[Fe(CN)₆]⁴⁻
[Fe(CN)₆]³⁻
[Ti(CN)₆]³⁻
[Ni(CN)₄]²⁻
[Co(CN)₆]³⁻
Among the given complexes, number of paramagnetic complexes is ____________.

Solution

Chemistry ma’am · 3 steps

Cyanide is a strong-field ligand, so pair up the d electrons first, then count what is left unpaired.

1. Find the d count. Each CN⁻ carries a charge of -1, so the metal's oxidation state is the complex charge plus the number of cyanides. From that, read off the number of d electrons.
2. Strong field pairs. CN⁻ gives a large splitting Δₒ, so the electrons fill the lower t_(2g) set completely before any go up to e_g. In square planar [Ni(CN)₄]²⁻, the 8 electrons pair up in the four lower orbitals and the high-energy d_(x²-y²) stays empty.

   Complex	Metal ion	d electrons	Arrangement	Unpaired
   [Fe(CN)₆]⁴⁻	Fe²⁺	d⁶	t_(2g)⁶ e_g⁰	0
   [Fe(CN)₆]³⁻	Fe³⁺	d⁵	t_(2g)⁵ e_g⁰	1
   [Ti(CN)₆]³⁻	Ti³⁺	d¹	t_(2g)¹ e_g⁰	1
   [Ni(CN)₄]²⁻	Ni²⁺	d⁸	square planar, d_(x²-y²) empty	0
   [Co(CN)₆]³⁻	Co³⁺	d⁶	t_(2g)⁶ e_g⁰	0

3. So. Only [Fe(CN)₆]³⁻ and [Ti(CN)₆]³⁻ have an unpaired electron, so the number of paramagnetic complexes is 2.

The common slip: 3. Treating [Ni(CN)₄]²⁻ as tetrahedral leaves 2 unpaired electrons on Ni²⁺ and gives 3. With CN⁻, Ni²⁺ is square planar (dsp²) and diamagnetic.

The idea to keep. With a strong-field ligand like CN⁻, an odd d count always leaves one electron unpaired. d⁶ octahedral and d⁸ square planar end up fully paired and diamagnetic.

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