Question

# How can I write the formula for aluminum oxide?

- Aluminum is the element of group 13, and its atomic number is 13 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{1}}$. Therefore, the number of valence electrons is 3 which it can lose to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ . It is written as $A{{l}^{3+}}$ and three electrons take place in the bond formation.
- Oxygen is the element of group 16, and its atomic number is 8 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$. Therefore, the number of valence electrons is 4 and it can gain two electrons to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$. It is written as ${{O}^{2-}}$.
Therefore, the formula will be $A{{l}_{2}}{{O}_{3}}$.