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Crystal field stabilization energy for high spin 4 octahedral complex is: [2010]
  • a)
    – 1.8 Δ0
  • b)
    – 1.6 Δ0 + P
  • c)
    – 1.2 Δ0
  • d)
    – 0.6 Δ0
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Crystal field stabilization energy for high spin 4 octahedral complex ...
4 in high spin octahedral complex
CFSE = [0.6 × 1] + [–0.4 × 3] = – 0.6 Δ0
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Crystal field stabilization energy for high spin 4 octahedral complex ...
6Δo
b) 4Δo
c) 2Δo
d) -2Δo

The crystal field stabilization energy (CFSE) for a high spin octahedral complex is given by:

CFSE = -0.4 * Δo * n

where Δo is the crystal field splitting energy and n is the number of unpaired electrons in the complex.

For a high spin 4 octahedral complex, there are 4 unpaired electrons. Therefore, the CFSE is:

CFSE = -0.4 * Δo * 4 = -1.6Δo

Since the CFSE is negative, the correct option is (d) -2Δo.
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Crystal field stabilization energy for high spin 4 octahedral complex is: [2010]a)– 1.8 Δ0b)– 1.6 Δ0 + Pc)– 1.2 Δ0d)– 0.6 Δ0Correct answer is option 'D'. Can you explain this answer?
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