For the redox reaction, 
${\mathrm{MnO}}_4^{-}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{2+}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$, the correct coefficients of the reactants  ${\mathrm{MnO}}_4^{-},{\mathrm{C}}_2{\mathrm{O}}_4^{-},{\mathrm{H}}^{+}$ for the balanced reaction are, respectively:

1. 2, 5, 16

2. 16, 3, 12

3. 15, 16, 12

4. 2, 16, 5

When KMnO₄ acts as an oxidizing agent and ultimately forms \(\left[\mathrm{MnO}_4\right]^{-2}, \mathrm{MnO}_2, \mathrm{Mn}_2 \mathrm{O}_3, \mathrm{Mn}^{+2}\)${}^{}$
then the number of electrons transferred in each case, respectively, are:

In a balanced equation
$H_{2}S{O}_4+xHI\to H_{2}S+y{I}_2+z{H}_{2}O$, the values of x, y, z are:

In the following equation, the products A and B are, respectively are:
\(\small Cr_2O^{2-}_7 (aq) + 3SO^{2-}_3 (aq) + 8H^+(aq) \rightarrow A + B + 4H_2O(l)\)

1. \(2 \mathrm{Cr}^{2+} ; 3 \mathrm{SO}_2 \)
2. \(2 \mathrm{Cr}^{+} ; \mathrm{S}_2 \mathrm{O}_7{ }^{2-} \)
3. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{SO}_4{ }^{2-} \)
4. \(2 \mathrm{Cr}^{3+} ; 3 \mathrm{HSO}_4{ }^{-}\)