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 38A

(a)

Interpretation Introduction

Interpretation:

Electronic configuration of Mg with Z=12 should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is Z then Z numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

(a)

Expert Solution
Check Mark

Answer to Problem 38A

Complete electronic configuration of Magnesium with Z=12 is as follows:

  1s22s22p63s2

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p< 6s<4f<5d<6p<7s<5f<6d<7p .

So complete electronic configuration of Magnesium with Z=12 is as follows:

  1s22s22p63s2

(b)

Interpretation Introduction

Interpretation:

Electronic configuration of Li with Z=3 should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is Z then Z numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

(b)

Expert Solution
Check Mark

Answer to Problem 38A

Complete electronic configuration of Li with Z=3 is as follows:

  1s22s1

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p< 6s<4f<5d<6p<7s<5f<6d<7p .

So complete electronic configuration of Li with Z=3 is as follows:

  1s22s1

(c)

Interpretation Introduction

Interpretation:

Electronic configuration of O with Z=8 should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is Z then Z numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

(c)

Expert Solution
Check Mark

Answer to Problem 38A

Complete electronic configuration of O with Z=8 is as follows:

  1s22s22p4

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p< 6s<4f<5d<6p<7s<5f<6d<7p .

So complete electronic configuration of O with Z=8 is as follows:

  1s22s22p4

(d)

Interpretation Introduction

Interpretation:

Electronic configuration of S with Z=16 should be determined.

Concept introduction:

Electronic configuration can be assigned to any elements in ground when they follow certain rules like Hund rule, Pauli Exclusion Principle and Aufbau rule. If atomic number of an element is Z then Z numbers of electrons are filled into the orbitals which are arranged in increasing order of energy.

No two electrons in an atom can have same group of four quantum numbers and this is Pauli Exclusion Principle.

While filling of orbital’s, electron first enters to each energy level with degenerate energy before paring of electron begins and this is Hund’s rules.

(d)

Expert Solution
Check Mark

Answer to Problem 38A

Complete electronic configuration of S with Z=16 is as follows:

  1s22s22p63s23p4

Explanation of Solution

As per Aufbau rule electrons are filled in lower energy orbitals that are closer to the nucleus before they are filled in higher energy ones. The order of orbital arranged in their increasing energies is as follows:

  1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p< 6s<4f<5d<6p<7s<5f<6d<7p .

So complete electronic configuration of S with Z=16 is as follows:

  1s22s22p63s23p4

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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