum number ( l ), magnetic quantum number ( ml ), and spin quantum number ( s ). Here, for any value of n, l=n1 and values of ml is from -l to + l . spin quantum number explains the spin of the electron in an atomic orbital. It can be either +1/2 or -1/2.

(a)

Expert Solution
Check Mark

Explanation of Solution

For the given principal quantum number (n), the azimuthal quantum number can be calculated as follows:

  l=0, 1, 2,.....n1

Since the principal quantum number is given 1, the value n1 will be:

  n1=11=0

Thus, l=0

There is only 1 value for an azimuthal quantum number, which is 1.

Thus, there is only 1 subshell in principal energy level 1 that is s subshell.

(b)

Interpretation Introduction

Interpretation: The number of sublevels in principal energy level 2 needs to be determined.

Concept introduction: An atomic orbital explains the position of an electron in an atom. An atomic orbital can be occupied by a maximum of two electrons with opposite spins. The four basic atomic orbitals are s, p, d, and f . The number of electrons in each orbital can be explained with the help of quantum numbers. There are four quantum numbers; principal quantum number ( n ), azimuthal quantum number ( l ), magnetic quantum number ( ml ), and spin quantum number ( s ). Here for any value of n, l=n1 and values of ml is from -l to + l . spin quantum number explains the spin of the electron in an atomic orbital. It can be either +1/2 or -1/2.

(b)

Expert Solution
Check Mark

Explanation of Solution

For the given principal quantum number (n), the azimuthal quantum number can be calculated as follows:

  l=0, 1, 2,.....n1

Since the principal quantum number is given 2, the value n1 will be:

  n1=21=1

Thus, l=0,1

There are two values for an azimuthal quantum number that is 0 and 1.

Thus, there are two subshells in principal energy level 2 that is s and p subshells.

(c)

Interpretation Introduction

Interpretation: The number of sublevels in principal energy level 3 needs to be determined.

Concept introduction: An atomic orbital explains the position of an electron in an atom. An atomic orbital can be occupied by a maximum of two electrons with opposite spins. The four basic atomic orbitals are s, p, d, and f . The number of electrons in each orbital can be explained with the help of quantum numbers. There are four quantum numbers; principal quantum number ( n ), azimuthal quantum number ( l ), magnetic quantum number ( ml ), and spin quantum number ( s ). Here, for any value of n, l=n1 and values of ml is from -l to + l . spin quantum number explains the spin of the electron in an atomic orbital. It can be either +1/2 or -1/2.

(c)

Expert Solution
Check Mark

Explanation of Solution

For the given principal quantum number (n), the azimuthal quantum number can be calculated as follows:

  l=0, 1, 2,.....n1

Since the principal quantum number is given 3, the value n1 will be:

  n1=31=2

Thus, l=0, 1, 2

There are two values for an azimuthal quantum number that is 0, 1, and 2.

Thus, there are three subshells in principal energy level 3 that is s , p, and d subshell.

(d)

Interpretation Introduction

Interpretation: The number of sublevels in principal energy level 4 needs to be determined.

Concept introduction: An atomic orbital explains the position of an electron in an atom. An atomic orbital can be occupied by a maximum of two electrons with opposite spins. The four basic atomic orbitals are s, p, d, and f . The number of electrons in each orbital can be explained with the help of quantum numbers. There are four quantum numbers; principal quantum number ( n ), azimuthal quantum number ( l ), magnetic quantum number ( ml ), and spin quantum number ( s ). Here, for any value of n, l=n1 and values of ml is from -l to + l . spin quantum number explains the spin of the electron in an atomic orbital. It can be either +1/2 or -1/2.

(d)

Expert Solution
Check Mark

Explanation of Solution

For the given principal quantum number (n), the azimuthal quantum number can be calculated as follows:

  l=0, 1, 2,.....n1

Since the principal quantum number is given 4, the value n1 will be:

  n1=41=3

Thus, l=0, 1, 2, 3

There are four values for an azimuthal quantum number that is 0, 1, 2, and 3.

