Total number of molecules/species from following which will be paramagnetic is __________.
O₂, O₂⁺, O₂⁻, NO, NO₂, CO, K₂[NiCl₄],[Co(NH₃)₆] Cl₃, K₂[Ni(CN)₄]

Solution

Chemistry ma’am · 4 steps

For every species, ask one question: is there even a single unpaired electron left?

1. The test. A species is paramagnetic if it has at least one unpaired electron. Any odd total electron count guarantees one. For even counts, use MO filling for the small molecules and crystal field splitting for the complexes.
2. Molecules. In O₂ the two π^*2p electrons sit in separate orbitals. O₂⁺ (15), O₂⁻ (17), NO (15) and NO₂ (23) all have odd electron counts. CO has 14 electrons, all paired.
3. Complexes. Cl⁻ is a weak-field ligand, so [NiCl₄]²⁻ is tetrahedral and keeps 2 unpaired electrons. NH₃ with Co³⁺ and CN⁻ with Ni²⁺ are strong-field pairs, so all their d electrons pair up.

   Species	Electron picture	Unpaired electrons	Paramagnetic?
   O₂	16 electrons, π^*2pₓ¹ π^*2p_y¹	2	yes
   O₂⁺, O₂⁻	15 and 17 electrons, odd	1 each	yes, both
   NO, NO₂	15 and 23 electrons, odd	1 each	yes, both
   CO	14 electrons, all bonding levels full	0	no
   K₂[NiCl₄]	Ni²⁺ 3d⁸, weak field, sp³ tetrahedral	2	yes
   [Co(NH₃)₆]Cl₃	Co³⁺ 3d⁶, low spin t_(2g)⁶	0	no
   K₂[Ni(CN)₄]	Ni²⁺ 3d⁸, strong field, dsp² square planar	0	no

4. So. The paramagnetic ones are O₂, O₂⁺, O₂⁻, NO, NO₂ and K₂[NiCl₄], which makes 6.

The common slip: 5. Treating [NiCl₄]²⁻ as square planar like [Ni(CN)₄]²⁻ makes it diamagnetic and gives 5. Cl⁻ is too weak to pair the d electrons.

The idea to keep. An odd electron count always means paramagnetic. For an even count, fill the MOs or split the d orbitals. The ligand decides whether Ni²⁺ is tetrahedral and magnetic or square planar and not.

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