Match List I with List II:

Choose the correct answer from the options given below:

List I (Species of the d- and f-block)             List II (Characteristic feature)

A. $$CeO_2$$ (cerium(IV) oxide)             I. Orange-coloured salt

B. $$KMnO_4$$ (permanganate ion)       II. Not classified as a transition element

C. $$K_2Cr_2O_7$$ (dichromate ion)      III. Powerful oxidising agent in acidic medium, reduced to $$Mn^{2+}$$

D. $$Zn$$ (zinc metal)                  IV. Exhibits the stable +4 oxidation state and is used for glass-polishing (variable $$+3/+4$$ states)

We match each species with its correct characteristic one by one.

$$CeO_2$$ belongs to the 4f-block (lanthanoid). Cerium commonly shows $$+3$$ and $$+4$$ oxidation states; the $$+4$$ form $$CeO_2$$ is used industrially for glass-polishing and as an oxidising agent. Hence A → IV.

In $$KMnO_4$$ manganese is in the $$+7$$ oxidation state. In acidic medium it acts as a very strong oxidising agent and gets reduced to $$Mn^{2+}$$. Hence B → III.

$$K_2Cr_2O_7$$ is orange in colour because of a charge-transfer transition in the dichromate ion $$Cr_2O_7^{2-}$$. Hence C → I.

Zinc has completely filled $$3d^{10}$$ subshell in both atomic ($$3d^{10}4s^{2}$$) and ionic ($$Zn^{2+}: 3d^{10}$$) states, so it does not show variable oxidation states and is therefore not considered a transition element. Hence D → II.

The complete matching is

A - IV, B - III, C - I, D - II.

Therefore the correct option is:
Option A which is: A-IV, B-III, C-I, D-II.