MCQ
Crystal field splitting energy for high spin $d^4$ octahedral complex is
- A$- 1.2 \Delta _o$
- ✓$- 0.6 \Delta _o$
- C$- 0.8 \Delta _o$
- D$- 1.6 \Delta _o$
✓
Answer
Correct option: B.
$- 0.6 \Delta _o$
b
in the case of high spin complex $\Delta_{0}$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy $d-$orbitals would be higher than that required to place the electrons in the higher $d-$orbital. Thus pairing does not occur.
in the case of high spin complex $\Delta_{0}$ is small. Thus, the energy required to pair up the fourth electron with the electrons of lower energy $d-$orbitals would be higher than that required to place the electrons in the higher $d-$orbital. Thus pairing does not occur.
For high spin $d^4$ octahedral complex,
therefore, Crystal field stabilisation energy
$=(-3 \times 0.4+1 \times 0.6) \Delta_{0}$
$=-(-1.2 \times 0.6) \Delta_{0}$
$=-0.6\Delta_{0}$
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