World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 11, Problem 66A

(a)

Interpretation Introduction

Interpretation:

From the atomic number of rubidium electronic configuration is to be determined. And then from electronic configuration valence electrons is to be determined.

Concept Introduction:

Electrons present in the outermost shell that can participate in the formation of a chemical bond or a molecule are called valence electrons.

The number of electrons in the atom is equal to the atomic number of the element. So, we can easily determine the electronic configuration by the atomic number.

(a)

Expert Solution
Check Mark

Answer to Problem 66A

Electronic configuration of rubidium (Z=37) is 1s22s22p63s23p63d104s24p65s1 or [Kr]5s1

So, Rb has one valence electron.

Explanation of Solution

The electrons are filled from lower to higher energy level following Aufbau principle.

Lower (n+l) values correspond to lower orbital energies. If two orbitals share equal (n+l) values, the orbital with the lower n value is said to have lower energy associated with it.

Where,

  • n= principle quantum number
  • l= azimuthal quantum number

The order in which the orbitals are filled with electrons is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.

Maximum number of electrons in an orbital is given by (4n2+2) rule.

For example, in case of p orbital l=1 so maximum number of electrons occupied are (4×12+2)=6 .

(b)

Interpretation Introduction

Interpretation:

From the atomic number of arsenic electronic configuration is to be determined. And then from electronic configuration valence electrons is to be determined.

Concept Introduction:

Electrons present in the outermost shell that can participate in the formation of a chemical bond or a molecule are called valence electrons.

The number of electrons in the atom is equal to the atomic number of the element. So, we can easily determine the electronic configuration by the atomic number.

(b)

Expert Solution
Check Mark

Answer to Problem 66A

Electronic configuration of arsenic (Z=33) is 1s22s22p63s23p63d104s24p3 or [Ar]3d104s24p3 .

Hence, there are five valence electrons in arsenic.

Explanation of Solution

The electrons are filled from lower to higher energy level following Aufbau principle.

Lower (n+l) values correspond to lower orbital energies. If two orbitals share equal (n+l) values, the orbital with the lower n value is said to have lower energy associated with it.

Where,

  • n= principle quantum number
  • l= azimuthal quantum number

The order in which the orbitals are filled with electrons is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.

Maximum number of electrons in an orbital is given by (4n2+2) rule.

For example, in case of p orbital l=1 so maximum number of electrons occupied are (4×12+2)=6 .

(c)

Interpretation Introduction

Interpretation:

From the atomic number of aluminium electronic configuration is to be determined. And then from electronic configuration valence electrons is to be determined.

Concept Introduction:

Electrons present in the outermost shell that can participate in the formation of a chemical bond or a molecule are called valence electrons.

The number of electrons in the atom is equal to the atomic number of the element. So, we can easily determine the electronic configuration by the atomic number.

(c)

Expert Solution
Check Mark

Answer to Problem 66A

Electronic configuration of aluminium (Z=13)

  1s22s22p63s23p1 or [Ne]3s23p1 .

Hence, there are three valence electrons.

Explanation of Solution

The electrons are filled from lower to higher energy level following Aufbau principle.

Lower (n+l) values correspond to lower orbital energies. If two orbitals share equal (n+l) values, the orbital with the lower n value is said to have lower energy associated with it.

Where,

  • n= principle quantum number
  • l= azimuthal quantum number

The order in which the orbitals are filled with electrons is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.

Maximum number of electrons in an orbital is given by (4n2+2) rule.

For example, in case of p orbital l=1 so maximum number of electrons occupied are (4×12+2)=6 .

(d)

Interpretation Introduction

Interpretation:

From the atomic number of nickel electronic configuration is to be determined. And then from electronic configuration valence electrons is to be determined.

Concept Introduction:

Electrons present in the outermost shell that can participate in the formation of a chemical bond or a molecule are called valence electrons.

The number of electrons in the atom is equal to the atomic number of the element. So, we can easily determine the electronic configuration by the atomic number.

(d)

Expert Solution
Check Mark

Answer to Problem 66A

Electronic configuration of nickel (Z=28) is 1s22s22p63s23p63d84s2 or [Ar]3d84s2 .

Hence, there are two valence electrons.

Explanation of Solution

The electrons are filled from lower to higher energy level following Aufbau principle.

