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प्रश्न
The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
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उत्तर
Mn(II) possesses a 3d5 electronic configuration. Water is a weak field ligand, resulting in a small Δ0. Therefore, the hexaaqua complex will be a high-spin complex with five unpaired electrons. On the other hand, CN− is a strong field ligand, resulting in a substantial Δ0. Therefore, in its cyano complex, the electrons are paired, resulting in a single unpaired electron.
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संबंधित प्रश्न
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| Column I (Complex ion) | Column II (Hybridisation, number of unpaired electrons) |
| A. \[\ce{[Cr(H2O)6]^{3+}}\] | 1. dsp2, 1 |
| B. \[\ce{[Co(CN)4]^{2-}}\] | 2. sp3d2, 5 |
| C. \[\ce{[Ni(NH3)6]^{2+}}\] | 3. d2sp3, 3 |
| D. \[\ce{[MnF6]^{4-}}\] | 4. sp3, 4 |
| 5. sp3d2, 2 |
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OR
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