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(D) The reaction $X_2S_n + H_2O 	o X(OH)_n + H_2S$ indicates that the sulfide $X_2S_n$ undergoes hydrolysis to produce $H_2S$ gas $(Y)$.$H_2S$ gas re

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$Mg^{2+}$

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(D) The reaction $X_2S_n + H_2O 	o X(OH)_n + H_2S$ indicates that the sulfide $X_2S_n$ undergoes hydrolysis to produce $H_2S$ gas $(Y)$.$H_2S$ gas reacts with lead acetate $Pb(CH_3COO)_2$ to form a black precipitate of lead sulfide ($PbS$,$Z$).This hydrolysis reaction occurs for cations whose hydroxides are insoluble in water,such as $Fe^{3+}$,$Al^{3+}$,and $Cr^{3+}$.$Mg^{2+}$ does not form a stable sulfide $Mg_2S_3$ or $MgS$ that undergoes hydrolysis to produce $H_2S$ in this manner,as $MgS$ is typically prepared in dry conditions and $Mg(OH)_2$ is a weak base.Therefore,$Mg^{2+}$ is the cation that does not follow this specific reaction pathway.

$X_2S_n + H_2O \to X(OH)_n \downarrow + Y \uparrow (Gas) \xrightarrow{Pb(CH_3COO)_2} Z \downarrow (Black \ ppt.)$
Then $(X)$ cation can not be

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$X_2S_n + H_2O \to X(OH)_n \downarrow + Y \uparrow (Gas) \xrightarrow{Pb(CH_3COO)_2} Z \downarrow (Black \ ppt.)$Then $(X)$ cation can not be