Final answer:
The oxidation number for aluminum (Al) in its ionized form (Al3+) is +3, reflecting the loss of three valence electrons from its neutral state. The Lewis dot diagram for aluminum would have three dots for the neutral atom and no dots with a +3 charge for the ionized atom.
Step-by-step explanation:
The oxidation number for aluminum (Al) is determined by its position in the periodic table and its tendency to lose electrons when it forms an ion. In the case of an aluminum ion (Al3+), the oxidation number is +3. This means that in the process of forming an ion, aluminum loses three electrons.
The electron configuration of a neutral aluminum atom with an atomic number of 13 is as follows: 1s2 2s2 2p6 3s2 3p1. The valence electrons, which are the electrons in the outermost shell that determine the chemical reactivity of the element, are underlined in the configuration. Aluminum has three valence electrons (the 3s2 3p1 electrons), which it loses when forming Al3+.
When writing a Lewis dot diagram for aluminum, three dots are placed around the symbol ‘Al’ to represent the three valence electrons. Loss of these three valence electrons in the aluminum atom during oxidation would be represented by no dots around ‘Al’ and the charge +3 written above it as ‘Al3+’.