MCQ
$'A'$ will be
- A$Pd + H_2$
- B$NaBH_4$
- C$LiAlH_4$
- ✓$Pd/BaSO_4+H_2$
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$\begin{array}{*{20}{c}}
O\\
{||}\\
{C{H_3} - C - C{H_3}}
\end{array}$ $ + \begin{array}{*{20}{c}}
{C{H_2}OH}\\
{\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_2}OH}
\end{array}$ $\overset{HCl}{\longleftrightarrow}$ ?
$(i)\, [CoCl_3 (NH_3)_3]$ $(ii)\, [Co(en)_3]Cl_3$
$(iii)\, [Co(C_2O_4)_2(NH_3)_2]^-$ $(iv)\, [CrCl_2(NH_3)_2(en)]^+$
$1.$ The metal rod ${M}$ is
$(A)$ $\mathrm{Fe}$ $(B)$ $\mathrm{Cu}$ $(C)$ $\mathrm{Ni}$ $(D)$ $\mathrm{Co}$
$2.$ The compound ${N}$ is
$(A)$ $\mathrm{AgNO}_3$ $(B)$ $\mathrm{Zn}\left(\mathrm{NO}_3\right)_2$
$(C)$ $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ $(D)$ $\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$
$3.$ The final solution contains
$(A)$ $\left[\mathrm{Pb}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$
$(B)$ $\left[\mathrm{Al}\left(\mathrm{NH}_3\right)_4\right]^{3+}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
$(C)$ $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
$(D)$ $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$and $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}$
Give the answer question $1,2$ and $3.$