Sometimes yellow turbidity appears on passing $H_2S$ gas even in the absence of the second group radicals. This happens because

The solution of $BiCl_{3}$ in dil. $HCl$ when diluted with water,forms a white precipitate which is:

Find the compound $A$ from the following reaction sequences:
$A$ $\xrightarrow{\text{aqua-regia}} B$ $\xrightarrow[(2) CH_3COOH]{(1) KNO_2 \mid NH_4OH} \text{yellow ppt}$

$A$ salt with a few drops of conc. $HCl$ gives an apple green colour in a flame test. The group precipitate of the salt is dissolved in acetic acid and treated with $K_{2}CrO_{4}$ to give a yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. $HNO_{3}$ and ammonium molybdate,it results in a canary yellow precipitate. The cation and anion present in the salt are respectively,

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

Cations	Group reaction
$P \rightarrow Pb^{2+}, Cu^{2+}$	$(i)$ $H_2S$ gas in presence of dilute $HCl$
$Q \rightarrow Al^{3+}, Fe^{3+}$	$(ii)$ $(NH_4)_2CO_3$ in presence of $NH_4OH$
$R \rightarrow Co^{2+}, Ni^{2+}$	$(iii)$ $NH_4OH$ in presence of $NH_4Cl$
$S \rightarrow Ba^{2+}, Ca^{2+}$	$(iv)$ $H_2S$ in presence of $NH_4OH$

Which of the following does not give a precipitate with $CrO_4^{2-}$?