JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 4)
The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives :
$$\mathrm{Zn}(\mathrm{OH})_{2}$$
$$\mathrm{ZnO}$$
$$\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$
$$\left[\mathrm{ZnO}_{2}\right]^{2-}$$
Explanation
Zinc dissolves in excess of aqueous alkali.
$$\mathrm{Zn}+2 \mathrm{OH}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}+\mathrm{H}_{2} \uparrow$$ (Tetrahydroxozincate(II) ion)
However, this reaction in NCERT is given as
$$\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} \uparrow$$
$$\mathrm{ZnO}_{2}^{2-}$$ is anhydrous form of $$\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$.
$$\mathrm{ZnO}_{2}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$
So in aqueous medium best answer of this question is $\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$.
$$\mathrm{Zn}+2 \mathrm{OH}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}+\mathrm{H}_{2} \uparrow$$ (Tetrahydroxozincate(II) ion)
However, this reaction in NCERT is given as
$$\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} \uparrow$$
$$\mathrm{ZnO}_{2}^{2-}$$ is anhydrous form of $$\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$.
$$\mathrm{ZnO}_{2}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$
So in aqueous medium best answer of this question is $\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$.
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