Match List I with List II
| List I | List II |
| A. \(KMnO_4\) | I. Orange |
| B. \(K_2Cr_2O_7\) | II. Green |
| C. \(K_2CrO_4\) | III. Yellow |
| D. \(K_2MnO_4\) | IV. Purple |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-I, C-III, D-II
A. \( \ce{KMnO4} \) — This matches IV. Purple. Dark purple crystals of \( \ce{MnO4-} \) (Mn +7).
B. \( \ce{K2Cr2O7} \) — This matches I. Orange. Yellow chromate turns into orange dichromate on acidification:
\( \displaystyle \ce{2CrO4^2- + 2H+ -> Cr2O7^2- + H2O} \)
C. \( \ce{K2CrO4} \) — This matches III. Yellow. The chromate ion \( \ce{CrO4^2-} \) is yellow.
D. \( \ce{K2MnO4} \) — This matches II. Green. Dark green manganate (Mn +6), formed when \( \ce{MnO2} \) is fused with KOH in air:
\( \displaystyle \ce{2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O} \)
The colours of \( \ce{MnO4-} \), \( \ce{CrO4^2-} \) and \( \ce{Cr2O7^2-} \) (all d0) come from charge transfer from oxygen to the metal.
NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7.2H2O can be crystallised. …
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. …
Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of KClO4.”