The compound that contains zero oxidation state of Fe is:

$$
a Zn + b NO₃^- + cH⁺ → d NH₄⁺ + e H₂O + f Zn⁺² 

In the reaction given above, the correct values of a, b, c, d, e, and f respectively are: 

If the oxidation numbers of A, B, and C are + 2, +5, and –2 respectively, then the possible formula of the compound is:

Standard electrode potentials are:

$F{e}^{+2}/Fe$ $E°$ $=$ $-0.\mathrm{44 }$

$\mathrm{ }F{e}^{+3}/F{e}^{+2} E°$ $=0.77$

If $F{e}^{+2},$ $F{e}^{+3}$, and Fe  block are kept together, then:

1. $F{e}^{+3}$ increases 

2. $F{e}^{+3}$ decreases 

3. $\frac{F{e}^{+2}}{F{e}^{+3}}$ remains unchanged 

4. $F{e}^{+2}$ decreases

A non-feasible reaction among the following is:

1. $2$ $\mathrm{KI}+{\mathrm{Br}}_2\to 2\mathrm{KBr}+{\mathrm{I}}_2$

2. $2$ $\mathrm{KBr}+{\mathrm{I}}_2\to 2\mathrm{KI}+{\mathrm{Br}}_2$

3. $2$ $\mathrm{KBr}+{\mathrm{Cl}}_2\to 2\mathrm{KCl}+{\mathrm{Br}}_2$

4. $2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{F}}_2\to 4\mathrm{HF}+{\mathrm{O}}_2$

The oxidation states(O.S.) of sulphur in the anions ${{\mathrm{SO}}_3}^{2-},$ ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}$ $\mathrm{and}$ ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$ follow the order:

1. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

2. ${{\mathrm{SO}}_3}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}$

3. ${\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{{\mathrm{SO}}_3}^{2-}$

4. ${\mathrm{S}}_2{{\mathrm{O}}_6}^{2-}<{\mathrm{S}}_2{{\mathrm{O}}_4}^{2-}<{{\mathrm{SO}}_3}^{2-}$

Best description of the behavior of bromine in the reaction given below is:

${\mathrm{H}}_2\mathrm{O}+{\mathrm{Br}}_2\to \mathrm{HOBr}+\mathrm{HBr}$

1. Both oxidized and reduced

2. Oxidized only

3. Reduced only

4. Proton acceptor only