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I understand that π-backbonding in this complex involves the σ-donation of carbonyl to an empty d orbital of the metal, and the π-donation of the d electrons of the metal to the π* MO of the carbonyl. But why exactly does this lead to the weakening of the C-O bond (relative to free C-O)?

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    $\begingroup$ Do you know what a $\pi^*$ orbital is? $\endgroup$
    – Andrew
    Mar 18, 2021 at 19:04
  • $\begingroup$ In simple words, the pi* orbital is antibonding in nature. So, putting electron density there will weaken the C-O pi bond. The CO bond contains both sigma and pi components. The pi component will be weakened. $\endgroup$
    – S R Maiti
    Mar 19, 2021 at 20:04

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