JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
Match items of List-I with those of List-II :
Choose the most appropriate answer from the options given below :
| List - I (Property) | List - II (Example) | ||
|---|---|---|---|
| (a) | Diamagnetism | (i) | MnO |
| (b) | Ferrimagnetism | (ii) | O₂ |
| (c) | Paramagnetism | (iii) | NaCl |
| (d) | Antiferromagnetism | (iv) | Fe₃O₄ |
- (a) (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
- (b) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
- (c) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
- (d) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
Answer and solution
Answer (c) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
Solution
Chemistry ma’am · 3 steps
First ask if there are unpaired electrons, then ask how those little magnets line up with each other.
- Any unpaired electrons. With no unpaired electrons at all, the substance is diamagnetic. If unpaired electrons are present, the next question is how their moments arrange.
- How they line up. Moments pointing randomly give paramagnetism. Equal moments locked antiparallel cancel to zero, which is antiferromagnetism. Unequal moments locked antiparallel leave a net moment, which is ferrimagnetism.
Substance Unpaired electrons Arrangement of moments Property NaCl none: Na⁺ and Cl⁻ have noble gas configurations nothing to align diamagnetism O₂ 2, in the π^*2p orbitals random, no ordering paramagnetism MnO 5 on each Mn²⁺ (d⁵) equal moments, antiparallel, cancel antiferromagnetism Fe₃O₄ 4 on Fe²⁺, 5 on Fe³⁺ unequal moments, antiparallel, net moment left ferrimagnetism - So. The match is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i), which is option (c).
The idea to keep. Antiferro and ferri both have antiparallel moments. Equal moments cancel (antiferro, like MnO); unequal moments leave a net moment (ferri, like Fe₃O₄).
Answer from the NTA final answer key. Solution by PrepWiser.