Question

What is the electronic configuration of $A{l^{3 + }}$ ?

Answer 1

We need to remember that cations are formed when an electron is lost. When this occurs there are less electron-electron repulsions and there is a greater net nuclear attraction per electron. So, the newly formed ion becomes a more condensed version of its neutral atom. Anions are formed when an electron is gained.

Complete answer:
As we know that aluminum is located in period three, group $13$ in the periodic table and has an atomic number equal to $13$.
The electron configuration of the neutral atom will be
$Al$: $1{s^2}2{s^2}2{p^6}3{s^2}3{p^1}$
When aluminum forms ${3^ + }$ cations $A{l^{3 + }}$, it loses three electrons from its outermost energy shell.
$A{l^{3 + }}$: $1{s^2}2{s^2}2{p^6}$
Aluminum is a chemical element with the symbol $Al$ and atomic number $13$. Aluminum has a density lower than those of other common metals, at approximately one third that of steel. It has a great affinity towards oxygen, and forms a protective layer of oxide on the surface when exposed to air. An aluminum atom has $13$ electrons, arranged in an electron configuration of $[Ne]3{s^2}3{p^1}$ with three electrons beyond a stable noble gas configuration. Accordingly, the combined first three ionization energies of aluminum are far lower than the fourth ionization energy alone. Such an electron configuration is shared with the other well-characterized members of its group, boron, gallium, indium, and thallium.

Note:
We have to remember that aluminum is an excellent thermal and electrical conductor, having around $60\%$ the conductivity of copper, both thermal and electrical, while having only $30\%$ of copper's density. Aluminum is capable of superconductivity as well. It combines characteristics of pre- and post-transition metals.