In an octahedral strong ligand field, each t2g electron lowers the energy by 0.4Δo and each eg electron raises it by 0.6Δo (pairing energy ignored). Match List-I (metal ion configuration in the strong ligand field) with List-II (crystal field stabilisation energy).
| List-I | List-II |
| (A) d4 | (p) −1.6Δo |
| (B) d5 | (q) −1.8Δo |
| (C) d6 | (r) −2.0Δo |
| (D) d7 | (s) −2.4Δo |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-p, B-r, C-s, D-q
In a strong ligand field (Δo > P) the electrons pair up in the lower t2g set before any electron enters eg. So the t2g set fills to six electrons first. CFSE = (−0.4 × number of t2g electrons + 0.6 × number of eg electrons)Δo.
(A) d4 — This matches (p) −1.6Δo. Filling: t2g4eg0; CFSE = 4(−0.4)Δo = −1.6Δo.
(B) d5 — This matches (r) −2.0Δo. Filling: t2g5eg0; CFSE = 5(−0.4)Δo = −2.0Δo.
(C) d6 — This matches (s) −2.4Δo. Filling: t2g6eg0; CFSE = 6(−0.4)Δo = −2.4Δo.
(D) d7 — This matches (q) −1.8Δo. Filling: t2g6eg1, since t2g is full; CFSE = 6(−0.4)Δo + 1(+0.6)Δo = −1.8Δo.
NCERT (2026-27 edition):
5.5.4 Crystal Field Theory – “(ii) If Δo > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g4eg0. Ligands which produce this effect are known as strong field ligands and form low spin complexes.”