Total number of unpaired electrons in the complex ions [Co(NH₃)₆]³⁺ and [NiCl₄]²⁻ is ________.

Solution

Chemistry ma’am · 4 steps

Fix each metal's d-count first, then let the ligand and the geometry decide the pairing.

1. Find d-electron counts. In [Co(NH₃)₆]³⁺, NH₃ is neutral, so cobalt is Co³⁺, which is 3d⁶. In [NiCl₄]²⁻, four Cl⁻ give -4, so nickel is Ni²⁺, which is 3d⁸.
2. Cobalt complex. NH₃ is a strong-field ligand for Co³⁺, so the octahedral splitting Δₒ is larger than the pairing energy. All six electrons pair up in t_(2g), giving t_(2g)⁶ e_g⁰ with 0 unpaired electrons, a diamagnetic complex.
3. Nickel complex. Cl⁻ is a weak-field ligand and [NiCl₄]²⁻ is tetrahedral, where Δₜ is always small. The eight electrons fill as e⁴ t₂⁴, leaving 2 unpaired electrons in the t₂ set, a paramagnetic complex.

   Complex	Metal ion	Field and shape	Unpaired
   [Co(NH₃)₆]³⁺	Co³⁺, d⁶	strong field, octahedral: t_(2g)⁶ e_g⁰	0
   [NiCl₄]²⁻	Ni²⁺, d⁸	weak field, tetrahedral: e⁴ t₂⁴	2

4. So. The total number of unpaired electrons is 0 + 2 = 2.

The common slip: 6. Treating [Co(NH₃)₆]³⁺ as high spin gives 4 unpaired on cobalt and a total of 6. With Co³⁺, NH₃ acts as a strong-field ligand and pairs all six electrons.

The idea to keep. Oxidation state gives the d-count, then the ligand and geometry decide pairing. Co³⁺ with NH₃ is low spin, and tetrahedral complexes are always high spin.

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