The d-orbital electronic configuration of the complex among [Co(en)₃]³⁺,[CoF₆]³⁻, [Mn(H₂ O)₆]²⁺ and [Zn(H₂ O)₆]²⁺ that has the highest CFSE is :

Solution

Chemistry ma’am · 3 steps

Find each metal's d count, then ask whether the ligand is strong enough to force pairing into t_(2g).

1. CFSE formula. In an octahedral field each t_(2g) electron is stabilised by 0.4Δₒ and each e_g electron is destabilised by 0.6Δₒ. The more electrons sit in t_(2g) and the fewer in e_g, the larger the CFSE.

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

   Complex	Metal ion	Ligand	Configuration	CFSE
   [Co(en)₃]³⁺	Co³⁺, d⁶	en, strong field, so low spin	t_(2g)⁶ e_g⁰	-2.4Δₒ
   [CoF₆]³⁻	Co³⁺, d⁶	F⁻, weak field, so high spin	t_(2g)⁴ e_g²	-0.4Δₒ
   [Mn(H₂O)₆]²⁺	Mn²⁺, d⁵	H₂O, weak field, so high spin	t_(2g)³ e_g²	0
   [Zn(H₂O)₆]²⁺	Zn²⁺, d¹⁰	H₂O	t_(2g)⁶ e_g⁴	0

2. Bigger splitting too. en is a strong-field chelating ligand, so Δₒ for [Co(en)₃]³⁺ is also larger than for the fluoride or aqua complexes. That makes its lead even bigger.
3. So. The highest CFSE belongs to [Co(en)₃]³⁺, with the configuration t_(2g)⁶ e_g⁰, option (a).

The idea to keep. A strong-field ligand on a d⁶ ion fills t_(2g) completely, giving the largest possible CFSE of -2.4Δₒ. A half-filled high-spin d⁵ ion and a full d¹⁰ ion both have zero CFSE.

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