According to Crystal Field Theory (CFT), what is the electronic configuration of a d4 transition metal ion in an octahedral crystal field when the crystal field splitting energy is greater than the pairing energy (Δo > P)?

Answer and explanation

Correct answer: D. t2g^4 eg^0

  1. Step 1: In an octahedral field, the five degenerate d-orbitals split into a lower-energy triply degenerate set () and a higher-energy doubly degenerate set ().
  2. Step 2: The first three electrons enter the orbitals singly with parallel spins () according to Hund's rule.
  3. Step 3: For the fourth electron, when (strong field ligand), the energetic cost to pair in a orbital is less than the energy required to promote the electron to the higher level.
  4. Step 4: Therefore, the fourth electron pairs up in the level, yielding a low-spin configuration of .
Answer
t2g^4 eg^0 (low-spin complex)
Why not the others: 't2g^3 eg^1 (high-spin complex)' occurs when (weak field ligands); 't2g^2 eg^2' violates Hund's rule and crystal field stabilization; 't2g^0 eg^4' reverses the relative orbital energies of an octahedral field.

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