The correct option$(s)$ regarding the complex $[Co(en)(NH_3)_3(H_2O)]^{3+}$ $(en = H_2NCH_2CH_2NH_2)$ is (are):
$A$. It has two geometrical isomers
$B$. It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands
$C$. It is paramagnetic
$D$. It absorbs light at longer wavelength as compared to $[Co(en)(NH_3)_4]^{3+}$

$A$ mixture $X$ of $0.02 \, mol$ 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 of $AgNO_3 \to Y$
$1 \, L$ of mixture $X +$ excess of $BaCl_2 \to Z$
Number of moles of $Y$ and $Z$ respectively are

The correct option(s) about entropy $(S)$ is(are)
$[R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, if $\frac{dE_{cell}}{dT} = \frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes in entropy and internal energy are temperature independent).
$(B)$ The cell reaction, $Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 0.01 \ M) \parallel H^{+}(aq, 0.1 \ M) \mid H_2(g, 1 \ bar) \mid Pt_{(s)}$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S > 0$.
$(D)$ $\Delta S > 0$, for $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ (where $en =$ ethylenediamine).

Which of the following complexes is diamagnetic and tetrahedral?

Which of the following has a yellow colour?

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