The IR stretching frequencies of free CO and | vedclass

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Question

(B) In metal carbonyls,the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of $CO$ depends on the electron dens

Vedclass · Accepted Answer

$C-O$ bond is strongest in the cation and weakest in the anion.

Vedclass · Answer

(B) In metal carbonyls,the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of $CO$ depends on the electron density on the metal.$1$. As the negative charge on the metal increases (e.g.,$[V(CO)_6]^-$),the metal becomes more electron-rich,leading to greater back-donation into the $CO$ $\pi^*$ orbitals.$2$. Increased back-donation weakens the $C-O$ bond (decreasing its bond order and $IR$ stretching frequency) and strengthens the $M-C$ bond.$3$. Conversely,in the cation $[Mn(CO)_6]^+$,the metal is electron-deficient,resulting in minimal back-donation. Thus,the $C-O$ bond remains closer to that of free $CO$ (stronger),and the $M-C$ bond is weaker.$4$. Comparing the frequencies: $[V(CO)_6]^-$ $(1860 \, cm^{-1})$ < $[Cr(CO)_6]$ $(2000 \, cm^{-1})$ < $[Mn(CO)_6]^+$ $(2090 \, cm^{-1})$. The $C-O$ bond is strongest in the cation and weakest in the anion.

The $IR$ stretching frequencies of free $CO$ and $CO$ in $[V(CO)_6]^-$,$[Cr(CO)_6]$ and $[Mn(CO)_6]^+$ are $2143 \, cm^{-1}$,$1860 \, cm^{-1}$,$2000 \, cm^{-1}$ and $2090 \, cm^{-1}$,respectively. Then the correct statement about metal carbonyls is:

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