MCQ
Prussian blue is due to the formation of
- ✓$F{e_4}{\left[ {Fe{{(CN)}_6}} \right]_3}$
- B$F{e_2}\left[ {Fe{{(CN)}_6}} \right]$
- C$F{e_3}\left[ {Fe{{(CN)}_6}} \right]$
- D$Fe{\left[ {Fe{{(CN)}_6}} \right]_3}$
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$\begin{array}{|c|c|c|c|}\hline {} &\text { Molisch's Test} & {\text { Barfoed Test}} & {\text { Biuret Test}} \\ \hline \text { A} & { Positive } & {\text { Negative }} & {\text { Negative }} \\ \hline \text { B } & {\text { Positive }} & {\text { Positive }} & {\text { Negative }} \\ \hline \text { C } & {\text { Negative }} & {\text { Negative }} & {\text { Positive }} \\ \hline\end{array}$
$A, B$ and $C$ are respectively

$\Delta U = 2.1\ kcal, \, \Delta S = 20\, cal\, K^{-1}$ at $300\, K$ Hence, $\Delta G$ is ............ $\mathrm{kcal}$
| Column $I$ (Molecule/ion) | Column $II$ (Hybridisation) |
| $(A)$ $H_3O^+$ ion | $(P)$ $sp$ |
| $(B)$ $NH_2^-$ ion | $(Q)$ $sp^2$ |
| $(C)$ $NO_3^-$ ion | $(R)$ $sp^3$ |
| $(D)$ $ClF_3$ | $(S)$ $sp^3d$ |