# Question #41f77

Nov 15, 2015

The most common oxide of aluminum is $A {l}_{2} {O}_{3}$.

#### Explanation:

An oxide is a binary compound with oxygen and another element (in this case, aluminum).

Oxygen typically has a charge of $- 2$ since it likes to gain $2$ electrons to fill its valence shell. Aluminum commonly has a charge of $+ 3$ since it likes to lose $3$ electrons to reach noble gas configuration.

In order for the compound to form, there needs to be the same amount of electrons gained and lost, so there needs to be a common multiple of electrons either gained or lost on each side: $6$.

Since oxygen has a $- 2$ charge, it will be denoted as ${O}_{3}$, showing that there are $3$ oxygen atoms total, which will gain $6$ electrons in all. Aluminum's $+ 3$ charge means it will be in the compound in the form $A {l}_{2}$, since the $2$ aluminum atoms will donate $3$ electrons each, $6$ total.

The two elements in the compound will be written together, starting with the more metallic element, or the element further left on the periodic table: $A {l}_{2} {O}_{3}$.

Because $A l$ sometimes forms $+ 1$ ions so that it will still have a full $2 s$ subshell, and for other exceptions, aluminum($I$) oxide ($A l O$) and aluminum($I I$) oxide ($A {l}_{2} O$) also form, but they only do so very rarely.

Nov 15, 2015

Al2O3

#### Explanation:

Al has oxidation state +3 and Oxygen has -2 i.e $A {l}^{+} 3 , {O}^{-} 2$. By combining the oxidation state are cross multiplied.