Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 8 - Basic Concepts of Chemical Bonding - Exercises - Page 334: 8.17a

Answer

The ionic compound formed would have the chemical formula $$AlF_3$$

Work Step by Step

The electron configuration of $Al$ is $$[Ne]3s^23p^1$$ The electron configuration of $F$ is $$1s^22s^22p^5$$ Both of these atoms would tend to lose, gain or share electrons so that it could reach an octet, that is to have 8 valence electrons, or in other words, to achieve the configuration of a noble gas. $F$ atom lacks 1 more electron to reach an octet $(2s^22p^6)$, while $Al$ atom, if loses 3 valence electrons, would achieve the configuration of noble gas $Ne$, with 8 valence electrons (an octet). Therefore, $Al$ atom is willing to lose its 3 valence electrons to achieve an octet. However, a $F$ atom only needs 1 electron to reach an octet. So after $F$ atom reaches an octet, 2 more electrons are still available for 2 other $F$ atoms, each gaining 1 electron to reach an octet. Eventually, there is one $Al$ atom and 3 $F$ atoms involved in the reaction so that any atoms can achieve an octet. That means the ionic compound formed would have the chemical formula $$AlF_3$$
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