The pair$(s)$ of ions where $BOTH$ the ions are precipitated upon passing $H_2S$ gas in the presence of dilute $HCl$,is(are):
$(A) Ba^{2+}, Zn^{2+}$
$(B) Bi^{3+}, Fe^{3+}$
$(C) Cu^{2+}, Pb^{2+}$
$(D) Hg^{2+}, Bi^{3+}$

$MgSO_4$ on reaction with $NH_4OH$ and $Na_2HPO_4$ forms a white crystalline precipitate. What is its formula?

When $H_2S$ gas is passed through an $HCl$ solution containing $CuCl_2, HgCl_2, BiCl_3$,and $CoCl_2$,which of the following will not precipitate?

Which of the following statement$(s)$ is/are correct with reference to $Fe^{2+}$ and $Fe^{3+}$ ions?
$(1)$ $Fe^{3+}$ gives brown colour with potassium ferricyanide
$(2)$ $Fe^{2+}$ gives blue colour with potassium ferricyanide
$(3)$ $Fe^{3+}$ gives red colour with potassium thiocyanate
$(4)$ $Fe^{2+}$ gives brown colour with ammonium thiocyanate

The only cations present in a slightly acidic solution are $Fe^{3+}$,$Zn^{2+}$,and $Cu^{2+}$. The reagent that when added in excess to this solution would identify and separate $Fe^{3+}$ in one step is

$Fe^{2+}$ ion can be distinguished from $Fe^{3+}$ ion by