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प्रश्न
Explain, why [Co(NH3)6]3+ ion is low spin? Calculate number of unpaired electrons and write the geometry of [Co(NH3)6]3+.
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उत्तर
NH3 is a strong field ligand. Strong field ligands cause larger splitting of d orbitals and pairing of electrons is favoured. Hence, [Co(NH3)6]3+ is a low spin complex.
- In [Co(NH3)6]3+ ion, oxidation state of cobalt is +3.
Valence shell electronic configuration of Co3+ is:
- Number of ammine ligands is 6. Therefore, the number of vacant metal ion orbitals required for bonding with ligands must be six. Complex is low spin, so the pairing of electrons will take place prior to hybridisation.
Electronic configuration after pairing would be:
- Six orbitals available for hybridisation are two 3d, one 4s, three 4p orbitals.
The orbitals for hybridization are decided from the number of ammine ligands which is six. Here, (n−1)d orbitals participate in hybridization since it is the low spin complex. - Electronic configuration after complex formation is:
- Number of unpaired electrons = 0
- Geometry of the complex ion = Octahedral
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\[\ce{[Ni(Cl)4]^{2-}}\]
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\[\ce{[Ni(CN)4]^2-}\]
