Question

The formula of Aluminium oxide is $$A{l}_2{O}_3$$. Find the valencies of Aluminium and Oxygen.

Answer 1

- Hint: We know that the atomic number of Aluminium is 13 and the atomic number of Oxygen is 8. Using the criss-cross method we can determine the valency of Aluminium and Oxygen in Aluminium oxide or $$A{l}_2{O}_3$$.

Complete step by step answer:
We know that according to Bohr’s atomic model, the number of electrons that can be present on the different shells is determined by the formula $$2(n^2)$$, where n denotes the number of the shell.
Since the atomic number of Aluminium is 13, using the formula $$2(n^2)$$ we get,
In the first shell, the number of electrons is $$2\times 1^2=2$$ electrons.
In the second shell, the number of electrons is $$2\times 2^2=8$$ electrons.
In the third shell, the number of valence electrons is = 3 electrons.

Since $$2+8+3=13$$ electrons, the atomic number of Aluminium is 13.
Now, the atomic number of Oxygen is 8. Using the formula $$2(n^2)$$ we get,
In the first shell, the number of electrons is $$2\times 1^2=2$$ electrons
In the second shell, the number of valence electrons = 6 electrons.
Since $$2+6=8$$ electrons, the atomic number of Oxygen is 8.
Therefore, to attain the octet configuration, Aluminium loses 3 electrons, and Oxygen gains 2 electrons.
Now, we have to write the formula of Aluminium Oxide using the criss-cross method.
We know that in the criss-cross method the numerical value of each of the ion charges is crossed over to become the subscript of the other ion.
The charge of Aluminium is +3 and the charge of Oxygen is -2.

Therefore, we can conclude that the valency of Aluminium in $$A{l}_2{O}_3$$ is 3 and the valency of Oxygen in $$A{l}_2{O}_3$$ is 2.

Note: Aluminium loses the 3 valence electrons to gain the noble gas configuration of Neon, which is 2,8.
Similarly, Oxygen gains 2 valence electrons to gain the noble gas configuration of Neon, that is 2,8.