The volume (in $mL$) of $0.125\; M\; AgNO_{3}$ required to quantitatively precipitate chloride ions in $0.3\; g$ of $[Co(NH_{3})_{6}]Cl_{3}$ is (Given: $M_{[Co(NH_{3})_{6}]Cl_{3}} = 267.46\; g/mol$,$M_{AgNO_{3}} = 169.87\; g/mol$).

Consider the following reaction and identify the correct statement:
$[M(gly)_2(NH_3)_2]^+ \xrightarrow[-2NH_3]{+gly} [M(gly)_3]$

Which of the following is a low spin complex?

Write down the $IUPAC$ name for each of the following complexes and indicate the oxidation state,electronic configuration,and coordination number. Also,give the stereochemistry and magnetic moment of the complex:
$(i)$ $K[Cr(H_2O)_2(C_2O_4)_2] \cdot 3H_2O$
$(ii)$ $[Co(NH_3)_5Cl]Cl_2$
$(iii)$ $CrCl_3(py)_3$
$(iv)$ $Cs[FeCl_4]$
$(v)$ $K_4[Mn(CN)_6]$

Match the complexes in Column-$I$ with their properties listed in Column-$II$.

Column-$I$	Column-$II$
$A$. $[Co(NH_3)_4(H_2O)_2]Cl_2$	$p$. geometrical isomers
$B$. $[Pt(NH_3)_2Cl_2]$	$q$. paramagnetic
$C$. $[Co(H_2O)_5Cl]Cl$	$r$. diamagnetic
$D$. $[Ni(H_2O)_6]Cl_2$	$s$. metal ion with $+2$ oxidation state

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:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$