MCQ
Alcohol which gives red colour with Victor Meyer test is
- ✓${C_2}{H_5}OH$
- B$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}} \\
{|\,\,\,\,\,} \\
{OH}
\end{array}$ - C$C{(C{H_3})_3}OH$
- DNone of these
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${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{ - 4}}\xrightarrow{{MnO_4^ - /{H^ + }}}F{e^{ + 3}} + C{O_2} + NO_3^ - $
Product $(A)$ is
${C_6}{H_5}N{H_2}\mathop {\xrightarrow{{NaN{O_2}/HCl}}}\limits_{0 - 5\,^o C} X\mathop {\xrightarrow{{HN{O_2}}}}\limits_{C{H_2}O} Y + {N_2} + HCl$
$X$ and $Y$ are respectively
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{O}_2^{-} \rightarrow \mathrm{O}_2+\mathrm{O}_2{ }^{2-}$ | $(p)$ redox reaction |
| $(B)$ $\mathrm{CrO}_4{ }^{2-}+\mathrm{H}^{+} \rightarrow$ | $(q)$ one of the products has trigonal planar structure |
| $(C)$ $\mathrm{MnO}_4^{-}+\mathrm{NO}_2^{-}+\mathrm{H}^{+} \rightarrow$ | $(r)$ dimeric bridged tetrahedral metal ion |
| $(D)$ $\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Fe}^{2+} \rightarrow$ | $(s)$ disproportionation |
Statement $II :$ The lone pair of electrons on nitrogen in pyridine makes it basic.
Choose the CORRECT answer from the options given below:
