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Question
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
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Solution 1
In both [Fe(H2O)6]3+ and [Fe(CN)6]3−, Fe exists in the +3 oxidation state i.e., in d5 configuration.
| ↑ | ↑ | ↑ | ↑ | ↑ |
Since CN− is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in the d-orbital.
| ↑↓ | ↑↓ | ↑ |
∴ μ = `sqrt (n(n+2))`
= `sqrt (1(1+2))`
= `sqrt 3`
= 1.732 BM
On the other hand, H2O is a weak field ligand. Therefore, it cannot cause the pairing of electrons. This means that the number of unpaired electrons is 5.
∴ μ = `sqrt (n(n+2))`
= `sqrt (5(5+2))`
= `sqrt 35`
≈ 5.92 BM
Thus, it is evident that [Fe(H2O)6]3+ is strongly paramagnetic, while [Fe(CN)6]3− is weakly paramagnetic.
Solution 2
In the presence of CN– (strong field ligand), the 3d-electron pair up, leaving only one unpaired electron. An inner orbital complex with d2sp3 hybridisation is formed. Hence, [Fe(CN)6]3− is weakly paramagnetic. In the presence of HO (weak field ligand), the 3d-electrons are not paired, i.e., the hybridisation is sp3 which forms an outer orbital complex having five unpaired electrons; hence, [Fe(H2O)6]3+ is strong paramagnetic.
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