Question

The CFSE for the \( d^8 \) octahedral ion is:

• A. -0.4 \( \Delta_o \)
• B. -0.6 \( \Delta_o \)
• C. 1.2 \( \Delta_o \)
• D. -1.2 \( \Delta_o \)

Hint:

For \( d^8 \) configuration in octahedral fields, the CFSE is calculated based on the number of electrons in lower and higher energy orbitals, giving a value of -0.4 \( \Delta_o \).

Answer 1

The Correct Option is A

Step 1: Understanding CFSE for Octahedral Complexes.
The Crystal Field Stabilization Energy (CFSE) for an octahedral complex is calculated using the formula: \[ \text{CFSE} = -0.4 \Delta_o \times \text{(number of electrons in lower energy orbitals)} + 0.6 \Delta_o \times \text{(number of electrons in higher energy orbitals)} \] where \( \Delta_o \) is the octahedral crystal field splitting energy.

Step 2: Applying the Formula to \( d^8 \) Configuration.
For a \( d^8 \) configuration in an octahedral field, the electron arrangement will have 2 electrons in the higher-energy \( e_g \)-orbitals and 6 electrons in the lower-energy \( t_{2g} \)-orbitals. The CFSE for such a configuration is: \[ \text{CFSE} = -0.4 \Delta_o \times 6 + 0.6 \Delta_o \times 2 = -2.4 \Delta_o + 1.2 \Delta_o = -0.4 \Delta_o \]

Step 3: Conclusion.
Thus, the CFSE for the \( d^8 \) octahedral ion is -0.4 \( \Delta_o \), corresponding to option (1).