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AnswersHow can I write the formula for aluminum oxide?
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Hint: As the name, aluminum oxide suggests that there are two atoms in the compound that is aluminum and oxygen. Aluminum is the element of group 13 and its valence electrons are 3, and oxygen is the element of group 16 and its valence electrons are 2.
Complete Solution :
- As the name aluminum oxide suggests, there are two atoms in the compound that is aluminum and oxygen.
- 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 ${{O}^{2-}}$.
- The atoms in the compound in such a way that the electrons donated by one atom must be equal to the number of electrons gained by the other atom. So, when two aluminum atoms donate 3 + 3 = 6 electrons then three oxygen atoms will gain 2 + 2 + 2 = 6 electrons.
Therefore, the formula will be $A{{l}_{2}}{{O}_{3}}$.
Note: The bond between the aluminum and oxygen in the aluminum oxide is ionic because the bond is formed between a metal and a nonmetal in which the aluminum is the metal and oxygen is the non-metal.
Complete Solution :
- As the name aluminum oxide suggests, there are two atoms in the compound that is aluminum and oxygen.
- 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 ${{O}^{2-}}$.
- The atoms in the compound in such a way that the electrons donated by one atom must be equal to the number of electrons gained by the other atom. So, when two aluminum atoms donate 3 + 3 = 6 electrons then three oxygen atoms will gain 2 + 2 + 2 = 6 electrons.
Therefore, the formula will be $A{{l}_{2}}{{O}_{3}}$.
Note: The bond between the aluminum and oxygen in the aluminum oxide is ionic because the bond is formed between a metal and a nonmetal in which the aluminum is the metal and oxygen is the non-metal.