Verified Solution • 2025–26 Session moderate

QJEEM22Ch7c38: JEE Main 2022 Chemistry - Class 12 JEEM22Ch7c38 Solution: The reaction of zinc with excess of aqueous alkali, evolves…

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

Multiple Choice Options

  • A Zn(OH)2\mathrm{Zn}(\mathrm{OH})_{2}
  • B ZnO\mathrm{ZnO}
  • C [Zn(OH)4]2−\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}
  • D [ZnO2]2−\left[\mathrm{ZnO}_{2}\right]^{2-}

Step-by-Step Verified Solution

Zinc dissolves in excess of aqueous alkali.

Zn+2OH−+2H2O⟶[Zn(OH)4]2−+H2↑\mathrm{Zn}+2 \mathrm{OH}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}+\mathrm{H}_{2} \uparrow (Tetrahydroxozincate(II) ion)

However, this reaction in NCERT is given as

Zn+2NaOH⟶Na2ZnO2+H2↑\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} \uparrow

ZnO22−\mathrm{ZnO}_{2}^{2-} is anhydrous form of [Zn(OH)4]2−\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-} .

ZnO22−+2H2O⇌[Zn(OH)4]2−\mathrm{ZnO}_{2}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}

So in aqueous medium best answer of this question is [Zn(OH)4]2−\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-} .

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