Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 17, Problem 91A
Interpretation Introduction

Interpretation:

The change in ionization energy needs to be explained as go from left to right side along the period 1-5 of the periodic table.

Concept introduction:

The ionization energy is the amount of energy which is required to remove the electron form the outermost shell.

Expert Solution & Answer
Check Mark

Answer to Problem 91A

On moving from left to right side along the period 1-5 of the periodic table, the ionization energy is increased.

Explanation of Solution

The ionization energy is minimum amount of energy which is required to remove the electrons from the outermost shell. It is represented in KJ/mol. Atomic radius is the distance between the nucleus and outermost shell electron of the element. As we go from left to right side along the period 1-5 of the periodic table, the ionization energy is increases because, due to the decrease in atomic radius, it becomes complicated to remove an electron. So, large energy is required to remove the electrons. Hence, the ionization energy is increased.

Conclusion

Along the period 1-5, as we going from left to right the ionization energy is increased.

Chapter 17 Solutions

Chemistry: Matter and Change

Ch. 17.1 - Prob. 11SSCCh. 17.1 - Prob. 12SSCCh. 17.2 - Prob. 13SSCCh. 17.2 - Prob. 14SSCCh. 17.2 - Prob. 15SSCCh. 17.2 - Prob. 16SSCCh. 17.2 - Prob. 17SSCCh. 17.3 - Prob. 18PPCh. 17.3 - Prob. 19PPCh. 17.3 - Prob. 20PPCh. 17.3 - Prob. 21PPCh. 17.3 - Prob. 22PPCh. 17.3 - Prob. 23PPCh. 17.3 - Prob. 24PPCh. 17.3 - Prob. 25PPCh. 17.3 - Prob. 26PPCh. 17.3 - Prob. 27SSCCh. 17.3 - Prob. 28SSCCh. 17.3 - Prob. 29SSCCh. 17.3 - Prob. 30SSCCh. 17.3 - Prob. 31SSCCh. 17.3 - Prob. 32SSCCh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Why are compounds such as sodium chloride usually...Ch. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 70ACh. 17 - Prob. 71ACh. 17 - Prob. 72ACh. 17 - Prob. 73ACh. 17 - Prob. 74ACh. 17 - Prob. 75ACh. 17 - Prob. 76ACh. 17 - Prob. 77ACh. 17 - Prob. 78ACh. 17 - Evaluate this statement: A low value for Keq means...Ch. 17 - Prob. 80ACh. 17 - Prob. 81ACh. 17 - Prob. 82ACh. 17 - Prob. 83ACh. 17 - Prob. 84ACh. 17 - Prob. 85ACh. 17 - Prob. 86ACh. 17 - Prob. 87ACh. 17 - Prob. 88ACh. 17 - Prob. 89ACh. 17 - Prob. 90ACh. 17 - Prob. 91ACh. 17 - Prob. 92ACh. 17 - Prob. 93ACh. 17 - Prob. 94ACh. 17 - Prob. 95ACh. 17 - Prob. 96ACh. 17 - Prob. 97ACh. 17 - Prob. 98ACh. 17 - Prob. 99ACh. 17 - Prob. 100ACh. 17 - Prob. 101ACh. 17 - Prob. 102ACh. 17 - Prob. 103ACh. 17 - Prob. 104ACh. 17 - Prob. 105ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STPCh. 17 - Prob. 8STPCh. 17 - Prob. 9STPCh. 17 - Prob. 10STPCh. 17 - Prob. 11STPCh. 17 - Prob. 12STPCh. 17 - Prob. 13STPCh. 17 - Prob. 14STPCh. 17 - Prob. 15STPCh. 17 - Prob. 16STPCh. 17 - Prob. 17STPCh. 17 - Prob. 18STP
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