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(A) In the analysis of group $V$ cations,$(NH_4)_2CO_3$ is used as a group reagent in the presence of $NH_4Cl$ and $NH_4OH$.$NH_4Cl$ suppresses the io

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$Mg^{2+}$ ions will also be precipitated

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(A) In the analysis of group $V$ cations,$(NH_4)_2CO_3$ is used as a group reagent in the presence of $NH_4Cl$ and $NH_4OH$.$NH_4Cl$ suppresses the ionization of $(NH_4)_2CO_3$ due to the common ion effect,keeping the concentration of $CO_3^{2-}$ low enough to exceed only the solubility product $(K_{sp})$ of $Ca^{2+}$,$Ba^{2+}$,and $Sr^{2+}$ carbonates.If $Na_2CO_3$ is used instead,it is a strong electrolyte and dissociates completely to provide a high concentration of $CO_3^{2-}$ ions.This high concentration of $CO_3^{2-}$ exceeds the $K_{sp}$ of $MgCO_3$ as well,causing $Mg^{2+}$ ions to precipitate along with the group $V$ cations.

List-$I$ (Ion)	List-$II$ (Group Number in Cation Analysis)
$A. Co^{2+}$	$I. Group-I$
$B. Mg^{2+}$	$II. Group-III$
$C. Pb^{2+}$	$III. Group-IV$
$D. Al^{3+}$	$IV. Group-VI$

Sodium Carbonate cannot be used in place of $(NH_4)_2CO_3$ for the identification of $Ca^{2+}$,$Ba^{2+}$,and $Sr^{2+}$ ions (in group $V$) during mixture analysis because

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Match List-$I$ with List-$II$:

List-$I$ (Ion) List-$II$ (Group Number in Cation Analysis)

$A. Co^{2+}$ $I. Group-I$

$B. Mg^{2+}$ $II. Group-III$

$C. Pb^{2+}$ $III. Group-IV$

$D. Al^{3+}$ $IV. Group-VI$

Choose the correct answer from the options given below:

In the fifth group of qualitative analysis,$(NH_4)_2CO_3$ is added to precipitate out the carbonates. We do not add $Na_2CO_3$ because

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