$A$ mixture $X$ containing $0.02 \ mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.02 \ mol$ of $[Co(NH_3)_5Br]SO_4$ was prepared in $2 \ L$ of solution.
$1 \ L$ of mixture $X +$ excess $AgNO_3 \to Y$.
$1 \ L$ of mixture $X +$ excess $BaCl_2 \to Z$.
The number of moles of $Y$ and $Z$ are:

Match the following complexes in List-$I$ with their colors in List-$II$:

List-$I$ (Complex)	List-$II$ (Color)
$A. [Ni(en)_3]^{2+}$	$I. \text{Green}$
$B. [Ni(H_2O)_4(en)]^{2+}$	$II. \text{Blue}$
$C. [Ni(H_2O)_6]^{2+}$	$III. \text{Pale blue}$
$D. [Ni(H_2O)_2(en)_2]^{2+}$	$IV. \text{Violet}$

Select the correct code about complex $[Cr(NO_2)(NH_3)_5][ZnCl_4]$:
$(I)$ $IUPAC$ name of compound is pentaamminenitrito$-N$-chromium$(III)$ tetrachlorozincate$(II)$
$(II)$ It shows geometrical isomerism
$(III)$ It shows linkage isomerism
$(IV)$ It shows coordination isomerism

The complexes given below show:

The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

Which will give $Fe^{3+}$ ions in solution?