In the formation of an ionic bond between aluminum and fluorine, aluminum loses three electrons, represented in orbital notation as \([ \uparrow \downarrow ] [ \uparrow \downarrow ] [ \uparrow \downarrow ]\), and fluorine gains one electron, depicted as \([ \uparrow \downarrow ] [ \uparrow \downarrow ] [ \uparrow \downarrow \uparrow ]\).
Orbital notation is typically used to represent the arrangement of electrons in the electron orbitals of an atom. Ionic bonding involves the transfer of electrons from one atom to another. In the case of aluminum (Al) and fluorine (F), aluminum tends to lose three electrons to achieve a stable electron configuration, while fluorine tends to gain one electron.
1. Aluminum (Al) has the electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^1\).
During ionic bonding, aluminum loses its three valence electrons (\(3s^2 3p^1\)) to achieve a stable \(1s^2 2s^2 2p^6\) configuration.
The orbital notation for aluminum losing its electrons would be: \([ \uparrow \downarrow ] [ \uparrow \downarrow ] [ \uparrow \downarrow ]\).
2. Fluorine (F) has the electron configuration: \(1s^2 2s^2 2p^5\).
Fluorine gains one electron to achieve a stable \(1s^2 2s^2 2p^6\) configuration.
The orbital notation for fluorine gaining an electron would be: \([ \uparrow \downarrow ] [ \uparrow \downarrow ] [ \uparrow \downarrow \uparrow ]\).
The resulting ionic bond involves the transfer of electrons from aluminum to fluorine. The final products are Al\(^{3+}\) and F\(^-\), forming an ionic compound \(AlF_3\). Note that the actual electron transfer occurs in a crystal lattice, and the notation simplifies the process for conceptual understanding.