- A$La (Z\, = 57)$
- B$Eu (Z\, = 63)$
- ✓$Ce (Z = 58)$
- D$Gd (Z= 64)$
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$1-$ Bromopropane is reacted with reagents in List $I$ to give product in List $II$
| List$I-$ Reagent | List$II-$Product |
| $A$ $KOH (\text { alc) }$ | $I$ Nitrile |
| $B$ $KCN \text { (alc) }$ | $II$ Ester |
| $C$ $AgNO _2$ | $III$ Alkene |
| $D$ $H _3 CCOOAg$ | $IV$ Nitroalkane |
|
List-$I$ Reaction |
List-$II$ Type of redox reaction |
| $(A)$ $\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}$ | $(I)$ Decomposition |
| $(B)$ $\begin{aligned} & 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{PbO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\end{aligned}$ | $(II)$ Displacement |
| $(C)$ $\begin{aligned} 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{NaOH}_{(\mathrm{aq} .)}+\mathrm{H}_{2(\mathrm{~g})}\end{aligned}$ | $(III)$ Disproportionation |
| $(D)$ $\begin{aligned} 2 \mathrm{NO}_{2(\mathrm{~g})}+2-\mathrm{OH}_{(\mathrm{aq})} \rightarrow \mathrm{NO}_{2(\mathrm{aq} .)}^{-}+\mathrm{NO}_{3(\mathrm{qq} .)}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\end{aligned}$ | $(IV)$ Combination |
Choose the correct answer from the options given below:
Statement $(I)$ : Fusion of $\mathrm{MnO}_2$ with $\mathrm{KOH}$ and an oxidising agent gives dark green $\mathrm{K}_2 \mathrm{MnO}_4$.
Statement $(II)$ : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.
In the light of the above statements, choose the correct answer from the options given below.

Statement $(I)$ : The $4 \mathrm{f}$ and $5 \mathrm{f}$ - series of elements are placed separately in the Periodic table to preserve the principle of classification.
Statement $(II)$ :$S$-block elements can be found in pure form in nature. In the light of the above statements, choose the most appropriate answer from the options given below :