- ✓$N_2$ in both cases
- B$N_2$ with ammonium dichromate and $NO$ with barium azide
- C$N_2O$ with ammonium dichromate and $N_2$ with barium azide
- D$N_2O$ with ammonium dichromate and $NO_2$ with barium azide
$\mathrm{Ba}\left(\mathrm{N}_{3}\right)_{2} \rightarrow 3 \mathrm{N}_{2}+\mathrm{Ba}$
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$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V}$
$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \mathrm{E}^{\circ}=-0.04 \mathrm{~V}$
$\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \mathrm{E}^{\circ}=-0.25 \mathrm{~V}$
$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \mathrm{E}^{\circ}=0.80 \mathrm{~V}$
$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \mathrm{E}^{\circ}=1.40 \mathrm{~V}$
Consider the given electrochemical reactions, The number of metal$(s)$ which will be oxidized be $\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}$, in aqueous solution is. . . . . .
(At. nos. $Mn = 25, Fe = 26, Co = 27, Ni = 28$)

$Cr_2O_7^{2-}$ ( $aq$ $1\,M$ ) $+ 14H^+(aq) + 6e^-$ $\rightarrow 2Cr^{+3} (aq,\,1\,M) + 7H_2O (l)$
$CH_3-CH_2-OH$ $\mathop {\xrightarrow{{(i)\,KMn{O_4}/\mathop O\limits^\Theta H/\Delta }}}\limits_{(ii)\,{H^ \oplus }} (A)\mathop {\xrightarrow{{(i)\,SOC{l_2}}}}\limits_{(ii)\,N{H_3}/\Delta } (B)$ $\xrightarrow{{B{r_2}/KOH}}(C)$
$(C)$ will be :