In which one of the following sets all species show disproportionation reaction?

First we recall the definition of a disproportionation reaction. In such a redox process the same element present in a single oxidation state is simultaneously oxidised to a higher oxidation state and reduced to a lower oxidation state. Therefore, a necessary condition for disproportionation is that the element must be in an oxidation state which is intermediate between at least one higher and one lower stable oxidation state. Mathematically, if the element is presently in oxidation state $$x$$ while higher and lower attainable states are $$x_{\text{high}}$$ and $$x_{\text{low}},$$ then

$$x_{\text{low}} < x < x_{\text{high}}$$

must hold. If the element is already in either its maximum or minimum oxidation state, it cannot undergo disproportionation.

Now we examine every species in each option and calculate the oxidation state of the central atom to judge whether the species can disproportionate under suitable conditions.

Option A contains $$\mathrm{ClO_4^-},\; \mathrm{MnO_4^-},\; \mathrm{ClO_2^-}$$ and $$\mathrm{F_2}.$$

For $$\mathrm{ClO_4^-}$$ we have

$$\text{Let the oxidation state of Cl be } x.$$

Since the overall charge is $$-1$$ and each oxygen is $$-2,$$

$$x + 4(-2) = -1 \;\;\Longrightarrow\;\; x = +7.$$

$$+7$$ is the maximum attainable oxidation state of chlorine; no higher state exists. Hence $$\mathrm{ClO_4^-}$$ cannot disproportionate. Therefore the whole set in Option A fails the test.

Option B contains $$\mathrm{MnO_4^{2-}},\; \mathrm{ClO_2},\; \mathrm{Cl_2}$$ and $$\mathrm{Mn^{3+}}.$$ We examine them one by one.

1. For $$\mathrm{MnO_4^{2-}}$$ (manganate ion):

$$x + 4(-2) = -2 \;\;\Longrightarrow\;\; x = +6.$$

Manganese exhibits both higher $$+7$$ (in $$\mathrm{MnO_4^-}$$) and lower $$+4,\,+3,\,+2$$ states. The well-known reaction

$$3\,\mathrm{MnO_4^{2-}} + 2\,\mathrm{H_2O} \;\longrightarrow\; 2\,\mathrm{MnO_4^-} + \mathrm{MnO_2} + 4\,\mathrm{OH^-}$$

shows it disproportionating to $$+7$$ and $$+4$$ states. So $$\mathrm{MnO_4^{2-}}$$ qualifies.

2. For $$\mathrm{ClO_2}$$ (chlorine dioxide):

$$x + 2(-2) = 0 \;\;\Longrightarrow\;\; x = +4.$$

Chlorine has higher $$+5$$ (in $$\mathrm{ClO_3^-}$$) and lower $$+3$$ (in $$\mathrm{ClO_2^-}$$) states. Indeed, in alkaline medium

$$2\,\mathrm{ClO_2} + \mathrm{OH^-} \;\longrightarrow\; \mathrm{ClO_3^-} + \mathrm{ClO_2^-} + \mathrm{H^+}$$

illustrates its disproportionation. Hence $$\mathrm{ClO_2}$$ also fits.

3. For $$\mathrm{Cl_2}$$ (molecular chlorine):

The oxidation state of each chlorine atom is $$0.$$ It can increase to $$+1$$ (in $$\mathrm{ClO^-}$$) and decrease to $$-1$$ (in $$\mathrm{Cl^-}$$), so in basic solution

$$\mathrm{Cl_2} + 2\,\mathrm{OH^-} \;\longrightarrow\; \mathrm{ClO^-} + \mathrm{Cl^-} + \mathrm{H_2O}$$

is a textbook disproportionation. Thus $$\mathrm{Cl_2}$$ meets the criterion.

4. For $$\mathrm{Mn^{3+}}$$:

The oxidation state is $$+3.$$ Manganese can rise to $$+4$$ (in $$\mathrm{MnO_2}$$) and fall to $$+2$$ (in $$\mathrm{Mn^{2+}}$$). The reaction

$$2\,\mathrm{Mn^{3+}} \;\longrightarrow\; \mathrm{Mn^{2+}} + \mathrm{Mn^{4+}}$$

proves its disproportionation capability. So $$\mathrm{Mn^{3+}}$$ also satisfies the condition.

Therefore, every species in Option B can undergo a disproportionation reaction.

Option C includes $$\mathrm{Cr_2O_7^{2-}},\; \mathrm{MnO_4^-},\; \mathrm{ClO_2^-}$$ and $$\mathrm{Cl_2}.$$ For $$\mathrm{Cr_2O_7^{2-}}$$ each chromium is $$+6,$$ which is the maximum common state for chromium; chromium cannot be oxidised further, hence $$\mathrm{Cr_2O_7^{2-}}$$ cannot disproportionate. Thus Option C fails.

Option D again contains $$\mathrm{Cr_2O_7^{2-}},$$ already shown incapable of disproportionation, so this option also fails.

Only Option B features a set in which all the listed species are able to undergo disproportionation.

Hence, the correct answer is Option B.