Amongst the following, the number of oxide(s) which are paramagnetic in nature is
Na₂ O, KO₂, NO₂, N₂ O, ClO₂, NO, SO₂, Cl₂ O

Solution

Chemistry ma’am · 3 steps

Count the valence electrons in each species. An odd total means an unpaired electron, so it is paramagnetic.

1. The test. A species is paramagnetic if it has at least one unpaired electron. For a neutral molecule, an odd total of valence electrons guarantees one. For an ion, count the electrons on the ion itself.
2. Look inside KO2. KO₂ is potassium superoxide, made of K⁺ and O₂⁻. The superoxide ion has 6 + 6 + 1 = 13 valence electrons, which leaves one unpaired electron in a π^* orbital.

   Species	Electrons counted	Odd or even	Magnetic nature
   Na₂O	O²⁻: 10 in total, all paired	even	diamagnetic
   KO₂	O₂⁻: 6 + 6 + 1 = 13	odd	paramagnetic
   NO₂	5 + 12 = 17	odd	paramagnetic
   N₂O	10 + 6 = 16	even	diamagnetic
   ClO₂	7 + 12 = 19	odd	paramagnetic
   NO	5 + 6 = 11	odd	paramagnetic
   SO₂	6 + 12 = 18	even	diamagnetic
   Cl₂O	14 + 6 = 20	even	diamagnetic

3. So. KO₂, NO₂, ClO₂ and NO each have an unpaired electron. That makes 4 paramagnetic species.

The common slip: 3. Treating KO₂ as an ordinary oxide gives 3. It is a superoxide, and O₂⁻ has 13 valence electrons, so one is unpaired.

The idea to keep. For quick magnetic checks, count valence electrons. An odd total means paramagnetic. For ionic oxides, judge the anion: oxide O²⁻ is diamagnetic, but superoxide O₂⁻ is paramagnetic.

Answer from the NTA final answer key. Solution by PrepWiser.