The conditions and consequence that favours the t_(2 g) ³ e_g ¹ configuration in a metal complex are :

Solution

Chemistry ma’am · 3 steps

The fourth electron skipped pairing and went up to e_g: that only happens when Δₒ is small.

1. Read the configuration. t_(2g)³ e_g¹ is a d⁴ ion. The first three electrons fill the three t_(2g) orbitals singly, and the fourth sits alone in an e_g orbital.
2. The fourth electron. It has two choices. It can pair up in t_(2g), which costs the pairing energy P, or it can go up to e_g, which costs the splitting energy Δₒ. It takes whichever path is cheaper.

   Ligand field	Energy comparison	Configuration	Spin
   Weak field	Δₒ < P	t_(2g)³ e_g¹, 4 unpaired	high spin
   Strong field	Δₒ > P	t_(2g)⁴ e_g⁰, 2 unpaired	low spin

3. So. A weak field ligand keeps Δₒ small, so the electron goes to e_g. The complex then has the most unpaired electrons, so it is high spin. The answer is (c).

The idea to keep. For d⁴ to d⁷, compare Δₒ with P. If Δₒ < P (weak field), electrons spread out and the complex is high spin. If Δₒ > P (strong field), they pair up and it is low spin.

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