Question

What is the chemical formula of Aluminum hydroxide?
A.$Al{\left( {OH} \right)_2}$
B.$Al{\left( {OH} \right)_3}$
C.$A{l_2}{\left( {OH} \right)_3}$
D.$A{l_2}\left( {OH} \right)$

Answer 1

Hint: We know that ionic compounds consist of cations and anions. The sum of the charges in an ionic compound is always zero (overall). Ionic bonds held the ions together to form ionic compounds. Due to the presence of ionic bonds between them, they have strong melting and boiling points.

Complete step by step answer:
We can now see the rules for determining the formula of an ionic compound.
Rules for determining the formula of an ionic compound:
The cation and anion is determined, and the periodic table is used to determine the charges.
The formula of the cation is written first, followed by the anion. If the anion and cation have the same charge, one of each ion is needed for an overall charge of zero.
If a compound has unequal charges, the charge on the ions tells us the number of oppositely charged ions required to balance the charge.
Now let us identify the cation and anion in aluminum hydroxide.
In aluminum hydroxide, the cation is aluminum $\left( {A{l^{3 + }}} \right)$.
The anion is a polyatomic ion containing hydrogen and oxygen, the name of the polyatomic ion is hydroxide $\left( {O{H^ - }} \right)$.
We can now write the charges of cation and anion.
The charge on aluminum cation is $ + 3.$
The charge on hydroxide anion is $ - 1.$
From the above statement regarding the charges, we can see that the compound is formed from ions of unequal charges.
The charges are not equal in magnitude, $ + 3$ and $ - 1$.
Three hydroxides ions are required to balance the charges.
$A{l^{3 + }} + O{H^ - }\xrightarrow{{}}Al{\left( {OH} \right)_3}$
The formula of aluminum hydroxide is $Al{\left( {OH} \right)_3}$.

$\therefore $ Option (B) is correct.

Note: We can use aluminum hydroxide as fire retardant filler. It is used as a precursor to aluminum compounds, and also used in pharmaceuticals. It has low toxicity. It is amphoteric. It acts as a Bronsted-Lowry base in acids. It acts as Lewis acids by binding hydroxide ions in bases. A salt is yield when aluminum hydroxide is neutralized with acid.
Example: Formation of sodium bromide.
Sodium is a metal and it forms the cation. Bromine is a non-metal and it forms the anion.
The charge on sodium is ${\text{ + 1}}\,\left( {{\text{N}}{{\text{a}}^{\text{ + }}}} \right){\text{,}}$ as it is located in group${\text{1A}}{\text{.}}$ Bromine is a main group element and is located in group ${\text{7A}}{\text{.}}$ The charge on bromine is ${\text{ - 1}}\,\left( {{\text{B}}{{\text{r}}^{\text{ - }}}} \right){\text{.}}$
One sodium ion and one bromide ion forms sodium bromide.
Each sodium ion needs one bromide ion to balance the charge and the charges are equal in magnitude.
${\text{N}}{{\text{a}}^{\text{ + }}}{\text{ + B}}{{\text{r}}^{\text{ - }}}\xrightarrow{{}}{\text{NaBr}}$
The formula of the ionic compound formed from sodium and bromine is ${\text{NaBr}}{\text{.}}$
Example: Formation of magnesium iodide.
Magnesium is a metal and it forms the cation. Iodine is a non-metal and it forms the anion.
The charge on magnesium is ${\text{ + 2}}\left( {{\text{M}}{{\text{g}}^{{\text{2 + }}}}} \right){\text{,}}$ as it is located in group${\text{2A}}{\text{.}}$ Iodine is a main group element and is located in group ${\text{7A}}{\text{.}}$ The charge on iodine is ${\text{ - 1}}\,\left( {{{\text{I}}^{\text{ - }}}} \right){\text{.}}$
Since the charges are unequal, their magnitudes are used to determine the relative number of each ion to give an overall charge of zero.
${\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{ + }}{{\text{I}}^{\text{ - }}}\xrightarrow{{}}{\text{Mg}}{{\text{I}}_2}$
Two iodine ions $\left( {{{\text{I}}^{\text{ - }}}} \right)$are needed for each magnesium ion$\left( {{\text{M}}{{\text{g}}^{{\text{2 + }}}}} \right){\text{.}}$
The formula of the ionic compound formed from magnesium and iodine is ${\text{Mg}}{{\text{I}}_{\text{2}}}{\text{.}}$