The figure shows the splitting of the d orbitals of a metal ion in an octahedral crystal field.
In the octahedral complex \( \ce{[Cr(H2O)6]^3+} \), Cr3+ has three d electrons (3d3). By how much is the total energy of these three electrons lowered below the barycentre in the ground state of the complex?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (6/5)Δo
In a d3 ion the three electrons occupy the three lower t2g orbitals singly, in accordance with Hund’s rule; none enters the eg level. The figure shows the t2g level (2/5)Δo below the barycentre, so each of the three electrons is lowered by (2/5)Δo.
Total lowering
\( \displaystyle 3 \times \frac{2}{5}\Delta_o = \frac{6}{5}\Delta_o \)
NCERT (2026-27 edition):
5.5.4 Crystal Field Theory – “…the energy separation is denoted by Δo (the subscript o is for octahedral) (Fig.5.8). Thus, the energy of the two eg orbitals will increase by (3/5) Δo and that of the three t2g will decrease by (2/5)Δo.”
“…Obviously, the single d electron occupies one of the lower energy t2g orbitals. In d2 and d3 coordination entities, the d electrons occupy the t2g orbitals singly in accordance with the Hund’s rule.”
Fig. 5.8: d orbital splitting in an octahedral crystal field