Among the following species
N₂, N₂⁺, N₂⁻, N₂²⁻, O₂, O₂⁺, O₂⁻, O₂²⁻
the number of species showing diamagnesim is _______________.

Solution

Chemistry ma’am · 3 steps

Count electrons, then look only at the π^* orbitals: diamagnetic means every electron is paired.

1. Odd counts first. A species with an odd number of electrons must have at least one unpaired electron, so it is paramagnetic. That removes N₂⁺ (13), N₂⁻ (15), O₂⁺ (15) and O₂⁻ (17) at once.
2. Even counts. An even count is not enough. Past 14 electrons, the extra ones go into the two degenerate π^*2p orbitals, and by Hund's rule 2 electrons there stay unpaired. Only 0 or 4 electrons in π^* leave everything paired.

   Species	Electrons	Electrons in π^*2p	Unpaired
   N₂	14	0	0, diamagnetic
   N₂⁺	13	0	1
   N₂⁻	15	1	1
   N₂²⁻	16	2 (one in each)	2
   O₂	16	2 (one in each)	2
   O₂⁺	15	1	1
   O₂⁻	17	3	1
   O₂²⁻	18	4	0, diamagnetic

3. So. Only N₂ and O₂²⁻ have all electrons paired, so the number of diamagnetic species is 2.

The common slip: 3. Counting N₂²⁻ as diamagnetic because 16 is even gives 3. It is isoelectronic with O₂, so it has two unpaired electrons.

The idea to keep. For diatomics of second-period elements, 14 and 18 electrons are diamagnetic. A 16-electron species is always paramagnetic like O₂, because its two π^* electrons sit unpaired.

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