The problem involves calculating the crystal field stabilization energy (CFSE) for the inner orbital complex [Fe(CN)6]3–. To solve this, follow these steps:
1. Identify the oxidation state of iron in [Fe(CN)6]3–. Given that CN– is a ligand with a charge of –1, the overall charge of the complex is –3:
Fe + 6(-1) = -3
Fe = +3. Thus, iron is Fe3+.
2. Determine the electronic configuration of Fe3+: The ground state of Fe is [Ar] 3d6 4s2. After losing three electrons, Fe3+ becomes [Ar] 3d5.
3. Arrange the electrons in d-orbitals under octahedral coordination: Cyanide is a strong field ligand causing pairing of electrons in the t2g orbitals:
(t2g) 3d5: (↑↓, ↑↓, ↑) and (eg): (0, 0).
4. Calculate CFSE: CFSE = [(Number of electrons in t2g) × (–0.4)Δ0] + [(Number of electrons in eg) × (+0.6)Δ0 = [(5) × (–0.4)Δ0] + [(0) × (+0.6)Δ0 = –2Δ0.
5. Verify that the calculated CFSE falls within the given range (2, 2): The calculated CFSE of –2Δ0 matches exactly with the required nearest integer value –2. Thus, it is valid.
Conclusion: The crystal field stabilization energy for [Fe(CN)6]3– is –2Δ0, confirmed within the expected range.
In [Fe(CN)6]3–, Fe is present in (+3) oxidation state Fe(III)
\(⇒\) inner orbital complex
\(⇒\) d 5(with pairing)
Configuration \(⇒\)\(t_{2g}^5\)
\(\text{CFSE} = \frac{5 \times (-2)}{5} \Delta_0\)
\(=\text{CFSE} = -2\Delta_0\)
So, the answer is 2.
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The reaction : \(A_2 \rightleftharpoons 2A\)

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The reaction : \(A_2 \rightleftharpoons 2A\)
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