Question

When metal $X$ is treated with sodium hydroxide, a white precipitate $(A)$ is obtained, which is soluble in excess of $NaOH$ to give soluble complex $(B)$ Compound $(A)$ is soluble in dilute $HCl$ to form compound $(C)$. The compound $(A)$ when heated strongly gives $(D)$, which is used to extract metal. The compound $(D)$ is

• A. $Al(OH{)}_3$
• B. $AlC{l}_3$
• C. $NaAl{O}_2$
• D. $A{l}_2{O}_3$

Answer 1

Answer: (D)

The given data suggests that the metal $X$ is aluminium
$\underset{(X)}{+}3NaOH\to \underset{\text{White gelatinous ppt. (A)}}{Al(OH{)}_3}\underset{NaOH}{\overset{\text{Excess}}{\to }}\underset{\text{Sodium tetrahydroxo aluminate (III) (soluble, B)}}{N{a}^{+}[Al(OH{)}_4{]}^{-}}$
$\underset{(A)}{Al(OH{)}_3}+3HCl⟶\underset{(c)}{AlC{l}_3}+3H_{2}O$
$\underset{(A)}{2Al(OH{)}_3}\stackrel{\Delta }{\to }\underset{(D)}{A{l}_2{O}_3}+3H_{2}O$