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, Problem 72A

Chapter 7, Problem 74A

Chapter 7 Solutions

Chemistry: Matter and Change

Ch. 7.1 - Prob. 1SSC Ch. 7.1 - Prob. 2SSC Ch. 7.1 - Prob. 3SSC Ch. 7.1 - Prob. 4SSC Ch. 7.1 - Prob. 5SSC Ch. 7.1 - Prob. 6SSC Ch. 7.2 - Prob. 7PP Ch. 7.2 - Prob. 8PP Ch. 7.2 - Prob. 9PP Ch. 7.2 - Prob. 10PP

Ch. 7.2 - Prob. 11PP Ch. 7.2 - Prob. 12SSC Ch. 7.2 - Prob. 13SSC Ch. 7.2 - Prob. 14SSC Ch. 7.2 - Prob. 15SSC Ch. 7.2 - Prob. 16SSC Ch. 7.2 - Prob. 17SSC Ch. 7.2 - Prob. 18SSC Ch. 7.3 - Prob. 19PP Ch. 7.3 - Prob. 20PP Ch. 7.3 - Prob. 21PP Ch. 7.3 - Prob. 22PP Ch. 7.3 - Prob. 23PP Ch. 7.3 - Prob. 24PP Ch. 7.3 - Prob. 25PP Ch. 7.3 - Prob. 26PP Ch. 7.3 - Prob. 27PP Ch. 7.3 - Prob. 28PP Ch. 7.3 - Prob. 29PP Ch. 7.3 - Prob. 30PP Ch. 7.3 - Prob. 31PP Ch. 7.3 - Prob. 32PP Ch. 7.3 - Prob. 33PP Ch. 7.3 - Prob. 34SSC Ch. 7.3 - Prob. 35SSC Ch. 7.3 - Prob. 36SSC Ch. 7.3 - Prob. 37SSC Ch. 7.3 - Prob. 38SSC Ch. 7.3 - Prob. 39SSC Ch. 7.4 - Prob. 40SSC Ch. 7.4 - Prob. 41SSC Ch. 7.4 - Prob. 42SSC Ch. 7.4 - Prob. 43SSC Ch. 7.4 - Prob. 44SSC Ch. 7.4 - Prob. 45SSC Ch. 7 - How do positive ions and negative ions form?Ch. 7 - Prob. 47A Ch. 7 - Prob. 48A Ch. 7 - Prob. 49A Ch. 7 - Prob. 50A Ch. 7 - Prob. 51A Ch. 7 - Prob. 52A Ch. 7 - Prob. 53A Ch. 7 - Prob. 54A Ch. 7 - Prob. 55A Ch. 7 - Prob. 56A Ch. 7 - Prob. 57A Ch. 7 - Prob. 58A Ch. 7 - Prob. 59A Ch. 7 - Prob. 60A Ch. 7 - Prob. 61A Ch. 7 - Prob. 62A Ch. 7 - Prob. 63A Ch. 7 - Prob. 64A Ch. 7 - Prob. 65A Ch. 7 - Prob. 66A Ch. 7 - Prob. 67A Ch. 7 - Prob. 68A Ch. 7 - Prob. 69A Ch. 7 - Prob. 70A Ch. 7 - Prob. 71A Ch. 7 - Prob. 72A Ch. 7 - Prob. 73A Ch. 7 - Prob. 74A Ch. 7 - Prob. 75A Ch. 7 - Prob. 76A Ch. 7 - Prob. 77A Ch. 7 - Prob. 78A Ch. 7 - Prob. 79A Ch. 7 - Prob. 80A Ch. 7 - Prob. 81A Ch. 7 - Prob. 82A Ch. 7 - Prob. 83A Ch. 7 - Prob. 84A Ch. 7 - Prob. 85A Ch. 7 - Prob. 86A Ch. 7 - Prob. 87A Ch. 7 - Prob. 88A Ch. 7 - Prob. 89A Ch. 7 - Prob. 90A Ch. 7 - Prob. 91A Ch. 7 - Prob. 92A Ch. 7 - Prob. 93A Ch. 7 - Prob. 94A Ch. 7 - Prob. 95A Ch. 7 - Prob. 96A Ch. 7 - Prob. 97A Ch. 7 - Prob. 98A Ch. 7 - Prob. 99A Ch. 7 - Prob. 100A Ch. 7 - Prob. 101A Ch. 7 - Prob. 102A Ch. 7 - Prob. 103A Ch. 7 - Prob. 104A Ch. 7 - Prob. 105A Ch. 7 - Prob. 106A Ch. 7 - Prob. 107A Ch. 7 - Prob. 108A Ch. 7 - Prob. 109A Ch. 7 - Prob. 110A Ch. 7 - Prob. 111A Ch. 7 - Prob. 112A Ch. 7 - Prob. 113A Ch. 7 - Prob. 114A Ch. 7 - Prob. 115A Ch. 7 - Prob. 116A Ch. 7 - Prob. 117A Ch. 7 - Prob. 118A Ch. 7 - Prob. 119A Ch. 7 - Prob. 120A Ch. 7 - Prob. 121A Ch. 7 - Prob. 122A Ch. 7 - Prob. 123A Ch. 7 - Prob. 124A Ch. 7 - Prob. 125A Ch. 7 - Prob. 126A Ch. 7 - Prob. 127A Ch. 7 - Prob. 128A Ch. 7 - Prob. 129A Ch. 7 - Prob. 130A Ch. 7 - Prob. 131A Ch. 7 - Prob. 132A Ch. 7 - Prob. 133A Ch. 7 - Prob. 134A Ch. 7 - Prob. 1STP Ch. 7 - Prob. 2STP Ch. 7 - Prob. 3STP Ch. 7 - Prob. 4STP Ch. 7 - Prob. 5STP Ch. 7 - Prob. 6STP Ch. 7 - Prob. 7STP Ch. 7 - Prob. 8STP Ch. 7 - Prob. 9STP Ch. 7 - Prob. 10STP Ch. 7 - Prob. 11STP Ch. 7 - Prob. 12STP Ch. 7 - Prob. 13STP Ch. 7 - Prob. 14STP Ch. 7 - Prob. 15STP Ch. 7 - Prob. 16STP Ch. 7 - Prob. 17STP Ch. 7 - Prob. 18STP Ch. 7 - Prob. 19STP Ch. 7 - Prob. 20STP

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