Lower (n+l) values correspond to lower orbital energies. If two orbitals share equal (n+l) values, the orbital with the lower n value is said to have lower energy associated with it.

Where,

  • n= principle quantum number
  • l= azimuthal quantum number

The order in which the orbitals are filled with electrons is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.

Maximum number of electrons in an orbital is given by (4n2+2) rule.

For example, in case of p orbital l=1 so maximum number of electrons occupied are (4×12+2)=6 .

Chapter 11 Solutions

World of Chemistry, 3rd edition

Ch. 11.2 - Prob. 4RQCh. 11.2 - Prob. 5RQCh. 11.2 - Prob. 6RQCh. 11.3 - Prob. 1RQCh. 11.3 - Prob. 2RQCh. 11.3 - Prob. 3RQCh. 11.3 - Prob. 4RQCh. 11.3 - Prob. 5RQCh. 11.3 - Prob. 6RQCh. 11.4 - Prob. 1RQCh. 11.4 - Prob. 2RQCh. 11.4 - Prob. 3RQCh. 11.4 - Prob. 4RQCh. 11.4 - Prob. 5RQCh. 11.4 - Prob. 6RQCh. 11.4 - Prob. 7RQCh. 11 - Prob. 1ACh. 11 - Prob. 2ACh. 11 - Prob. 3ACh. 11 - Prob. 4ACh. 11 - Prob. 5ACh. 11 - Prob. 6ACh. 11 - Prob. 7ACh. 11 - Prob. 8ACh. 11 - Prob. 9ACh. 11 - Prob. 10ACh. 11 - Prob. 11ACh. 11 - Prob. 12ACh. 11 - Prob. 13ACh. 11 - Prob. 14ACh. 11 - Prob. 15ACh. 11 - Prob. 16ACh. 11 - Prob. 17ACh. 11 - Prob. 18ACh. 11 - Prob. 19ACh. 11 - Prob. 20ACh. 11 - Prob. 21ACh. 11 - Prob. 22ACh. 11 - Prob. 23ACh. 11 - Prob. 24ACh. 11 - Prob. 25ACh. 11 - Prob. 26ACh. 11 - Prob. 27ACh. 11 - Prob. 28ACh. 11 - Prob. 29ACh. 11 - Prob. 30ACh. 11 - Prob. 31ACh. 11 - Prob. 32ACh. 11 - Prob. 33ACh. 11 - Prob. 34ACh. 11 - Prob. 35ACh. 11 - Prob. 36ACh. 11 - Prob. 37ACh. 11 - Prob. 38ACh. 11 - Prob. 39ACh. 11 - Prob. 40ACh. 11 - Prob. 41ACh. 11 - Prob. 42ACh. 11 - Prob. 43ACh. 11 - Prob. 44ACh. 11 - Prob. 45ACh. 11 - Prob. 46ACh. 11 - Prob. 47ACh. 11 - Prob. 48ACh. 11 - Prob. 49ACh. 11 - Prob. 50ACh. 11 - Prob. 51ACh. 11 - Prob. 52ACh. 11 - Prob. 53ACh. 11 - Prob. 54ACh. 11 - Prob. 55ACh. 11 - Prob. 56ACh. 11 - Prob. 57ACh. 11 - Prob. 58ACh. 11 - Prob. 59ACh. 11 - Prob. 60ACh. 11 - Prob. 61ACh. 11 - Prob. 62ACh. 11 - Prob. 63ACh. 11 - Prob. 64ACh. 11 - Prob. 65ACh. 11 - Prob. 66ACh. 11 - Prob. 67ACh. 11 - Prob. 68ACh. 11 - Prob. 69ACh. 11 - Prob. 70ACh. 11 - Prob. 71ACh. 11 - Prob. 72ACh. 11 - Prob. 73ACh. 11 - Prob. 74ACh. 11 - Prob. 75ACh. 11 - Prob. 76ACh. 11 - Prob. 77ACh. 11 - Prob. 1STPCh. 11 - Prob. 2STPCh. 11 - Prob. 3STPCh. 11 - Prob. 4STPCh. 11 - Prob. 5STPCh. 11 - Prob. 6STPCh. 11 - Prob. 7STPCh. 11 - Prob. 8STPCh. 11 - Prob. 9STPCh. 11 - Prob. 10STPCh. 11 - Prob. 11STPCh. 11 - Prob. 12STP
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