Why does the MnO₄²⁻ ion undergo disproportionation in an acidic medium, whereas the MnO₄⁻ ion does not?

1.	Due to manganese being in its highest oxidation state in MnO₄²⁻.
2.	Due to manganese being in its highest oxidation state in MnO₄⁻
3.	Because the disproportionation reaction of MnO₄²⁻ is endothermic.
4.	Because the disproportionation reaction of MnO₄²⁻ is exothermic.

Nitric acid reacts with PbO but does not react with PbO₂ , because -

1. PbO is a base while PbO₂ is a strong oxidizing reagent

2. PbO is a base while PbO₂ is a weak oxidizing reagent

3. PbO is neutral while PbO₂ is a strong oxidizing reagent

4. PbO is acid while PbO₂ is a strong oxidizing reagent

The oxidation state of P in ${\mathrm{HPO}}_3^{2-}$ is-

the oxidation state of P in ${\mathrm{PO}}_4^{3-}$?

1. +2

2. +3

3. +4

4. +5

The oxidation state of two S-atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is:

1. +2 and +4

2. +3 and -2

3. +4 and -2

4. +6 and -2

Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements is: 

In the given reaction, what is the name of the species that bleaches the substances due to its oxidising action?

Cl₂(g) + 2OH^-(aq) → ClO^-(aq) + Cl^-(aq) + H₂O(l)

1. ClO^- 

2. Cl₂

3. Cl^-

4. Both ClO^- and Cl^-

KMnO₄ (mol. wt=158) oxidizes oxalic acid in acidic medium to CO₂ and water as follows.

$5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{MnO}}_4^{-}+16{\mathrm{H}}^{+}\to 10{\mathrm{CO}}_2+2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}$

What is the equivalent weight of KMnO₄?

1. 158

2. 31.6

3. 39.5

4. 79

An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

1. 4 mL

2. 20 mL

3. 40 mL

4. 60 mL

The equivalent weight of MnSO₄ is half its molecular weight when it is converted to

1. MnO

2. ${\mathrm{MnO}}_4^{2-}$

3. ${\mathrm{MnO}}_2$

4. ${\mathrm{MnO}}_4^{-}$