Question

Study the given flow-sheet involving reaction of a metal and answer the question at the end.

A metal M and its compounds can given the following observable changes in a sequence of reaction :

$M\underset{HCl}{\overset{dilute}{\to }}\underset{A}{\left[\text{Colorless solution}\right]}\underset{NaOH}{\overset{aqueous}{\to }}\underset{B}{\left[\text{White precipitate}\right]}\underset{NaO{H}_{\left(aq\right)}}{\overset{excess}{\to }}\underset{C}{\left[\text{Colorless solution}\right]}\underset{NaO{H}_{\left(aq\right)}}{\overset{H_{2}S}{\to }}\underset{D}{\left[\text{White precipitate}\right]}$

Metal $M$ on reaction with $NaOH$ forms:

• A. $H_2$ gas
• B. $O_2$ gas
• C. $M{H}_2$ (hydride)
• D. $M{H}_3$ (hydride)

Answer 1

Answer: (A)

$H_2$ gas

Here, the metal is zinc.

$Zn\underset{HCl}{\stackrel{dilute}{⟶}}ZnC{l}_2+HCl\left[\text{Colorless solution}\right]\underset{NaOH}{\stackrel{aqueous}{⟶}}Zn(OH{)}_2\left[\text{White precipitate}\right]$

$Zn(OH{)}_2\left[\text{White precipitate}\right]\underset{NaO{H}_{(aq.)}}{\stackrel{excess}{⟶}}N{a}_{2}Zn{O}_2\left[\text{Colorless solution}\right]\underset{NaO{H}_{(aq.)}}{\stackrel{H_{2}S}{⟶}}ZnS\left[\text{White ppt}\right]$

Zinc reacts with sodium hydroxide to produce zincate sodium and evolve hydrogen gas.

Hence, option A is correct.