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प्रश्न
Draw figure to show the splitting of d orbitals in an octahedral crystal field.
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उत्तर
- Let the six ligands approach the central metal ion symmetrically along the Cartesian axes (x, y, z), with the metal ion positioned at the origin.
- As the ligands approach, there is an initial elevation in the energy of d-orbitals compared to that of the free ion, analogous to the behaviour observed in a spherical field.
- The (`d_(z^2)` and `d_(x^2 - y^2)`) orbitals point directly along the axes (in the direction of the approaching ligands). They experience stronger repulsion and are raised in energy. The dxy, dyz and dzx orbitals point between the axes. They experience less repulsion and remain lower in energy.
- In a spherical field, no splitting happens at all. All five d-orbitals remain completely equal in energy (degenerate). The splitting into higher- and lower-energy groups happens only because of the octahedral crystal field.
- Therefore, the degenerate set of d-orbitals splits into two groups: the low energy orbital group t2g and the high energy orbital group eg separated by energy Δo.
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संबंधित प्रश्न
On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.
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(1) XeF6
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(i) \[\ce{[MnCl6]^{3-}}\]
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(iii) \[\ce{[CoF6]^{3-}}\]
(iv) \[\ce{[Ni(NH3)6]^{2+}}\]
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\[\ce{[Cr(Cl)6]^{3-}, [Cr(CN)6]^{3-}, [Cr(NH3)6]^{3+}}\].
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\[\ce{[CoF6]^{3-}, [Co(H2O)6]^{2+}, [Co(Cn)6]^{3-}}\]
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- Ni(II) complex can very rarely below spin.
- With strong field Ligands, Mn(II) complexes can be low spin.
- Aqueous solution of Mn (II) ions is yellow in colour.
The correct statements are:
On the basis of crystal field theory, write the electronic configuration for the d5 ion with a weak ligand for which Δ0 < P.
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- [Co(NH3)6]3+
- [Co(CN)6]3−
- [Cu(H2O)4]2+
- [Ti(H2O)6]3+
Choose the correct answer from the options given below:
