$$A \xrightarrow{4KOH, O_2} 2B_{(Green)} + 2H_2O$$; $$B \xrightarrow{4HCl} 2C_{(Purple)} + MnO_2 + 2H_2O$$; $$2C \xrightarrow{H_2O, KI} 2A + KOH + D$$. In the above sequence of reactions, A and D, respectively, are:

We start from the first change

$$A \;+\;4KOH\;+\;O_2 \;\longrightarrow\;2B\;(green)\;+\;2H_2O$$

Whenever the oxide of manganese, $$\mathrm{MnO_2}$$, is fused with aqueous alkali in the presence of dioxygen it is oxidised to green potassium manganate, $$\mathrm{K_2MnO_4}$$. The well-known laboratory equation is

$$2\;{\color{blue}{MnO_2}} \;+\;4KOH\;+\;O_2 \;\longrightarrow\;2\;{\color{blue}{K_2MnO_4}}\;(green)\;+\;2H_2O$$

Comparing this with the general form given to us, we can put

$$A = MnO_2 \qquad\text{and}\qquad B = K_2MnO_4$$

Now we move to the second change

$$B\;+\;4HCl \;\longrightarrow\;2C\;(purple)+MnO_2+2H_2O$$

Potassium manganate is converted by dilute acid to purple potassium permanganate, and some $$\mathrm{MnO_2}$$ is precipitated at the same time. The textbook equation is

$$2\;{\color{blue}{K_2MnO_4}} \;+\;4HCl \;\longrightarrow\;2\;{\color{blue}{KMnO_4}}\;(purple)\;+\;MnO_2+2H_2O+2KCl$$

Because the spectator salt $$2KCl$$ is not shown in the statement, the skeletal form supplied to us fits perfectly, so

$$C = KMnO_4$$

Finally, the third change is

$$2C \;+\;H_2O\;+\;KI \;\longrightarrow\;2A+KOH+D$$

The reaction of permanganate with iodide ion in neutral medium is a classic redox reaction in which

• $$\mathrm{MnO_4^-}$$ is reduced from $$+7$$ to $$+4$$, giving $$\mathrm{MnO_2}$$.
• $$\mathrm{I^-}$$ is oxidised from $$-1$$ to $$+5$$, giving $$\mathrm{IO_3^-}$$.

First we write the ionic form and then insert potassium ions:

$$2MnO_4^- + I^- + H_2O \;\longrightarrow\;2MnO_2 + 2OH^- + IO_3^-$$

Adding the potassium ions gives

$$2\,\underbrace{KMnO_4}_{C}\;+\;\underbrace{KI}_{I^-}\;+\;H_2O\; \longrightarrow\;2\,\underbrace{MnO_2}_{A}\;+\;2KOH\;+\;\underbrace{KIO_3}_{D}$$

The statement in the problem shows only one $$\mathrm{KOH}$$ instead of two, but the identity of the substances is unequivocal:

$$A = MnO_2 \qquad\text{and}\qquad D = KIO_3$$

Thus, in the list of alternatives, the pair “MnO2 and KIO3” appears in Option B.

Hence, the correct answer is Option B.