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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Question
Chapter 7, Problem 73A
Interpretation Introduction
Interpretation:
The feasibility of formation of compounds like CaKr, Na2S, BaCl3, MgF needs to be explained.
Concept introduction:
Molecules are bound together by different types of bonds namely ionic and covalent. Ionic bonds are formed when there is complete transfer of electrons from one atom to another. These atoms either lose or gain electrons to become negatively or positively charged ions. The forces of attraction between these ions cause the ionic bond formation. Ionic compounds are formed only when there is a balance between their charges.
Expert Solution & Answer
Answer to Problem 73A
The compounds CaKr, BaCl3, MgF will not be formed while compound Na2S will be formed.
Explanation of Solution
Ionic bonds involve the transfer of electrons from one atom to the other. The metal atoms, an electrons donor, loses electrons to form a positively charged ion while the non-metal atoms, electron acceptor, gains electrons to form negatively charged ions. During the process of donating or gaining electrons, the atoms follow octet rule to attain a stable noble gas configuration.
Considering CaKr, it can be noted that Ca belongs to Group 2 and Kr belongs to Group 18. Group 2 elements will donate 2 of its outermost electrons while Group 18 elements are Nobel gas. Elements belonging to Nobel gas will not accept or donate electrons. Hence, Compound CaKr will not form.
Considering Na2S, it can be seen that Na belongs to Group 1 and S belongs to Group 16.
Group 1 elements will have only one electron in its outermost shell and donate only one electron. But S will have 6 electrons in its outermost shell and needs two electrons to get octet. Therefore, if 2 Na ions combine with one S compound Na2S is formed.
Considering BaCl3, it can be seen that Ba belongs to Group 2 and Cl belongs to Group 17. Ba will donate two electrons and forms Barium ion. Chlorine has 7 electrons and requires only 1 electron. But as Barium will give 2 electrons and Cl needs only one electron, combining these two elements will not bring an electrical neutral compound. Hence, the compound BaCl3 will not be formed.
Considering MgF, it can be seen that Mg belongs to Group 2 and F belongs to Group 17 or Halogen group. Mg will be ready to donate 2 electrons while F having 7 electrons in its valence shell will accept only electron. Hence, combining Mg and F will not result in MgF.
Conclusion
The compounds CaKr, will not be formed as Kr is a Noble gas element, compounds BaCl3, MgF will not be formed as both does not have electrical neutrality. Compound Na2S will be formed.
Chapter 7 Solutions
Chemistry: Matter and Change
