MHT CET · Chemistry · s Block Elements
When Zn is treated with excess of \(\mathrm{NaOH}\), the product obtained is
- A \(\mathrm{Zn}(\mathrm{OH})_{2}\)
- B \(\mathrm{ZnOH}\)
- C \(\mathrm{Na}_{2} \mathrm{ZnO}_{2}\)
- D None of the above
Answer & Solution
Correct Answer
(C) \(\mathrm{Na}_{2} \mathrm{ZnO}_{2}\)
Step-by-step Solution
Detailed explanation
When \(\mathrm{Zn}\) is reacted with excess of \(\mathrm{NaOH}\), sodium zincate is obtained.
\(
\mathrm{Zn}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]\) \(+~\mathrm{H}_{2}
\)
soluble
Or
\(
\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}
\)
\(
\mathrm{Zn}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]\) \(+~\mathrm{H}_{2}
\)
soluble
Or
\(
\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2}
\)
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