JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
Number of paramagnetic oxides among the following given oxides is ____________.
Li₂O, CaO, Na₂O₂, KO₂, MgO and K₂O
- (a) 1
- (b) 2
- (c) 3
- (d) 0
Answer and solution
Answer (a) 1
Solution
Chemistry ma’am · 3 steps
Name the anion in each oxide: only the superoxide ion carries an unpaired electron.
- Three kinds of anion. The oxide ion O²⁻ has 10 electrons, all paired. The peroxide ion O₂²⁻ has 18 electrons, all paired. The superoxide ion O₂⁻ has 17 electrons, an odd count, so one is left unpaired.
- Cations do not count. Li⁺, Na⁺, K⁺, Mg²⁺ and Ca²⁺ all have noble-gas configurations with no unpaired electrons. Any paramagnetism must therefore come from the anion.
Oxide Anion Unpaired electrons Magnetic nature Li₂O O²⁻ (oxide) 0 diamagnetic CaO O²⁻ (oxide) 0 diamagnetic Na₂O₂ O₂²⁻ (peroxide) 0 diamagnetic KO₂ O₂⁻ (superoxide) 1 paramagnetic MgO O²⁻ (oxide) 0 diamagnetic K₂O O²⁻ (oxide) 0 diamagnetic - So. Only KO₂ is paramagnetic, so the count is 1, option (a).
The idea to keep. For s-block oxides, check the oxygen species: oxide and peroxide are diamagnetic, while superoxide O₂⁻ has one unpaired electron and is paramagnetic.
Answer from the NTA final answer key. Solution by PrepWiser.