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
Question
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Chapter 7, Problem 71A
Interpretation Introduction

Interpretation:

The ionic bond formed between Barium and Nitrogen needs to be explained using electronic configuration diagram

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.

Electronic configuration of atoms depicts the distribution of electrons in its energy shells following certain set of rules laid down by Aufbau’s Principle.

Expert Solution & Answer
Check Mark

Answer to Problem 71A

The ionic compound Barium  Ba3N2 will be formed when 3 Ba2+ combines with 2 N3.

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.

Barium has atomic number of 56 with electronic configuration as 1s22s22p63s23p63d104s24p64d105s25p66s2 or [Xe] 6s2. There are two outermost electrons which barium will lose to form a Barium ion, Ba2+.

Nitrogen has atomic number of 7 with electronic configuration as 1s22s22p3 or [He]2s22p3. It has 3 valence electrons and requires 3 more electron to get octet structure.

To form compound between Barium and Nitrogen, it can be seen that Barium ion is formed by losing two electrons to get Xenon configuration while Nitrogen ion is formed by gaining 3 electrons to get Neon configuration. To bring electrical neutrality, 3 atoms of Barium ion will combine with 2 atoms of Nitrogen ion to form a stable compound. In other words, 6 electrons donated by Barium will be taken up by 6 vacancies in Nitrogen.

Chemistry: Matter and Change, Chapter 7, Problem 71A

Conclusion

The ionic compound Barium  Ba3N2 will be formed when 3 Ba2+ combines with 2 N3.

Chapter 7 Solutions

Chemistry: Matter and Change

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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