The correct order of N-compounds in its decreasing order of oxidation states is :

1. ${\mathrm{HNO}}_3,$ $\mathrm{NO},$ ${\mathrm{N}}_2,$ ${\mathrm{NH}}_4\mathrm{Cl}$

2. ${\mathrm{HNO}}_3, \mathrm{NO}, {\mathrm{NH}}_4\mathrm{Cl}, {\mathrm{N}}_2$

3. ${\mathrm{HNO}}_3, {\mathrm{NH}}_4\mathrm{Cl}, \mathrm{NO}, {\mathrm{N}}_2$

4. ${\mathrm{NH}}_4\mathrm{Cl}, {\mathrm{N}}_2, \mathrm{NO}, {\mathrm{HNO}}_3$

For the redox reaction,
 a${\mathrm{MnO}}_4^{-}+{\mathrm{bC}}_2{\mathrm{O}}_4^{2-}+{\mathrm{cH}}^{+}\to {\mathrm{dMn}}^{2+}+{\mathrm{eCO}}_2+{\mathrm{fH}}_2\mathrm{O}$
the correct stoichiometric coefficients of the reactants a, b, and c respectively for the balanced equation are: 

1. 16, 5, 2
2. 2, 5, 16
3. 2, 16, 5
4. 5, 16, 2

(a) H₂O₂ + O₃ → H₂O + 2O₂ 

(b) H₂O₂ + Ag₂O → 2Ag + H₂O + O₂ 

The role of hydrogen peroxide in the above reactions is respectively:

The change in oxidation number of chlorine when Cl₂ gas reacts with hot and concentrated sodium hydroxide solution is:

1. Zero to +1 and Zero to –5

2. Zero to –1 and Zero to +5

3. Zero to –1 and Zero to +3

4. Zero to +1 and Zero to –3

The highest oxidation number of nitrogen among the following compounds is:

Oxidation numbers of P in ${\mathrm{PO}}_4^{3-}$, of S in ${\mathrm{SO}}_4^{2-}$ and that of Cr in ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$are respectively:
1. +5, +6 and +6
2. +3, +6 and +5
3. +5, +3 and +6
4. -3, +6 and +6

 ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.35$ $\mathrm{V}$
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.77$ $\mathrm{V}$
The strongest oxidizing agent in the above equation is:

1. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$

2. ${\mathrm{Fe}}^{2+}$

3. ${\mathrm{Fe}}^{3+}$

4. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

The number of moles of KMnO₄ that will be needed to react with one mole of sulphite ion in an acidic solution is:

Disproportionation reactions are:

(a) $2{\mathrm{Cu}}^{+}$ $\to$ ${\mathrm{Cu}}^{2+}+{\mathrm{Cu}}^0$

(b) $3{\mathrm{MnO}}_4^{2-}+4{\mathrm{H}}^{+}$ $\to$ $2{\mathrm{MnO}}_4^{-}+{\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}$

(c) $2{\mathrm{KMnO}}_4$ $\stackrel{∆}{\to }$ ${\mathrm{K}}_2{\mathrm{MnO}}_4+{\mathrm{MnO}}_2+{\mathrm{O}}_2$

(d) $2{\mathrm{MnO}}_4^{-}+3{\mathrm{Mn}}^{2+}+2{\mathrm{H}}_2\mathrm{O}$ $\to$ $5{\mathrm{MnO}}_2+4{\mathrm{H}}^{\oplus }$

1. (a) and (d) only

2. (a) and (b) only

3. (a), (b) and (c)

4. (a), (c) and (d)

The oxidation state of Cr in ${\mathrm{CrO}}_6$ is -