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  4. A metal (X) dissolves both in dil. HCl and dil. NaOH to liberate H2. Addition of NH4Cl and excess NH4OH to an HCl solution of X produces Y as a precipitate. Y is also produced by adding NH4Cl to the NaOH solution of X. The species X and Y, respectively, are

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Q. 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

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Solution:

$Al(X)$ 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 $Al$
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