Question

The dissolution of ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ by a solution of${\text{NaOH}}$ results in the formation of:
A. ${\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{4}}}{{\left( {{\text{OH}}} \right)}_{\text{2}}}} \right]^{\text{ + }}}$
B. $\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{3}}}{{\left( {{\text{OH}}} \right)}_{\text{3}}}} \right]$
C. ${\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{2}}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }$
D. $\left[ {Al{{\left( {{H_2}O} \right)}_6}} \right]{\left( {OH} \right)_3}$

Answer 1

Hint: ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$is forming a coordination compound, and to know about which ofoption is correct, we need to know about the coordination number of Al (Aluminium).

Complete step by step solution:
First, we will see what happens when ${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ is reacting with ${\text{NaOH}}$, for that we are going to write the balanced chemical reaction,
${\text{NaOH}}$ will dissociate into ${\text{O}}{{\text{H}}^{\text{ - }}}$ ion and ${\text{N}}{{\text{a}}^ + }$ ion, the ${\text{O}}{{\text{H}}^{\text{ - }}}$ will react with${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}}$ and forms,
${\text{Al}}{\left( {{\text{OH}}} \right)_{\text{3}}} + {\text{O}}{{\text{H}}^{\text{ - }}} \to {\left[ {{\text{Al}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }$
We know that the coordination number of Aluminium is 6 (i.e., Aluminium can accommodate 6 ligands in its inner shells)
As the solution mixture has water in it, therefore the rest of the 2 ligands will be ${{\text{H}}_2}{\text{O}}$.
This results in the formation of ${\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{2}}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }$

Hence, the correct option is (C)i.e., ${\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{2}}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]^ - }$.

Additional information: The coordination number or ligancy of a metal ion is defined as the number of atoms, ions, or molecules bonded to the metal ion.
Ligands are the lewis bases, which donates electrons to the metal ion, and forms a special bond known as coordination covalent bond. Which is denoted by an arrow. L$ \to $M, here L represents ligand and M represents a central metal atom

Note: Only the ligands present in the inner shell (i.ie, inside the parentheses’[]’ ) of a metal ion, will contribute to the coordination number, if a ligand is outside the parentheses then it is not considered while calculating the coordination number. Also, notice that the given options are in ionic form, that’s why we have used the ionic form of NaOH. Otherwise, the answer could also be written as ${\text{Na}}\left[ {{\text{Al}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{2}}}{{\left( {{\text{OH}}} \right)}_{\text{4}}}} \right]$.