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Q. Define high spin octahedral complex?
Craig showed that highly electronegative ligands tend to form spin-free complexes. Although the valence bond theory suggests two correct alternatives in low spin and high spin, yet it does not help in making the choice. However, it indirectly suggests that the low spin complexes should be relatively less reactive. It has also been shown that a d6 high spin octabedral Complexes will be more stable than four coordinated complexes which do not require outer d-orbitals for hybridisation.
Turning our attention to systems having d7 to dl0 electronic configuration, we find that even forcing all the electrons to pair up in the inner orbitals it still does not leave two vacant d-orbitals for hybridisation: hence d2sp3 hybridisation is not possible in such cases as shown below:
In complexes of this type, we are left with just one choice of forming high spin octahedral complex or else metal ion should form complexes with lower coordination number.
The configuration 1s 2 2s 2 2p 5 3s 1 shows: (1) Ground state of fluorine atom (2) Excited state of fluorine atom (3) Excited state of neon atom (4) Excited state
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