Aluminum atoms have 3 electrons in the outermost layer, and fluoride atoms have 7 electrons in the outermost layer. When forming a compound, aluminum atoms lose 3 electrons to form an aluminum ion with a positive charge of 3 units ($Al ^ {3 + }$）， its electronic formula (Dot Diagram) can be expressed as $Al ^ {3 + }$ ， there is no point around (because the outermost electron is lost). The fluorine atom will gain 1 electron to form a fluoride ion with a negative charge of 1 unit ($F ^ - $）， its electronic formula is $:\ ddot {F }:^ - $ ， There are 8 points around (including the obtained 1 electron). In aluminum fluoride ($AlF_3 $), three fluorine ions surround one aluminum ion and are bonded by ionic bonds. The electronic formula of aluminum fluoride (Dot Diagram) can be expressed as follows:

$[:\ ddot {F }:^ - ]_ 3 Al ^ {3 + }$

This electronic formula reflects the transfer and distribution of electrons between aluminum and fluorine when forming compounds, and clearly shows the bonding relationship between each atom. The aluminum ion is in the center position, and the three fluorine ions are evenly distributed around it, forming a stable ionic compound structure through electrostatic action.