Principles of Chemistry: A Molecular Approach (3rd Edition)
Principles of Chemistry: A Molecular Approach (3rd Edition)
3rd Edition
ISBN: 9780321971944
Author: Nivaldo J. Tro
Publisher: PEARSON
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Chapter 8, Problem 8.1P
Interpretation Introduction

Interpretation: The electronic configuration for each of the given elements is to be stated.

Concept introduction: The electrons of an atom or molecule are distributed in molecular or atomic orbitals. This distribution of electrons is known as the electronic configuration of the atom or molecule.

To determine: The electronic configuration for each of the given elements.

Expert Solution & Answer
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Answer to Problem 8.1P

Solution: The electronic configuration of Clis [Ne]3s23p5 . The electronic configuration of Siis [Ne]3s23p2 . The electronic configuration of Sris [Kr]5s2 . The electronic configuration of Ois [He]2s22p4 .

(a)

Explanation of Solution

Atomic number is the total number of electrons present in a neutral atom.

On basis of the periodic table of elements, the atomic number (Z)of chlorine is 17 .

So, chlorine has 17 electrons which are distributed in respective orbitals.

The electronic configuration of element Clis [Ne]3s23p5 .

Where, [Ne]represents the inner electrons this is the nearest noble gas and the electronic configuration for [ Ne]is 1s22s22p6 .

(b)

Atomic number is the total number of electrons present in a neutral atom.

On basis of the periodic table of elements, the atomic number (Z)of silicon is 14 .

So, silicon has 14electrons which are distributed in respective orbitals.

The electronic configuration of element Siis [Ne]3s23p2 .

Where, [Ne]represents the inner electrons this is the nearest noble gas and the electronic configuration for [Ne]is 1s22s22p6 .

(c)

Atomic number is the total number of electrons present in a neutral atom.

On basis of the periodic table of elements, the atomic number (Z)of strontium is 38 .

So, strontium has 38electrons which are distributed in respective orbitals.

The electronic configuration of element Sris [Kr]5s2 .

Where, [Kr]represents the inner electrons this is the nearest noble gas and the electronic configuration for [Kr]is [Ar]3d104s24p6 .

(d)

Atomic number is the total number of electrons present in a neutral atom.

On basis of the periodic table of elements, the atomic number (Z)of oxygen is 8 .

So, oxygen has 8electrons which are distributed in their respective orbitals.

The electronic configuration of element Ois [He]2s22p4 .

Where, [He]represents the inner electrons this is the nearest noble gas and the electronic configuration for [He]is 1s2 .

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Chapter 8 Solutions

Principles of Chemistry: A Molecular Approach (3rd Edition)

Ch. 8 - Prob. 8.8PCh. 8 - Prob. 8.8MPCh. 8 - Prob. 8.9PCh. 8 - Prob. 8.9MPCh. 8 - Prob. 1SAQCh. 8 - Prob. 2SAQCh. 8 - Prob. 3SAQCh. 8 - Prob. 4SAQCh. 8 - Prob. 5SAQCh. 8 - Prob. 6SAQCh. 8 - Prob. 7SAQCh. 8 - Prob. 8SAQCh. 8 - Prob. 9SAQCh. 8 - Prob. 10SAQCh. 8 - Prob. 11SAQCh. 8 - Prob. 12SAQCh. 8 - Prob. 13SAQCh. 8 - Prob. 14SAQCh. 8 - Prob. 15SAQCh. 8 - Prob. 1ECh. 8 - Prob. 2ECh. 8 - Prob. 3ECh. 8 - Prob. 4ECh. 8 - Prob. 5ECh. 8 - Prob. 6ECh. 8 - Prob. 7ECh. 8 - Prob. 8ECh. 8 - Prob. 9ECh. 8 - Prob. 10ECh. 8 - Prob. 11ECh. 8 - Prob. 12ECh. 8 - Prob. 13ECh. 8 - Prob. 14ECh. 8 - Prob. 15ECh. 8 - Prob. 16ECh. 8 - Which electrons experience a greater effective...Ch. 8 - Prob. 18ECh. 8 - Prob. 19ECh. 8 - Prob. 20ECh. 8 - Prob. 21ECh. 8 - Prob. 22ECh. 8 - Prob. 23ECh. 8 - Prob. 24ECh. 8 - Prob. 25ECh. 8 - Prob. 26ECh. 8 - Prob. 27ECh. 8 - Prob. 28ECh. 8 - Prob. 29ECh. 8 - Prob. 30ECh. 8 - Prob. 31ECh. 8 - Prob. 32ECh. 8 - Prob. 33ECh. 8 - Prob. 34ECh. 8 - Prob. 35ECh. 8 - Prob. 36ECh. 8 - Prob. 37ECh. 8 - Consider this set of successive ionization...Ch. 8 - Prob. 39ECh. 8 - Prob. 40ECh. 8 - Prob. 41ECh. 8 - Prob. 42ECh. 8 - Prob. 43ECh. 8 - Prob. 44ECh. 8 - Prob. 45ECh. 8 - Prob. 46ECh. 8 - Prob. 47ECh. 8 - Prob. 48ECh. 8 - Prob. 49ECh. 8 - Prob. 50ECh. 8 - Prob. 51ECh. 8 - Prob. 52ECh. 8 - Prob. 53ECh. 8 - Prob. 54ECh. 8 - Prob. 55ECh. 8 - Prob. 56ECh. 8 - Prob. 57ECh. 8 - Prob. 58ECh. 8 - Prob. 59ECh. 8 - Prob. 60ECh. 8 - Prob. 61ECh. 8 - Prob. 62ECh. 8 - Prob. 63ECh. 8 - The first ionization energy of sodium is 496...Ch. 8 - Prob. 65ECh. 8 - Prob. 66ECh. 8 - Consider the series of elements: B, C, N, O, F. a....Ch. 8 - Prob. 68ECh. 8 - Prob. 69ECh. 8 - Prob. 70ECh. 8 - Prob. 71ECh. 8 - Prob. 72ECh. 8 - Prob. 73ECh. 8 - Prob. 74ECh. 8 - Prob. 75ECh. 8 - Prob. 76ECh. 8 - Prob. 78ECh. 8 - Prob. 79ECh. 8 - Prob. 80ECh. 8 - Prob. 81ECh. 8 - Prob. 82ECh. 8 - Prob. 83ECh. 8 - Prob. 84ECh. 8 - Prob. 85E
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