The number of paramagnetic species from the following is _____________.
[Ni(CN)₄]²⁻,[Ni(CO)₄],[NiCl₄]²⁻
[Fe(CN)₆]⁴⁻,[Cu(NH₃)₄]²⁺
[Fe(CN)₆]³⁻ and [Fe(H₂O)₆]²⁺

Solution

Chemistry ma’am · 3 steps

Find each metal's d count, then let the ligand decide: CN and CO pair electrons, Cl and water do not.

1. Method. Work out the oxidation state of the metal and its d-electron count. Then check the ligand: strong-field ligands like CN⁻ and CO force electrons to pair up, while weak-field ligands like Cl⁻ and H₂O leave them spread out. Any unpaired electron makes the species paramagnetic.

   Species	Metal, d count	Ligand field and shape	Unpaired electrons
   [Ni(CN)₄]²⁻	Ni²⁺, d⁸	strong, square planar	0, diamagnetic
   [Ni(CO)₄]	Ni(0), 3d¹⁰ after pairing	strong, tetrahedral	0, diamagnetic
   [NiCl₄]²⁻	Ni²⁺, d⁸	weak, tetrahedral	2, paramagnetic
   [Fe(CN)₆]⁴⁻	Fe²⁺, d⁶	strong, low spin t_(2g)⁶	0, diamagnetic
   [Cu(NH₃)₄]²⁺	Cu²⁺, d⁹	square planar	1, paramagnetic
   [Fe(CN)₆]³⁻	Fe³⁺, d⁵	strong, low spin t_(2g)⁵	1, paramagnetic
   [Fe(H₂O)₆]²⁺	Fe²⁺, d⁶	weak, high spin t_(2g)⁴ e_g²	4, paramagnetic

2. Odd counts. A d⁹ or d⁵ ion always has at least one unpaired electron, whatever the ligand. So [Cu(NH₃)₄]²⁺ and [Fe(CN)₆]³⁻ are paramagnetic even though their ligands are fairly strong.
3. So. The paramagnetic species are [NiCl₄]²⁻, [Cu(NH₃)₄]²⁺, [Fe(CN)₆]³⁻ and [Fe(H₂O)₆]²⁺, which makes 4.

The common slip: 5. Treating [Ni(CN)₄]²⁻ as tetrahedral like [NiCl₄]²⁻ leaves 2 unpaired electrons on it and gives 5. Cyanide is strong field, so this d⁸ complex is square planar and diamagnetic.

The idea to keep. Same metal ion can flip its magnetism with the ligand: d⁸ and d⁶ pair up fully with strong-field ligands. An odd d count can never be fully paired, so it is always paramagnetic.

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