Thus, there are four subshells in principal energy level 4 that is s , p , d, and f subshells.

Chapter 5 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 5.2 - Prob. 11LCCh. 5.2 - Prob. 12LCCh. 5.2 - Prob. 13LCCh. 5.2 - Prob. 14LCCh. 5.3 - Prob. 15SPCh. 5.3 - Prob. 16SPCh. 5.3 - Prob. 17SPCh. 5.3 - Prob. 18SPCh. 5.3 - Prob. 19LCCh. 5.3 - Prob. 20LCCh. 5.3 - Prob. 21LCCh. 5.3 - Prob. 22LCCh. 5.3 - Prob. 23LCCh. 5.3 - Prob. 24LCCh. 5.3 - Prob. 25LCCh. 5.3 - Prob. 26LCCh. 5 - Prob. 27ACh. 5 - Prob. 28ACh. 5 - Prob. 29ACh. 5 - Prob. 30ACh. 5 - Prob. 31ACh. 5 - Prob. 32ACh. 5 - Prob. 33ACh. 5 - Prob. 34ACh. 5 - Prob. 35ACh. 5 - Prob. 36ACh. 5 - Prob. 37ACh. 5 - Prob. 38ACh. 5 - Prob. 39ACh. 5 - Prob. 40ACh. 5 - Prob. 41ACh. 5 - Prob. 42ACh. 5 - Prob. 43ACh. 5 - Prob. 44ACh. 5 - Prob. 45ACh. 5 - Prob. 46ACh. 5 - Prob. 47ACh. 5 - Prob. 48ACh. 5 - Prob. 49ACh. 5 - Prob. 50ACh. 5 - Prob. 51ACh. 5 - Prob. 52ACh. 5 - Prob. 53ACh. 5 - Prob. 54ACh. 5 - Prob. 55ACh. 5 - Prob. 56ACh. 5 - Prob. 57ACh. 5 - Prob. 58ACh. 5 - Prob. 59ACh. 5 - Prob. 60ACh. 5 - Prob. 61ACh. 5 - Prob. 62ACh. 5 - Prob. 63ACh. 5 - Prob. 64ACh. 5 - Prob. 65ACh. 5 - Prob. 66ACh. 5 - Prob. 67ACh. 5 - Prob. 68ACh. 5 - Prob. 69ACh. 5 - Prob. 70ACh. 5 - Prob. 71ACh. 5 - Prob. 72ACh. 5 - Prob. 73ACh. 5 - Prob. 74ACh. 5 - Prob. 75ACh. 5 - Prob. 77ACh. 5 - Prob. 78ACh. 5 - Prob. 79ACh. 5 - Prob. 80ACh. 5 - Prob. 81ACh. 5 - Prob. 82ACh. 5 - Prob. 83ACh. 5 - Prob. 85ACh. 5 - Prob. 86ACh. 5 - Prob. 88ACh. 5 - Prob. 89ACh. 5 - Prob. 90ACh. 5 - Prob. 91ACh. 5 - Prob. 92ACh. 5 - Prob. 93ACh. 5 - Prob. 94ACh. 5 - Prob. 95ACh. 5 - Prob. 96ACh. 5 - Prob. 97ACh. 5 - Prob. 98ACh. 5 - Prob. 99ACh. 5 - Prob. 100ACh. 5 - Prob. 101ACh. 5 - Prob. 102ACh. 5 - Prob. 103ACh. 5 - Prob. 104ACh. 5 - Prob. 105ACh. 5 - Prob. 106ACh. 5 - Prob. 1STPCh. 5 - Prob. 2STPCh. 5 - Prob. 3STPCh. 5 - Prob. 4STPCh. 5 - Prob. 5STPCh. 5 - Prob. 6STPCh. 5 - Prob. 7STPCh. 5 - Prob. 8STPCh. 5 - Prob. 9STPCh. 5 - Prob. 10STPCh. 5 - Prob. 11STPCh. 5 - Prob. 12STPCh. 5 - Prob. 13STPCh. 5 - Prob. 14STPCh. 5 - Prob. 15STP
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