MCQ
Crystal field stabilization energy for high spin $d^4$ octahedral complex is
- A$- 1.8 \Delta _o$
- B$- 1.6 \Delta _o + P$
- C$- 1.2 \Delta _o$
- ✓$- 0.6 \Delta _o$
✓
Answer
Correct option: D.
$- 0.6 \Delta _o$
d
Key ldea In case of high spin complex, $\Delta_{0}$ is small. Thus, the energy required to pair up the fourth electron with the
Key ldea In 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 Crystal fleld stabilisation energy
$=(-3 \times 0.4+1 \times 0.6) \Delta_{0}$
$=(-1.2+0.6) \Delta_{0}=-0.6 \Delta_{0}$
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