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प्रश्न
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
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उत्तर १
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))`
= `sqrt3`
= 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))`
= `sqrt35`
= 6 BM
Thus it is evident that [Fe(H2O)6]3+ is strongly paramagnetic, while [Fe(CN)6]3− is weakly paramagnetic.
उत्तर २
In the presence of CN– (strong field ligand), the 3d-electrons pair up leaving only one unpaired electron. An inner orbital complex with d2sp3 hybridisation is formed. Hence, [Fe(CN)6]3− is weak 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.
Notes
Students can refer to the provided solutions based on their preferred marks.
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संबंधित प्रश्न
For the complex [Fe(H2O)6]+3, write the hybridisation, magnetic character and spin of the complex. (At, number : Fe = 26)
For the complex [Fe(CN)6]3–, write the hybridization type, magnetic character and spin nature of the complex. (At. number : Fe = 26).
Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
Write the hybridization and shape of the following complexe : [Ni(CN)4]2–
(Atomic number : Co = 27, Ni = 28)
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Write the hybridization and magnetic character of the following complexes:
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(ii) sp3d2 hybridisation
(iii) paramagnetic
(iv) diamagnetic
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