MCQ
Second ionisation potential of $Li, Be, B$ is in the order
- A$Li > Be > B$
- ✓$Li > B > Be$
- C$Be > Li > B$
- D$B > Be > Li$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,C{H_3}}\\
{\,\,\,|\,\,}\\
{C{H_3} - C{H_2} - CH - C - N{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,}
\end{array}$ $\xrightarrow[\Delta ]{{B{r_2}/KOH}}\left( A \right)\xrightarrow{\begin{subarray}{l}
(1)\,\,C{H_3}I\,{\text{(excess)}} \\
(2)\,AgOH/\Delta \,
\end{subarray} }$ $(B)$
$Fe ^{2+} \rightarrow Fe ^{3+} + e ^{-} \quad E _{ Fe ^{3+} / Fe ^{2+}}=0.77 \,V$
$2 I ^{-} \rightarrow I _{2}+2 e ^{-} \quad E _{ I _{2} / I ^{-}}^{0}=0.54 \,V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \,V 298$ $K$. The value of $x$ is .... (Nearest Integer)
Reason : $NaOH$ is strong alkali.
Consider the above reaction and identify the missing reagent/chemical.