JEE MAIN - Chemistry (2020 - 5th September Morning Slot - No. 12)
The values of the crystal field stabilization
energies for a high spin d6 metal ion in
octahedral and tetrahedral fields, respectively,
are :
–0.4$$\Delta $$0
and –0.27$$\Delta $$t
–1.6$$\Delta $$0
and –0.4$$\Delta $$t
–0.4$$\Delta $$0
and –0.6$$\Delta $$t
–2.4$$\Delta $$0
and –0.27$$\Delta $$t
Explanation
In octahedral
_5th_September_Morning_Slot_en_12_1.png)
CFSE = [-0.4nt2g + 0.6neg]$$\Delta $$0
= [-0.4$$ \times $$4 + 0.6$$ \times $$2]$$\Delta $$0
= -0.4 $$\Delta $$0
In tetrahedral fields_5th_September_Morning_Slot_en_12_2.png)
CFSE = [-0.6neg + 0.4nt2g]$$\Delta $$t
= (– 0.6 × 3 + 0.4 × 3)$$\Delta $$t
= (–1.8 + 1.2)$$\Delta $$t
= –0.6$$\Delta $$t
CFSE = [-0.4nt2g + 0.6neg]$$\Delta $$0
= [-0.4$$ \times $$4 + 0.6$$ \times $$2]$$\Delta $$0
= -0.4 $$\Delta $$0
In tetrahedral fields
CFSE = [-0.6neg + 0.4nt2g]$$\Delta $$t
= (– 0.6 × 3 + 0.4 × 3)$$\Delta $$t
= (–1.8 + 1.2)$$\Delta $$t
= –0.6$$\Delta $$t
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