The complex compound formed when aluminum dissolves in aqueous sodium hydroxide is

Analyze the question: Aluminium is amphoteric, so it reacts with a strong base like NaOH. The product is a complex ion called sodium aluminate (sodium tetrahydroxoaluminate).

Concept: Aluminium forms a complex ion with hydroxide ions. In water, the common form is the tetrahydroxoaluminate(III) ion, [Al(OH)4]⁻. The charge is worked out as:

• Al = +3
• 4 OH⁻ = −4
• Net charge = +3 − 4 = −1

So the ion is [Al(OH)4]⁻, and it needs only one Na⁺ to balance the charge.

Solution:

Equation: 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)4] + 3H₂

(Often written as 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂, where NaAlO₂ is the dehydrated form.)

Checking the options:

• A. Na₃Al(OH)₄: the charge would be +3 + 3(−... ) → Al³⁺ + 4OH⁻ gives −1, so three Na⁺ would not balance. ✗
• B. NaAl(OH)₄: Na⁺ + [Al(OH)₄]⁻ is neutral. ✓
• C. NaAl(OH)₃: Na⁺ + Al³⁺ + 3OH⁻ gives a net +1. ✗
• D. Na₂Al(OH)₃: net charge is +2. ✗

Answer: B. NaAl(OH)₄

Tip: For amphoteric metals (Al, Zn, Pb), remember:

• Al with excess NaOH gives Na[Al(OH)₄], with Al having 4 OH groups, so the ion has a −1 charge.
• Zn with NaOH gives Na₂[Zn(OH)₄], with a −2 charge.

Always check that the charges balance to zero. That is the quickest way to eliminate wrong options.