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p-Block Elements - I
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A metal $(X)$ dissolves both in dil. $HCl$ and dil. $NaOH$ to liberate $H _2$. Addition of $NH _4 Cl$ and excess $NH _4 OH$ to an $HCl$ solution of $X$ produces $Y$ as a precipitate. $Y$ is also produced by adding $NH _4 Cl$ to the $NaOH$ solution of $X$. The species $X$ and $Y$, respectively, are
A$Zn$ and $Zn ( OH )_2$
B$Al$ and $Al ( OH )_3$
C$Zn$ and $Na _2 ZnO _2$
D$Al$ and $NaAlO _2$
(KVPY-2015)
Solution
(b)
$Al (X)$ metal dissolves both in dil. $HCl$ and dil. $NaOH$ to liberate $H _2$. Addition of $NH _4 Cl$ and excess $NH _4 OH$ to an $HCl$ solution of $Al$ produces $Al ( OH )_3(Y)$ as an precipitate. $Al ( OH )_3$ is also produced by adding $NH _4 Cl$ to the $NaOH$ solution of $Al$.
$\underset{(X)}{ Al }+ HCl \longrightarrow AlCl _3+ H _2$
$\underset{(X)}{ Al }+ NaOH \longrightarrow NaAlO _2+ H _2$
$\underset{(X)}{ Al } \xrightarrow{N{{H}_{4}}Cl}Al ( OH )_3 \longleftarrow AlCl _3+ NH _4 Cl + NH _4OH$(excess)
$Al (X)$ metal dissolves both in dil. $HCl$ and dil. $NaOH$ to liberate $H _2$. Addition of $NH _4 Cl$ and excess $NH _4 OH$ to an $HCl$ solution of $Al$ produces $Al ( OH )_3(Y)$ as an precipitate. $Al ( OH )_3$ is also produced by adding $NH _4 Cl$ to the $NaOH$ solution of $Al$.
$\underset{(X)}{ Al }+ HCl \longrightarrow AlCl _3+ H _2$
$\underset{(X)}{ Al }+ NaOH \longrightarrow NaAlO _2+ H _2$
$\underset{(X)}{ Al } \xrightarrow{N{{H}_{4}}Cl}Al ( OH )_3 \longleftarrow AlCl _3+ NH _4 Cl + NH _4OH$(excess)
Standard 11
Chemistry
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