- ✓$NF_3$
- B$NCl_3$
- C$NBr_3$
- D$NI_3$
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$\mathop { - CH{{(C{H_3})}_2}}\limits_1 \,\,\,\,\,\,\,\,\,\,\,\mathop { - C{H_2}C{H_2}Br}\limits_2 \,\,\,\,\,\,\,\,\,\,\mathop{ - C{H_2}Br}\limits_3 \,\,\,\,\,\,\,\,\,\mathop{ - C{{(C{H_3})}_3}}\limits_4$
$I = II = III = IV$
Reason : $Zn$ is deposited at anode, and $Cu$ is deposited at cathode.
Compound $\rightarrow$ O.State of $C.M.I.$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,}\\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3} - C{H_2} - C = C - CH - C - C{H_2} - C{H_2} - C{H_3}}\\
{|\,\,\,\,\,\,\,\,\,}\\
{{C_2}{H_5}}
\end{array}$