Among the following metal carbonyls,the C-O b | vedclass

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(D) The $C-O$ bond order in metal carbonyls decreases as the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of

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$\left[V(CO)_6\right]^{-}$

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(D) The $C-O$ bond order in metal carbonyls decreases as the extent of back-bonding from the metal $d$-orbitals to the $\pi^*$ antibonding orbitals of $CO$ increases.Back-bonding increases as the electron density on the metal center increases (i.e.,as the negative charge on the complex increases or the oxidation state of the metal decreases).Let us analyze the electron density on the metal center for each complex:$(A)$ $\left[Mn(CO)_6\right]^{+}$: $Mn$ is in $+1$ oxidation state. Total valence electrons $= 7 - 1 + 12 = 18$.$(B)$ $\left[Fe(CO)_5\right]$: $Fe$ is in $0$ oxidation state. Total valence electrons $= 8 + 10 = 18$.$(C)$ $\left[Cr(CO)_6\right]$: $Cr$ is in $0$ oxidation state. Total valence electrons $= 6 + 12 = 18$.$(D)$ $\left[V(CO)_6\right]^{-}$: $V$ is in $-1$ oxidation state. Total valence electrons $= 5 + 1 + 12 = 18$.The metal center with the lowest oxidation state (highest negative charge) has the highest electron density available for back-bonding.Comparing the oxidation states: $Mn(+1) > Cr(0) = Fe(0) > V(-1)$.Since $\left[V(CO)_6\right]^{-}$ has the lowest oxidation state $(-1)$,it has the highest electron density and thus the strongest back-bonding to $CO$,resulting in the lowest $C-O$ bond order.

Among the following metal carbonyls,the $C-O$ bond order is lowest in

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