How can I write the formula for aluminum oxide?

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How can I write the formula for aluminum oxide?

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Answer 1 · 2014-07-03T06:19:11

The formula for aluminum oxide is $A l_{2} O_{3}$ $Al_2O_3$ .

Explanation:

The correct answer is $A l_{2} O_{3}$ $Al_2O_3$ .

Let us see how we got the answer; Look at the electronic arrangement of Al and O atoms.

Al ( Z= 13) has 13 electrons with following electronic configuration. 1 $s^{2}$ $s^2$ 2 $s^{2}$ $s^2$ 2 $p^{6}$ $p^6$ 3 $s^{2}$ $s^2$ 3 $p^{1}$ $p^1$

It loses three electron in its 3s and 3p subshell to achieve stability and forms ion $A l^{3 +}$ $Al^(3+)$ .

$A l^{3 +}$ $Al^(3+)$ = 1 $s^{2}$ $s^2$ 2 $s^{2}$ $s^2$ 2 $p^{6}$ $p^6$

O ( Z=8) on the other hand has eight electrons and wants to gain two electrons to achieve stable noble gas configuration.Oxygen atom on gaining two electrons forms negative oxide ion, $O^{2 -}$ $O^(2-)$ ion.

O ( Z=7) = 1 $s^{2}$ $s^2$ 2 $s^{2}$ $s^2$ 2 $p^{4}$ $p^4$

$O^{2 -}$ $O^(2-)$ = 1 $s^{2}$ $s^2$ 2 $s^{2}$ $s^2$ 2 $p^{6}$ $p^6$

Three Oxygen atoms gains two electrons each ( total of six) from two Al atoms, each Al atom loses three electrons (total of six) to three oxygen atom , in this process each Al atom becomes $A l^{3 +}$ $Al^(3+)$ ion and Oxygen atom after gaining two electrons becomes $O^{2 -}$ $O^(2-)$ ion.

![http://gallery4share.com/a/http://aluminum-ion-lewis-dot-structure.html](https://useruploads.socratic.org/fdem7yigQvykudaJaOs1_Ionic_compound_of_O_and_Al.jpg)

2 $A l^{3 +}$ $Al^(3+)$ and 3 $O^{2 -}$ $O^(2-)$

or the formula is $A l_{2} O_{3}$ $Al_2O_3$ .

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How can I write the formula for aluminum oxide?

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