Question

Which of the following complexes shows the highest crystal field stabilization energy (CFSE) in octahedral geometry?

• A. \( d^3 \)
• B. \( d^5 \) high spin
• C. \( d^8 \)
• D. \( d^{10} \)

Hint:

Configurations with electrons concentrated in lower-energy \( t_{2g} \) orbitals show higher crystal field stabilization.

Answer 1

The Correct Option is A

Concept:
In octahedral complexes, d-orbitals split into: \[ t_{2g} \text{ and } e_g \] CFSE depends on electron distribution. Each electron in: \[ t_{2g} \rightarrow -0.4\Delta_o \] \[ e_g \rightarrow +0.6\Delta_o \]

Step 1: Calculate CFSE for each configuration.
For \( d^3 \): \[ t_{2g}^3 e_g^0 \] CFSE: \[ 3(-0.4\Delta_o)=-1.2\Delta_o \] For high-spin \( d^5 \): \[ t_{2g}^3 e_g^2 \] CFSE: \[ 3(-0.4)+2(+0.6)=0 \] For \( d^8 \): \[ t_{2g}^6 e_g^2 \] CFSE: \[ 6(-0.4)+2(+0.6) \] \[ =-2.4+1.2=-1.2\Delta_o \] However pairing energy considerations reduce stability comparison. For \( d^{10} \): \[ CFSE=0 \]

Step 2: Determine maximum stabilization.
Among common stable octahedral arrangements: \[ d^3 \] shows exceptionally high stability because no electron occupies antibonding \(e_g\) orbitals. Thus: \[ \boxed{d^3} \]