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 45A

(a)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of phosphorus with Z=15 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 45A

Valence electronic configuration using previous noble gas of phosphorus with Z=15 is as follows:

  [Ne]3s23p3

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 phosphorus with Z=15 is as follows:

  1s22s22p63s23p3

Hence valence electronic configuration using previous noble gas of phosphorus with Z=15 is as follows:

  [Ne]3s23p3

(b)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of chlorine with Z=17 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 45A

Valence electronic configuration using previous noble gas of chlorine with Z=17 is as follows:

  [Ne]3s23p5

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 chlorine with Z=17 is as follows:

  1s22s22p63s23p5

Hence valence electronic configuration using previous noble gas of chlorine with Z=17 is as follows:

  [Ne]3s23p5

(c)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of magnesium 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.

(c)

Expert Solution
Check Mark

Answer to Problem 45A

Valence electronic configuration using previous noble gas of magnesium with Z=12 is as follows:

  [Ne]3s2

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

Hence valence electronic configuration using previous noble gas of magnesium with Z=12 is as follows:

  [Ne]3s2

(d)

Interpretation Introduction

Interpretation:

Valence electronic configuration using previous noble gas of zinc with Z=30 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 45A

Valence electronic configuration using previous noble gas of zinc with Z=30 is as follows:

  [Ar]3d104s2

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 zinc with Z=30 is as follows:

  1s22s22p63s23p63d104s2

Hence valence electronic configuration using previous noble gas of zinc with Z=30 is as follows:

  [Ar]3d104s2

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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