- ✓$I_2$
- B$NaIO_3$
- C$ICl$
- D$IO_4^-$
$HCl$ gets oxidised by $KMnO _{4}$ into $Cl _{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${C_2}{H_5}OH + SOC{l_2}\xrightarrow{{{\text{Pyridine}}}}{C_2}{H_5}Cl + S{O_2} + HCl$
$R - \mathop {\mathop {C\,N{H_2} \to RC{H_2}N{H_2}}\limits^{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} }\limits^{O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} $
(use $R =0.083 L\,bar\,mol ^{-1}\, K ^{-1}$ )
$\mathrm{H}_3 \mathrm{PO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_4 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{CO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 $
$ \mathrm{H}_3 \mathrm{BO}_3 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_2 \ \ \ \ \mathrm{H}_2 \mathrm{CrO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_3 $