Match List-I with List-II.
| List-I (Oxoanion) | List-II (Property) |
|---|---|
| (A) \( \ce{CrO4^2-} \) | (I) Orange coloured |
| (B) \( \ce{Cr2O7^2-} \) | (II) Yellow coloured |
| (C) \( \ce{MnO4^2-} \) | (III) Paramagnetic, with one unpaired electron |
| (D) \( \ce{MnO4-} \) | (IV) Diamagnetic, with no unpaired electron |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-I, C-III, D-IV
(A) CrO42− — This matches (II) Yellow coloured. Sodium chromate obtained from chromite ore gives a yellow solution.
(B) Cr2O72− — This matches (I) Orange coloured. Acidifying the yellow chromate solution gives orange dichromate; chromium is +6 in both ions:
\( \displaystyle \ce{2CrO4^2- + 2H+ -> Cr2O7^2- + H2O} \)
(C) MnO42− — This matches (III) Paramagnetic, with one unpaired electron. Mn is +6 here: [Ar] 3d5 4s2 minus six electrons leaves 3d1, one unpaired electron.
(D) MnO4− — This matches (IV) Diamagnetic, with no unpaired electron. Mn is +7 here, which leaves 3d0, so no electron is unpaired.
NCERT (2026-27 edition):
4.4.1 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.”
“The manganate and permanganate ions are tetrahedral; the π- bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese. The green manganate is paramagnetic because of one unpaired electron but the permanganate is diamagnetic due to the absence of unpaired electron.”