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 41A

(a)

Interpretation Introduction

Interpretation:

Valence electrons of sodium with Z=11 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 41A

Since 3s is the valence shell therefore valence electron of sodium with Z=11 is 1.

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 sodium with Z=11 is as follows:

  1s22s22p63s1

Since 3s is the valence shell therefore valence electron of sodium with Z=11 is 1.

(b)

Interpretation Introduction

Interpretation:

Valence electrons of calcium with Z=20 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 41A

Since 4s2 is the valence shell therefore valence electron of calcium with Z=20 is 2.

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 calcium with Z=20 is as follows:

  1s22s22p63s13p64s2

Since 4s2 is the valence shell therefore valence electron of calcium with Z=20 is 2.

(c)

Interpretation Introduction

Interpretation:

Valence electrons of iodine with Z=53 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 41A

Since 5s5p is the valence shell therefore valence electron of iodine with Z=53 is 7.

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 iodine with Z=53 is as follows:

  1s22s22p63s23p63d104s24p64d105s25p5

Since 5s5p is the valence shell therefore valence electron of iodine with Z=53 is 7.

(d)

Interpretation Introduction

Interpretation:

Valence electrons of nitrogen with Z=7 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 41A

Since 2s2p is the valence shell therefore valence electron of nitrogen with Z=7 is 5.

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 nitrogen with Z=7 is as follows:

  1s22s22p3

Since 2s2p is the valence shell therefore valence electron of nitrogen with Z=7 is 5.

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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