Ch. 7.1 - Prob. 1SSCCh. 7.1 - Prob. 2SSCCh. 7.1 - Prob. 3SSCCh. 7.1 - Prob. 4SSCCh. 7.1 - Prob. 5SSCCh. 7.1 - Prob. 6SSCCh. 7.2 - Prob. 7PPCh. 7.2 - Prob. 8PPCh. 7.2 - Prob. 9PPCh. 7.2 - Prob. 10PP
Ch. 7.2 - Prob. 11PPCh. 7.2 - Prob. 12SSCCh. 7.2 - Prob. 13SSCCh. 7.2 - Prob. 14SSCCh. 7.2 - Prob. 15SSCCh. 7.2 - Prob. 16SSCCh. 7.2 - Prob. 17SSCCh. 7.2 - Prob. 18SSCCh. 7.3 - Prob. 19PPCh. 7.3 - Prob. 20PPCh. 7.3 - Prob. 21PPCh. 7.3 - Prob. 22PPCh. 7.3 - Prob. 23PPCh. 7.3 - Prob. 24PPCh. 7.3 - Prob. 25PPCh. 7.3 - Prob. 26PPCh. 7.3 - Prob. 27PPCh. 7.3 - Prob. 28PPCh. 7.3 - Prob. 29PPCh. 7.3 - Prob. 30PPCh. 7.3 - Prob. 31PPCh. 7.3 - Prob. 32PPCh. 7.3 - Prob. 33PPCh. 7.3 - Prob. 34SSCCh. 7.3 - Prob. 35SSCCh. 7.3 - Prob. 36SSCCh. 7.3 - Prob. 37SSCCh. 7.3 - Prob. 38SSCCh. 7.3 - Prob. 39SSCCh. 7.4 - Prob. 40SSCCh. 7.4 - Prob. 41SSCCh. 7.4 - Prob. 42SSCCh. 7.4 - Prob. 43SSCCh. 7.4 - Prob. 44SSCCh. 7.4 - Prob. 45SSCCh. 7 - How do positive ions and negative ions form?Ch. 7 - Prob. 47ACh. 7 - Prob. 48ACh. 7 - Prob. 49ACh. 7 - Prob. 50ACh. 7 - Prob. 51ACh. 7 - Prob. 52ACh. 7 - Prob. 53ACh. 7 - Prob. 54ACh. 7 - Prob. 55ACh. 7 - Prob. 56ACh. 7 - Prob. 57ACh. 7 - Prob. 58ACh. 7 - Prob. 59ACh. 7 - Prob. 60ACh. 7 - Prob. 61ACh. 7 - Prob. 62ACh. 7 - Prob. 63ACh. 7 - Prob. 64ACh. 7 - Prob. 65ACh. 7 - Prob. 66ACh. 7 - Prob. 67ACh. 7 - Prob. 68ACh. 7 - Prob. 69ACh. 7 - Prob. 70ACh. 7 - Prob. 71ACh. 7 - Prob. 72ACh. 7 - Prob. 73ACh. 7 - Prob. 74ACh. 7 - Prob. 75ACh. 7 - Prob. 76ACh. 7 - Prob. 77ACh. 7 - Prob. 78ACh. 7 - Prob. 79ACh. 7 - Prob. 80ACh. 7 - Prob. 81ACh. 7 - Prob. 82ACh. 7 - Prob. 83ACh. 7 - Prob. 84ACh. 7 - Prob. 85ACh. 7 - Prob. 86ACh. 7 - Prob. 87ACh. 7 - Prob. 88ACh. 7 - Prob. 89ACh. 7 - Prob. 90ACh. 7 - Prob. 91ACh. 7 - Prob. 92ACh. 7 - Prob. 93ACh. 7 - Prob. 94ACh. 7 - Prob. 95ACh. 7 - Prob. 96ACh. 7 - Prob. 97ACh. 7 - Prob. 98ACh. 7 - Prob. 99ACh. 7 - Prob. 100ACh. 7 - Prob. 101ACh. 7 - Prob. 102ACh. 7 - Prob. 103ACh. 7 - Prob. 104ACh. 7 - Prob. 105ACh. 7 - Prob. 106ACh. 7 - Prob. 107ACh. 7 - Prob. 108ACh. 7 - Prob. 109ACh. 7 - Prob. 110ACh. 7 - Prob. 111ACh. 7 - Prob. 112ACh. 7 - Prob. 113ACh. 7 - Prob. 114ACh. 7 - Prob. 115ACh. 7 - Prob. 116ACh. 7 - Prob. 117ACh. 7 - Prob. 118ACh. 7 - Prob. 119ACh. 7 - Prob. 120ACh. 7 - Prob. 121ACh. 7 - Prob. 122ACh. 7 - Prob. 123ACh. 7 - Prob. 124ACh. 7 - Prob. 125ACh. 7 - Prob. 126ACh. 7 - Prob. 127ACh. 7 - Prob. 128ACh. 7 - Prob. 129ACh. 7 - Prob. 130ACh. 7 - Prob. 131ACh. 7 - Prob. 132ACh. 7 - Prob. 133ACh. 7 - Prob. 134ACh. 7 - Prob. 1STPCh. 7 - Prob. 2STPCh. 7 - Prob. 3STPCh. 7 - Prob. 4STPCh. 7 - Prob. 5STPCh. 7 - Prob. 6STPCh. 7 - Prob. 7STPCh. 7 - Prob. 8STPCh. 7 - Prob. 9STPCh. 7 - Prob. 10STPCh. 7 - Prob. 11STPCh. 7 - Prob. 12STPCh. 7 - Prob. 13STPCh. 7 - Prob. 14STPCh. 7 - Prob. 15STPCh. 7 - Prob. 16STPCh. 7 - Prob. 17STPCh. 7 - Prob. 18STPCh. 7 - Prob. 19STPCh. 7 - Prob. 20STP
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