Which of the following will have maximum stabilization due to crystal field?

Solution

Chemistry ma’am · 3 steps

Big splitting and a full lower set win together, and low-spin Co³⁺ with cyanide has both.

1. What decides CFSE. Crystal field stabilization depends on two things. The first is how many electrons sit in the lower set compared with the upper set. The second is the size of the splitting, which grows with a stronger-field ligand and a higher metal charge.

   CFSE (octahedral) = (-0.4 n_(t_(2g)) + 0.6 n_(e_g))Δₒ

   Complex	Metal ion, ligand	Configuration	CFSE
   [Ti(H₂O)₆]³⁺	Ti³⁺, d¹, H₂O	t_(2g)¹	-0.4Δₒ
   [Co(H₂O)₆]²⁺	Co²⁺, d⁷, weak H₂O (high spin)	t_(2g)⁵ e_g²	-0.8Δₒ, with a small Δₒ
   [Co(CN)₆]³⁻	Co³⁺, d⁶, strong CN⁻ (low spin)	t_(2g)⁶ e_g⁰	-2.4Δₒ, with a very large Δₒ
   [Cu(NH₃)₄]²⁺	Cu²⁺, d⁹, NH₃, square planar	only one vacancy in the d set	small: the upper orbitals are almost full

2. Why cyanide wins. In [Co(CN)₆]³⁻ all six electrons are in t_(2g), which is the largest multiple of Δₒ in the list. The CN⁻ ligand and the +3 charge also give the largest Δₒ of the four, so the splitting easily pays the pairing energy.
3. So. [Co(CN)₆]³⁻, option (c), has the maximum crystal field stabilization.

The idea to keep. For maximum CFSE, look for a strong-field ligand, a high metal charge and a low-spin d⁶ ion. Those six electrons fill t_(2g) completely and nothing goes into e_g.

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