MCQ
$C{H_2} = CHCl$ reacts with $HCl$ to form
- A$C{H_2}Cl - C{H_2}Cl$
- ✓$C{H_3} - CHC{l_2}$
- C$C{H_2} = CHCl.HCl$
- DNone of these

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| List-$I$ (Compound) | List-$II$ (Colour) |
| $A$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH}_2 \mathrm{O}$ | $I$ Violet |
| $B$ $\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-}$ | $II$ Blood Red |
| $C$ $[\mathrm{Fe}(\mathrm{SCN})]^{2+}$ | $III$ Prussian Blue |
| $D$ $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$ | $IV$ Yellow |
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