In the conversion, ${\mathrm{H}}_2{\mathrm{SO}}_4<mspace\; width="1em"></mspace>\to {\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_8$ which process occurs?

1.  Oxidation

2.  Reduction

3.  Oxidation as well as a reduction

4.  Neither oxidation nor reduction 

1 mole of ${\mathrm{N}}_2{\mathrm{H}}_4$ loses 10 moles of electrons to form a new compound y. Assuming that all nitrogen appears in the new compound, what is the oxidation state of nitrogen in y?

(There is no change in the oxidation state of hydrogen)

1. +3

2. -3

3. -1

4. +5

The number of peroxide linkages in $K_{3}Cr{O}_8$ is/are 

1. 1

2. 2

3. 3

4. 4

Which acts as a reducing agent only?

1.  $H_{2}S$

2.  $S{O}_2$

3.  $HN{O}_3$

4.  $KMn{O}_4$

In which reaction: $H_2{O}_2<mspace\; width="1em"></mspace>acts<mspace\; width="1em"></mspace>as<mspace\; width="1em"></mspace>reducing<mspace\; width="1em"></mspace>agent?$

1. $A{g}_{2}O+H_2{O}_2\to 2Ag+H_{2}O+O_2$

2. $2KI+<mspace\; width="1em"></mspace>H_2{O}_2\to 2KOH<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>I_2$

3. $Pbs+4<mspace\; width="1em"></mspace>H_2{O}_2\to Pbs{O}_4+4H_{2}O$

4.$H_2{O}_2+S{O}_2\to H_{2}S{O}_4$

In the equation, 

$N{O}_2^{-}+H_{2}O\to {\mathrm{NO}}_3^{-}+2H^{+}<mspace\; width="1em"></mspace>+n{e}^{-}$, n stands for: 

1. 1

2. 2

3. 3

4. 4

The reaction , 

$3Cl{O}^{-}\left(aq\right)\to Cl{O}_3^{-}\left(aq\right)<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2C{l}^{-}\left(aq\right)<mspace\; width="1em"></mspace>is<mspace\; width="1em"></mspace>an<mspace\; width="1em"></mspace>example<mspace\; width="1em"></mspace>of:<mspace\; width="1em"></mspace>$

1. oxidation reaction 

2. reduction reaction 

3. disproportionation reaction 

4. decomposition reaction 

The oxidation number of Pt in 

${\left[Pt\left(C_2{H}_4\right)C{l}_3\right]}^{-}$is :

1. +1

2. +2 

3. +3

4. +4

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:

The oxidation state of Cu in  ${\mathrm{YBa}}_2{\mathrm{Cu}}_3{\mathrm{O}}_7$  is -

(Yttrium = +3)