${\mathrm{MnO}}_4^{2-}$ undergoes disproportionation reaction in acidic medium but ${\mathrm{MnO}}_4^{-}$ does not. It is:

1.	Due to the highest oxidation state of Mn in  MnO42-
2.	Due to the highest oxidation state of Mn in  MnO4-
3.	Due to the endothermic nature of the disproportionation reaction.
4.	Due to the exothermic nature of the disproportionation reaction.

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

The element that does not show a disproportionation tendency is/are:

1. Cl

2. Br

3. F

4. I

Given below are two statements: 

The incorrect statement regarding the rule to find the oxidation number among the following is-

The largest oxidation number exhibited by an element depends on its electronic configuration. With which of the following electronic configurations will the element exhibit the largest oxidation number?

1. 3d¹4s²

2. 3d³4s²

3. 3d⁵4s¹

4. 3d⁵4s²

Given below are two statements: 

What is the alteration in the oxidation state of carbon in the given reaction?

\(\mathrm{{CH_4}_{(g)} + 4{Cl_2}_{(g)} \rightarrow {CCl_4}_{(l)} + 4 HCl_{(g)}}\)

1. 0 to +4

2. –4 to +4

3. 0 to –4

4. +4 to +4

Among the following the correct order of acidity is-

1. HClO < HClO₂< HClO₃< HClO₄
2. HClO₂< HClO < HClO₃< HClO₄
3. HClO₄< HClO₂< HClO < HClO₃
4. HClO₃< HClO₄< HClO₂< HClO

Standard reduction potentials of the half-reactions are given below: 

$F_{2\left(g\right)}$ $+$ $2e^{-}\to 2{F^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+2.85$ $V$ $$

$C{l}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{C{l}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.36$ $V$ $$

$B{r}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{B{r}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.06$ $V$ $$

$I_{2\left(g\right)}$ $+$ $e^{-}\to 2{I^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+0.53$ $V$ $$

The strongest oxidizing and reducing agents, respectively, are: 

1. $B{r}_2$ and $C{l}^{-}$

2. $C{l}_2$ and $B{r}^{-}$

3. $C{l}_2$ and $I_2$

4. $F_2$ and $l^{-}$