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.

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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