Question

Crystal field stabilization energy for high spin d 4 octahedral
complex is:
(a) – 1.8 ∆₀ (b) – 1.6 ∆₀ + P
(c) – 1.2 ∆₀ (d) – 0.6 ∆₀

Answer 1

The crystal field crystallization energy for high spin d4 octahedral complex is -0.6∆₀

1)

we know the formula of crystal field stabilization energy is

n1 × (-0.4∆₀) + n2(0.6∆₀) + total paired orbital × p

where p stands for pair

n1 = number of electrons in triple degenratory(t2g)

n2= number of electrons in double degenratory (eg)

2) so according to the given question 3 electrons will be in t2g and 1 electrons in eg

So

3(-0.4∆₀) + 1 (0.6 ∆₀ ) + 0

-0.12∆₀ +0.6∆₀

-0.6∆₀

3) so option (d) is the right answer

Answer 2

The Crystal field stabilization energy for high spin d 4 octahedral  complex is - 0.6 ∆₀

Option (D) is correct.

Explanation:

For high spin d4 octahedral complex the splitting is given by:

• CFSE = [ - 0.4 x 3 + 0.6 x 1] ∆₀
• CFSE = [ -1.2 + 0.6 ] ∆₀
• CFSE =  - 0.6 ∆₀

Thus the Crystal field stabilization energy for high spin d 4 octahedral  complex is - 0.6 ∆₀

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Define crystal field splitting energy.  ?

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