Oxidation state of oxygen in hydrogen peroxide is :

1. -1

2. +1

3. 0

4. -2

$\mathrm{}$ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{O}_2+6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Z}+5 \mathrm{O}_2+8 \mathrm{H}_2 \mathrm{O}\).In this reaction, Z is :

1. ${\mathrm{Mn}}^{2+}$

2. ${\mathrm{Mn}}^{4+}$

3. ${\mathrm{MnO}}_2$

4. Mn

Which of the following is the most powerful oxidizing agent ?

1. ${\mathrm{F}}_2$

2. ${\mathrm{Cl}}_2$

3. ${\mathrm{Br}}_2$

4. ${\mathrm{I}}_2$

Nitrogen show different oxidation states in the range :

1. 0 to +5

2. -3 to +5

3. -5 to +3

4. -3 to +3

How many moles of potassium dichromate (K₂Cr₂O₇) are required to reduce 1 mole of  Sn ²⁺ in an acidic medium? 

1.  1/3

2. 1/6

3. 2/3

4. 1

${\mathrm{MnO}}_4^{2-}$ (1 mole ) in neutral aqueous medium will disproportionate to :

1. 2/3 Mole of ${\mathrm{MnO}}_4^{-}$ and 1/3 mole of ${\mathrm{MnO}}_2$

2. 1/3 Mole of ${\mathrm{MnO}}_4^{-}$ and 2/3 mole of ${\mathrm{MnO}}_2$

3. 1/3 Mole of ${\mathrm{Mn}}_2{\mathrm{O}}_7$ and 1/3 mole of ${\mathrm{MnO}}_2$

4. 2/3 Mole of ${\mathrm{Mn}}_2{\mathrm{O}}_7$ and 1/3 mole of ${\mathrm{MnO}}_2$

The oxidation number of oxygen in ${\mathrm{KO}}_3$ $\mathrm{and}$ ${\mathrm{Na}}_2{\mathrm{O}}_2$ is respectively -

1. 3, and 2
2. 1, and 0
3. 0, and 1
4. -0.33, and -1

The oxidation number of phosphorus in \(Ba \left(H_{2} PO_{2}\right)_{2}\) is-

1. -1

2. +1

3. +2

4. +3

2 mole of ${\mathrm{N}}_2{\mathrm{H}}_4$ loses 16 mole of electron is being converted to a new compound X. Assuming that all of the N appears in the new compound. What is the oxidation state of 'N' in X ?

1. -1

2. -2

3. +2

4. +4

Given the following reaction: 

\(K_2Cr_2O_7 \ \rightarrow K_2CrO_4\)

The change in the oxidation state of chromium in the above conversion is:
1. 0
2. 6
3. 4
4. 3