Question

When metal XX is treated with sodium hydroxide, a white ppt. of AA is obtained which is soluble in excess of NaOHNaOH to form soluble complex BB. Compound AA is soluble in dilute HClHCl to form compound CC. The compound AA when strongly heated gives compound DD which is used to extract metals.

Compound $A$ is :

• A. $Al(OH{)}_3$
• B. $Cu(OH{)}_2$
• C. $Hg(OH{)}_2$
• D. $Mg(OH{)}_2$

Answer 1

The correct option is A $Al(OH{)}_3$

$X+NaOH⟶A$

(white ppt.)

Compound A is $Al(OH{)}_3$, metal $X$ is $Al$ because:

$Al+3NaOH⟶Al(OH{)}_3+3N{a}^{+}$

(white ppt.)

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

(soluble complex)

$Al(OH{)}_3+3HCl⟶AlC{l}_3+3H_{2}O$

(dil) (Compound C)

$2Al(OH{)}_3\underset{Heating△}{\overset{strong}{\to }}A{l}_2{O}_3+3H_{2}O$

$\downarrow$

(used to extract metals)

Compound $D=A{l}_2{O}_3$

Compound $A=Al(OH{)}_3$

Compound $B=N{a}^{+}\left[Al\right(OH{)}_4{]}^{-}$

Compound $C=AlC{l}_3$

Compound $D=A{l}_2{O}_3$