Which of the following basic radicals will not be precipitated by $H_2S$ gas in the presence of $NH_3$?

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

$A$ salt on treatment with dil. $HCl$ gives a pungent smelling gas and a yellow precipitate. The salt gives a green flame when tested. The salt solution gives a yellow precipitate with potassium chromate. The salt is:

$A$ mixture of two salts is used to prepare a solution $S$,which gives the following results:
White precipitate$(s)$ $\xleftarrow{\text{Dilute } NaOH(aq.) \text{ at Room temperature}}$ $S$ (aq. solution of the salts) $\xrightarrow{\text{Dilute } HCl(aq.) \text{ at Room temperature}}$ White precipitate$(s)$ only
The correct option$(s)$ for the salt mixture is(are):
$A$. $Pb(NO_3)_2$ and $Zn(NO_3)_2$
$B$. $Pb(NO_3)_2$ and $Bi(NO_3)_3$
$C$. $AgNO_3$ and $Bi(NO_3)_3$
$D$. $Pb(NO_3)_2$ and $Hg(NO_3)_2$

$A$ mixture of chlorides of $Cu, Cd, Cr$ and $Fe$ was dissolved in water acidified with dil. $HCl$ and $H_2S$ gas was passed. It was filtered,and boiled later adding two drops of $HNO_3$. To this solution $NH_4Cl$ and $NH_4OH$ (or $NaOH$ in excess) were added. The filtrate shall give a test for :-

$A$ compound is soluble in water. If ammonia is added,a red precipitate appears which is soluble in dilute $HCl$. The compound contains: