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 77A
Interpretation Introduction

Interpretation : The error in each diagram is to be identified and corrected.

Concept Introduction : Electronic configurations refer to the arrangements of electrons in different orbitals around atoms' nuclei. The Aufbau principle, the Pauli exclusion principle, and Hund's rule are three rules that can be used to determine the electronic configuration of atoms.

Expert Solution
Check Mark

Answer to Problem 77A

The pairing of electrons should not take place until four electrons are present in p orbitals.

The correct diagram of the nitrogen atom is given as:

  1s22s22p3

Explanation of Solution

The energy levels and symbols for each sub-level occupied by an electron are written for an atom's electrical configuration.

The given element is the nitrogen ground state.

The electronic configuration is 1s22s22p3 .

According to Hund’s rule, until every orbital in that subshell has one electron in it, or until it is completely occupied, pairs of electrons cannot exist in orbitals that belong to the same subshell (p, d, or f).

The pairing of electrons will begin in the p orbitals with the entry of four electrons since there are three p orbitals.

The given diagram is wrong since the pairing of electrons takes place when there are only three electrons in p orbitals.

Since in a nitrogen atom, there are only three electrons so no pairing of electrons takes place.

The correct orbitals picture of the nitrogen atom is given as:

  1s22s22p3

Interpretation Introduction

Interpretation : The error in each diagram is to be identified and corrected.

Concept Introduction : Electronic configurations refer to the arrangements of electrons in different orbitals around atoms' nuclei. The Aufbau principle, the Pauli exclusion principle, and Hund's rule are three rules that can be used to determine the electronic configuration of atoms.

Expert Solution
Check Mark

Answer to Problem 77A

The orbital filling of electrons in magnesium is wrong since there are only ten electrons are present.

The correct diagram of the magnesium atom is given as:

  1s22s22p33s2

Explanation of Solution

The energy levels and symbols for each sub-level occupied by an electron are written for an atom's electrical configuration.

The given element is the magnesium ground state.

The electronic configuration is 1s22s22p63s2 . There are total twelve electrons in magnesium.

In the given orbitals filling of magnesium, there are only ten electrons present so it is wrong.

The correct orbitals picture of the magnesium atom is given as:

  1s22s22p33s2

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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