First, let's make the electronic distribution for neutral atom of Al.
13Al: 1s² 2s² 2p⁶ 3s² 3p¹
Al3+ lost 3 electrons, so we need to take 3 electrons of the electronic distribution of neutral Al.
13Al3+: 1s² 2s² 2p⁶
Answer: 1s2 2s2 2p6
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