Question

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

Metal $X$ is:

• A. $Al$
• B. $Cu$
• C. $Hg$
• D. $Mg$

Answer 1

Answer: (A)

$Al$

$Al+3NaOH⟶Al(OH{)}_3\downarrow +3N{a}^{+}$

$\left(X\right)$ $\left(ppt\right)$

$2Al(OH{)}_3+2NaOH⟶2N{a}^{+}[Al(OH{)}_4{]}^{-}$

$\left(A\right)$ $\left(Excess\right)$ Sodium tetrahydroxoaluminate(III)

$\left(B\right)$

$Al(OH{)}_3+3HCl(aq)⟶AlC{l}_3+3H_{2}O$

$\left(A\right)$ $\left(C\right)$

$2Al(OH{)}_3\underrightarrow{△}A{l}_2{O}_3+3H_{2}O$

$\left(A\right)$ $\left(D\right)$

Metal $X$ on treatment with sodium hydroxide gives white precipitate which dissolves in excess of $NaOH$ to give soluble complex$\left(B\right)$, therefore, the metal $X$ is $Al$.