A (Orange color compound) xrightarrow{(B) + t | vedclass

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(B) The reaction sequence describes the chromyl chloride test.$1$. $A$ is a dichromate salt (e.g.,$K_2Cr_2O_7$),which is orange.$2$. When $A$ reacts w

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$NH_4Cl, NH_3$

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(B) The reaction sequence describes the chromyl chloride test.$1$. $A$ is a dichromate salt (e.g.,$K_2Cr_2O_7$),which is orange.$2$. When $A$ reacts with $B$ (a chloride salt,$NH_4Cl$) and concentrated $H_2SO_4$,it produces $C$ (Chromyl chloride,$CrO_2Cl_2$),which is a red gas.$3$. $CrO_2Cl_2$ reacts with $NaOH$ and $AgNO_3$ to form $Ag_2CrO_4$ (Red precipitate,$D$).$4$. $Ag_2CrO_4$ dissolves in $NH_3$ to form a yellow solution.$5$. $B$ $(NH_4Cl)$ reacts with $NaOH$ upon heating to release $F$ ($NH_3$ gas),which gives white fumes with $HCl$ due to the formation of $NH_4Cl$ smoke.Therefore,$B$ is $NH_4Cl$ and $F$ is $NH_3$.

$LIST$-$I$ (Precipitating reagent and conditions)	$LIST$-$II$ (Cation)
$A$. $NH_4Cl + NH_4OH$	$I$. $Mn^{2+}$
$B$. $NH_4OH + (NH_4)_2CO_3$	$II$. $Pb^{2+}$
$C$. $NH_4OH + NH_4Cl + H_2S$ gas	$III$. $Al^{3+}$
$D$. Dilute $HCl$	$IV$. $Sr^{2+}$

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