Inorganic



  • IMG_20200607_193751353.jpg

    44
    Please explain statement 4 and 5

    ....don't know the answer



  • @Sanket-Biradar-0 In a metal carbonyl, the metal-carbon bond has the characteristics of both σ and π bonds. The bond between the carbonyl molecule and the metal becomes stronger by the synergic effect that the metal-ligand bond produces

    The formation of a metal-carbon σ bond takes place because of the donation of electrons by the carbonyl molecules.They donate it to the vacant orbitals of the metal. This is one way of formation of the metal carbonyls
    The other form is the creation of a metal-carbon π bond because of the donation of a pair of electrons from a filled d orbital metal into the vacant antibonding π* orbital of carbonyl ligand.

    Ferrocene is one of the most stable organometallic compounds with a sandwich-shaped structure and the most useful one among metallocenes due to a reversible redox properties and two rotatory coplanar cyclopentadienyl (Cp) rings.



  • @jai-d-gr8
    Why ∆° increases?



  • @Sanket-Biradar-0 The net effect of the π* orbitals is to increase the magnitude of 10 Dq (the splitting between the
    t2g and eg levels by lowering t*
    2g to a level lower in energy than when no π* orbitals are involved.
    Consequently, the complexes are predicted to be more stable when the ligands have π and π*
    orbitals available for bonding. The ligand CO may be predicted to bond increasingly strongly
    with electron releasing metal atoms. The bond order of CO decreases progressively as the π*
    orbitals are increasingly populated by d→ π* donation.
    As discussed above, the low-lying empty π* orbitals on CO allow back bonding from the metal d
    electrons to the ligand. It has a very pronounced effect on the coordinated C-O bond order.

    Since there are electrons present in the d orbital, there will be back - bonding. So, the electron density around the central atom should decrease.
    So, the value if ∆o must change



  • @jai-d-gr8
    Please explain statement 5 also


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