Question

The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives:

• A. Zn(OH)₂
• B. ZnO
• C. [Zn(OH)₄]^2–
• D. [ZnO₂]^2–

Answer 1

The Correct Option is D

When zinc reacts with excess aqueous alkali, such as with a strong base like NaOH, hydrogen gas is evolved, and an anionic complex forms. This reaction can be represented by the chemical equation:

\(2\text{Zn} + 2\text{NaOH} + 2\text{H}_2\text{O} \rightarrow 2\text{Na}[ZnO_2] + 3\text{H}_2\uparrow\)

In this reaction: 

1. Metallic zinc reacts with sodium hydroxide, a strong alkali.
2. The water present in an aqueous medium helps to facilitate the reaction, resulting in the formation of sodium zincate and hydrogen gas.
3. The complex formed, sodium zincate, contains the anion \([\text{ZnO}_2]^{2-}\).

To understand why \([\text{ZnO}_2]^{2-}\) is formed rather than other options:

• Zn(OH)₂: Zinc hydroxide is typically formed when zinc reacts with limited alkali (e.g., not in excess), resulting in an insoluble hydroxide which does not occur in this scenario.
• ZnO: Zinc oxide would require a different reaction condition such as heating zinc with oxygen or insufficient alkali; it is not produced in this aqueous alkaline scenario.
• [Zn(OH)₄]^2–: Though [Zn(OH)₄]^2– is a tetrahydroxo complex, the correct answer in the context of the reaction with excess alkali typically leads to the formation of [ZnO₂]^2–, the sodium zincate ion.

Therefore, the correct product formed when zinc reacts with excess aqueous alkali is the complex ion \([\text{ZnO}_2]^{2-}\), making the answer: [ZnO₂]^2–.

Answer 2

Zinc reacting with an excess of aqueous alkali to produce hydrogen gas.
\(Zn + 2NaOH → Na_2ZnO_2 + H_2\)
                                  ^{Sodium Zincate    Hydrogen gas}
Thus along with H₂ it gives Na₂ZnO₂ or ZnO₂^2-.
So, the correct option is (D): [ZnO₂]^2–