(i.e $en$ and $C_2O_4$ ), so coordination number of $E$ is $6$
$(2 \times 2 + 1 \times 2 = 6)$
Let the oxidation state of $E$ in complex be $x$, then
$[x + (-2) = 1]$ or $x -2 = 1$
or $x = + 3$, so its oxidation state is $+ 3$ Thus option $(d)$ is correct.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,C{H_3}}\\
{\,\,\,|\,\,}\\
{C{H_3} - C{H_2} - CH - C - N{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,}
\end{array}$ $\xrightarrow[\Delta ]{{B{r_2}/KOH}}\left( A \right)\xrightarrow{\begin{subarray}{l}
(1)\,\,C{H_3}I\,{\text{(excess)}} \\
(2)\,AgOH/\Delta \,
\end{subarray} }$ $(B)$
