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
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Chapter 20.2, Problem 18LC
Interpretation Introduction

Interpretation: The use of the charge of polyatomic ions in assigning oxidation numbers is to be determined.

Concept introduction: The zero, positive or negative numbers that are assigned to an atom for indicating its degree of oxidation and reduction is the oxidation number. This number indicates the number of electrons that are either gained or lost during the bond formation in a chemical compound.

Expert Solution & Answer
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Answer to Problem 18LC

The charge of polyatomic ions can be used for obtaining the oxidation number of elements present in the compound by taking the total sum of oxidation numbers as the charge of that polyatomic ion.

Explanation of Solution

The oxidation number of elements in a polyatomic ion can be calculated with the help of the total charge present in that compound. For a polyatomic ion, the total sum of oxidation numbers of elements is taken as the charge of the whole ion.

For example, the polyatomic ion, NO3 the oxidation number of nitrogen can be calculated by using this rule. And also the oxidation number of oxygen is -2.

Let x be the oxidation number of nitrogen in NO3 ,

  x+1+2×2=1x6=1x=61x=5

Therefore, the oxidation number of nitrogen is +5 and is calculated from the charge of the polyatomic ion.

Chapter 20 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 20.2 - Prob. 11SPCh. 20.2 - Prob. 12SPCh. 20.2 - Prob. 13SPCh. 20.2 - Prob. 14SPCh. 20.2 - Prob. 15SPCh. 20.2 - Prob. 16LCCh. 20.2 - Prob. 17LCCh. 20.2 - Prob. 18LCCh. 20.2 - Prob. 19LCCh. 20.2 - Prob. 20LCCh. 20.3 - Prob. 21SPCh. 20.3 - Prob. 22SPCh. 20.3 - Prob. 23SPCh. 20.3 - Prob. 24SPCh. 20.3 - Prob. 25SPCh. 20.3 - Prob. 26SPCh. 20.3 - Prob. 27LCCh. 20.3 - Prob. 28LCCh. 20.3 - Prob. 29LCCh. 20.3 - Prob. 30LCCh. 20.3 - Prob. 31LCCh. 20 - Prob. 32ACh. 20 - Prob. 33ACh. 20 - Prob. 34ACh. 20 - Prob. 35ACh. 20 - Prob. 36ACh. 20 - Prob. 37ACh. 20 - Prob. 38ACh. 20 - Prob. 39ACh. 20 - Prob. 40ACh. 20 - Prob. 41ACh. 20 - Prob. 42ACh. 20 - Prob. 43ACh. 20 - Prob. 44ACh. 20 - Prob. 45ACh. 20 - Prob. 46ACh. 20 - Prob. 47ACh. 20 - Prob. 48ACh. 20 - Prob. 49ACh. 20 - Prob. 50ACh. 20 - Prob. 51ACh. 20 - Prob. 52ACh. 20 - Prob. 53ACh. 20 - Prob. 54ACh. 20 - Prob. 55ACh. 20 - Prob. 56ACh. 20 - Prob. 57ACh. 20 - Prob. 58ACh. 20 - Prob. 59ACh. 20 - Prob. 60ACh. 20 - Prob. 61ACh. 20 - Prob. 62ACh. 20 - Prob. 63ACh. 20 - Prob. 64ACh. 20 - Prob. 65ACh. 20 - Prob. 66ACh. 20 - Prob. 67ACh. 20 - Prob. 68ACh. 20 - Prob. 69ACh. 20 - Prob. 70ACh. 20 - Prob. 71ACh. 20 - Prob. 72ACh. 20 - Prob. 73ACh. 20 - Prob. 74ACh. 20 - Prob. 75ACh. 20 - Prob. 76ACh. 20 - Prob. 77ACh. 20 - Prob. 78ACh. 20 - Prob. 79ACh. 20 - Prob. 80ACh. 20 - Prob. 81ACh. 20 - Prob. 82ACh. 20 - Prob. 83ACh. 20 - Prob. 84ACh. 20 - Prob. 85ACh. 20 - Prob. 86ACh. 20 - Prob. 87ACh. 20 - Prob. 88ACh. 20 - Prob. 89ACh. 20 - Prob. 90ACh. 20 - Prob. 91ACh. 20 - Prob. 92ACh. 20 - Prob. 93ACh. 20 - Prob. 94ACh. 20 - Prob. 95ACh. 20 - Prob. 96ACh. 20 - Prob. 97ACh. 20 - Prob. 98ACh. 20 - Prob. 99ACh. 20 - Prob. 100ACh. 20 - Prob. 1STPCh. 20 - Prob. 2STPCh. 20 - Prob. 3STPCh. 20 - Prob. 4STPCh. 20 - Prob. 5STPCh. 20 - Prob. 6STPCh. 20 - Prob. 7STPCh. 20 - Prob. 8STPCh. 20 - Prob. 9STPCh. 20 - Prob. 10STPCh. 20 - Prob. 11STP
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