Question:
In the following unbalanced reaction, the product formed is
\[
Al_2O_3 + NaOH(aq) + H_2O \rightarrow
\]
In the following unbalanced reaction, the product formed is
\[ Al_2O_3 + NaOH(aq) + H_2O \rightarrow \]
\[ Al_2O_3 + NaOH(aq) + H_2O \rightarrow \]
Show Hint
Aluminium oxide is amphoteric and dissolves in aqueous alkali to form aluminate complexes:
\[
Al_2O_3 + NaOH + H_2O
\rightarrow
\text{Sodium aluminate complex}
\]
In excess aqueous \(NaOH\), the commonly represented product is
\[
Na_3[Al(OH)_6].
\]
Updated On: Jun 26, 2026
- \(Na_3[Al(OH)_6]\)
- \(Na_3[Al(OH)_4]\)
- \(Na_2[Al(OH)_5]\)
- \(Na[Al(OH)_6]\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Identify the nature of \(Al_2O_3\).
Aluminium oxide is an amphoteric oxide.
It reacts with both acids and bases.
In the presence of aqueous sodium hydroxide, aluminium oxide behaves as an acidic oxide and dissolves to form a complex aluminate ion.
Step 2: Write the reaction with aqueous sodium hydroxide.
When \(Al_2O_3\) reacts with excess aqueous \(NaOH\) in the presence of water, the hexahydroxoaluminate(III) complex is formed: \[ Al_2O_3 + 6NaOH + 3H_2O \rightarrow 2Na_3[Al(OH)_6] \] Thus, the product formed is \[ Na_3[Al(OH)_6] \]
Step 3: Verify the complex ion formed.
The complex anion present is \[ [Al(OH)_6]^{3-} \] Since its charge is \(-3\), three sodium ions are required for electrical neutrality.
Therefore, the compound formed is \[ Na_3[Al(OH)_6] \]
Step 4: Analyze the other options.
\[ Na_3[Al(OH)_4] \] is not electrically neutral because \[ [Al(OH)_4]^- \] requires only one \(Na^+\).
\[ Na_2[Al(OH)_5] \] and \[ Na[Al(OH)_6] \] do not correspond to the stable aluminate complex formed in aqueous alkali.
Step 5: Final conclusion.
Hence, the product formed is \[ \boxed{Na_3[Al(OH)_6]} \] Therefore, the correct option is \[ \boxed{(1)} \]
Aluminium oxide is an amphoteric oxide.
It reacts with both acids and bases.
In the presence of aqueous sodium hydroxide, aluminium oxide behaves as an acidic oxide and dissolves to form a complex aluminate ion.
Step 2: Write the reaction with aqueous sodium hydroxide.
When \(Al_2O_3\) reacts with excess aqueous \(NaOH\) in the presence of water, the hexahydroxoaluminate(III) complex is formed: \[ Al_2O_3 + 6NaOH + 3H_2O \rightarrow 2Na_3[Al(OH)_6] \] Thus, the product formed is \[ Na_3[Al(OH)_6] \]
Step 3: Verify the complex ion formed.
The complex anion present is \[ [Al(OH)_6]^{3-} \] Since its charge is \(-3\), three sodium ions are required for electrical neutrality.
Therefore, the compound formed is \[ Na_3[Al(OH)_6] \]
Step 4: Analyze the other options.
\[ Na_3[Al(OH)_4] \] is not electrically neutral because \[ [Al(OH)_4]^- \] requires only one \(Na^+\).
\[ Na_2[Al(OH)_5] \] and \[ Na[Al(OH)_6] \] do not correspond to the stable aluminate complex formed in aqueous alkali.
Step 5: Final conclusion.
Hence, the product formed is \[ \boxed{Na_3[Al(OH)_6]} \] Therefore, the correct option is \[ \boxed{(1)} \]
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