If Pb^{2+}, Cu^{2+}, Al^{3+}, Cd^{2+}, Zn^{2+ | vedclass

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(B) $1$. When $H^{+}/H_2S$ is added,$Pb^{2+}$,$Cu^{2+}$,and $Cd^{2+}$ precipitate as their respective sulfides ($PbS$,$CuS$,$CdS$) because they belong

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Clear solution in which two radicals are present as a soluble complex

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(B) $1$. When $H^{+}/H_2S$ is added,$Pb^{2+}$,$Cu^{2+}$,and $Cd^{2+}$ precipitate as their respective sulfides ($PbS$,$CuS$,$CdS$) because they belong to Group $II$ of the qualitative analysis scheme.$2$. The remaining ions in solution $(X)$ are $Al^{3+}$ and $Zn^{2+}$.$3$. When excess $NaOH$ is added to solution $(X)$ containing $Al^{3+}$ and $Zn^{2+}$,both form soluble complexes.$4$. $Al^{3+} + 4OH^{-} \rightarrow [Al(OH)_4]^{-}$ (soluble aluminate).$5$. $Zn^{2+} + 4OH^{-} \rightarrow [Zn(OH)_4]^{2-}$ (soluble zincate).$6$. Thus,a clear solution is obtained containing two radicals as soluble complexes.

If $Pb^{2+}, Cu^{2+}, Al^{3+}, Cd^{2+}, Zn^{2+}$ radicals are present in an aqueous solution,and it reacts with $H^{+}/H_2S$,a precipitate is produced. The solution $(X)$ obtained after the removal of the precipitate is treated with excess $NaOH$. What is the result?

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